Answer:
No, it doesn’t.
Explanation:
The second law of thermodynamics states that in a natural thermodynamic process, the sum of the entropies of the interacting thermodynamic systems increases. Equivalently, machines that spontaneously convert thermal energy into mechanical work are impossible.
If you combine milk and coffee, the entropy will rise until the mixture is entirely homogenous and you can no longer differentiate between the two substances. At that point, the mixture will be a single, dull hue.
But in the process of mixing up coffee, before it’s fully mixed together but after you have started mixing, you might notice some complex swirl patterns appear for a brief moment in the chaos before vanishing away.
That’s what human life is.
We’re not violating thermodynamics because if you take the system as a whole, including the sun and the earth, entropy is still increasing. The sun will eventually run out of fuel and die out. Eventually all suns will die out and the whole universe will be homogeneous and we will have heat death as the expanding universe rips complex atoms apart.
But there can be brief pockets of complexity within that system, that exists for a brief period of time, before eventually and inevitably fading away. It does not violate thermodynamics because entropy is still increasing in the system as a whole.
You are the process engineer at Corvallis Automobiles Inc., and you have received an order to turn a cylindrical bar on an engine lathe to the dimensions specified in Fig. 1. For this order you will use cylindrical bar stock that is 48-inches long and 4-inches in diameter. The 48-inch length bar will be chucked in the lathe and supported at the opposite end using a live center. You are planning to complete the operation in one pass using a cutting speed of 400 ft./min. and a feed of 0.010 in./rev. Determine the following: a) The required depth of cut (in inches) b) The material removal rate (in cubic inches per minute)
c) The time required to complete the cutting pass (in minutes)
a. the depth of cut is 0.625 inches.
b. the material removal rate is 0.003125 cubic inches per minute.
c. the time required to complete the cutting pass is 20 minutes.
How do we calculate?a) The required depth of cut can be determined by :
DOC = (4 in - 2.75 in)/2 = 0.625 in
Therefore, the depth of cut is 0.625 inches.
b) The material removal rate can be found by applying:
MRR = DOC x Width of cut x Feed rate
assuming we are using a standard carbide insert tool with a width of cut of 0.5 inches.
MRR = 0.625 in x 0.5 in x 0.010 in/rev = 0.003125 cubic inches per minute
c) The time required to complete the cutting pass is determined by:
Time = Length of cut / (Cutting speed x Width of cut x Feed rate)
Time = 48 in / (400 ft/min x (0.5 in) x (0.010 in/rev) x (1/12 ft/in)) = 20 minutes
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How does the star formation in spirals compare to the star formation of elliptical galaxies?
The spiral galaxies are characterized by the arms winding out from a central nucleus while the elliptical galaxies are characterized by their lack of structure or a central bulge.
Star formation refers to the process by which dense areas within molecular clouds in interstellar space, typically lasting tens of millions of years, form newborn stars. It takes a long time for stars to form, and this process is not well understood.
In comparison to spiral galaxies, elliptical galaxies have low star formation.
Furthermore, elliptical galaxies are made up of stars with a wide range of ages, indicating that the star formation process was rapid and early on in their history.
Spiral galaxies have more gas and dust in their disks than elliptical galaxies, and these are the sites of intense star formation.
The arms are believed to be regions of higher density of stars and interstellar material, as well as more significant gravitational interactions among stars, gas, and dust than in the rest of the disk.
Thus, spiral galaxies are sites of ongoing star formation while elliptical galaxies are mainly populated by old and evolved stars that no longer form. Therefore, spiral galaxies have a higher rate of star formation than elliptical galaxies.
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: suppose a planet has a mass of 10 times that of the Earth and a radius that is 100 times that of the Earth. The acceleration of gravity on the surface of the planet, expressed in units of the Earth's acceleration of gravity, g, is 1000 g. g/1000 10 g.
The acceleration of gravity on the surface of the planet, expressed in units of the Earth's acceleration of gravity, g, is g/1000.
Given,Mass of the planet, m = 10m_Earth Radius of the planet, r = 100r_Earth Acceleration of gravity on the surface of the planet, g' = 1000 g_Earth To find, the acceleration of gravity on the surface of the planet, expressed in units of the Earth's acceleration of gravity, g.
Assuming the planet to be a perfect sphere, the acceleration due to gravity on the surface of the planet, g' is given by,g' = GM / r²where M is the mass of the planet, r is the radius of the planet and G is the gravitational constant.We know that, the acceleration of gravity on the surface of the Earth, g_Earth is given by,g_Earth = GM_Earth / r_Earth²Thus, we have the ratio of g' and g_Earth,g' / g_Earth = GM / r² × r_Earth² / GM_Earth= r_Earth / r = 1 / 100∴ g' = 1 / 100 × g_Earth = g_Earth / 1000.Hence, the acceleration of gravity on the surface of the planet, expressed in units of the Earth's acceleration of gravity, g is g/1000. Therefore, the answer is g/1000.
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a particle moving along the x axis in simple harmonic motion starts from its equilibrium position, the ori- gin, at t 5 0 and moves to the right. the amplitude of its motion is 2.00 cm, and the frequency is 1.50 hz. (a) find an expression for the position of the particle as a function of time. determine (b) the maximum speed of the particle and (c) the earliest time (t . 0) at which the particle has this speed. find (d) the maxi- mum positive acceleration of the particle and (e) the earliest time (t . 0) at which the particle has this accel- eration. (f) find the total distance traveled by the par- ticle between t 5 0 and t 5 1.00 s.
(a) The position of the particle as a function of time is given by:
x(t) = A cos(2πft)
where A is the amplitude (2.00 cm), f is the frequency (1.50 Hz), and cos is the cosine function.
Substituting the given values, we get:
x(t) = 2.00 cos(3πt)
(b) The maximum speed of the particle occurs at the equilibrium position, where the displacement is zero. At this point, the velocity is maximum and is given by:
vmax = Aω
where ω is the angular frequency and is equal to 2πf. Substituting the given values, we get:
vmax = 2.00 × 2π × 1.50 = 18.85 cm/s
(c) The earliest time at which the particle has this speed is when it passes through the equilibrium position. This happens at t = 0, so the earliest time is t = 0.
(d) The maximum positive acceleration of the particle occurs at the ends of its motion, where the displacement is maximum. At these points, the acceleration is given by:
amax = Aω^2
Substituting the given values, we get:
amax = 2.00 × (2π × 1.50)^2 = 282.74 cm/s^2
(e) The earliest time at which the particle has this acceleration is when it reaches the maximum displacement. This happens at t = 1/4T, where T is the period of the motion. The period is given by:
T = 1/f = 2/3 s
So, t = 1/4T = 1/4 × 2/3 = 0.33 s
(f) The total distance traveled by the particle between t = 0 and t = 1.00 s is equal to one complete cycle of its motion. The distance traveled in one complete cycle is equal to four times the amplitude, or:
4A = 8.00 cm
Therefore, the total distance traveled is:
8.00
a researcher is studying the distribution of auxin in roots and stems exposed to sunlight. he notices that more auxin collects in the sides of stems and roots that are not exposed to light. why?
The researcher's observation that more auxin collects in the sides of stems and roots that are not exposed to light is likely due to the phenomenon of phototropism.
In the process of phototropism, light influences the direction and rate of growth of plant cells. In particular, light induces the cells on one side of a stem or root to create less auxin than the cells on the shaded side. Less auxin is produced on the lighted side and more auxin is produced on the shaded side as a result. The hormone auxin is essential for controlling the growth and development of plants. Auxin generally promotes cell growth and elongation at greater concentrations while inhibiting cell elongation at lower concentrations. Since the cells on the lighted side of the stem or root will contain less auxin when there is light.
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why do nuclear reactors have three separate water loops instead of just a single one that runs from the water source, through the reactor, then back to the cooling tower?
Nuclear reactors have three separate water loops instead of just a single one that runs from the water source, through the reactor, then back to the cooling tower because the water running through the reactor is highly radioactive.
What are nuclear reactors?A nuclear reactor is a device that controls and maintains a sustained nuclear chain reaction for the purpose of generating heat or power, as well as the materials that make up a nuclear reactor.
The water running through the reactor is highly radioactive, which means that it cannot be released into the atmosphere or allowed to come into touch with humans or the environment. As a result, nuclear reactors are designed with three separate water loops.
The first loop circulates ordinary water that passes through the reactor and generates heat. The second loop, which is a separate circuit, brings this water to a steam turbine. The third loop, which is also a closed circuit, recovers the cooling water after it has passed through the turbine and transports it back to the reactor's inlet.In summary, nuclear reactors have three separate water loops instead of a single one that runs from the water source, through the reactor, and back to the cooling tower because the water running through the reactor is highly radioactive.
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A car has an intial velocity of 50 km hr after 5 h, its final velocity is 70 km hr. solve for the car acceleration
Answer:
4 km/hr^2
Explanation:
We can use the formula for acceleration:
a = (v_f - v_i) / t
where:
a = acceleration
v_f = final velocity
v_i = initial velocity
t = time taken
Substituting the given values, we get:
a = (70 km/hr - 50 km/hr) / 5 hr
a = 20 km/hr / 5 hr
a = 4 km/hr^2
Review your answer to part c. In addition, reread the portion of your physics text that discusses Newton's third law. Then consider a book on a level table: e. Which force completes the Newton's third law (or action-reaction) force pair with the normal force exerted on the book by the table?
In this case, the normal force exerted by the table on the book is the action force and the reaction force is the force that the book exerts on the table. This force is equal in magnitude to the normal force and acts in the opposite direction.
Newton's third law states that for every action, there is an equal and opposite reaction. This means that when one object exerts a force on another object, the second object exerts a force back on the first object that is equal in magnitude and opposite in direction.
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three forces applied to a trunk that moves leftward by 3.010 m over a frictionless floor. The force magnitudes are F1 = 5.86 N, F2 = 9.180 N, and F3 = 3.850 N, and the indicated angle is θ = 67.8°. During the displacement, what is the net work done on the trunk by the three forces? (Note that there are other forces acting on the block, but we only care about the net work done by these three forces.) And by how much does the kinetic energy of the trunk increase (enter a positive value) or decrease (negative value)?
The kinetic energy of the trunk increases by ½ mvf² = ½ m(10.65 m/s)²= 71.44 J during the displacement.
Net work = ΔK
W = Fd cosθ
W1 = F1d cosθ = (5.86 N)(3.010 m) cos(67.8°) = 6.99 J
W2 = F2d cosθ = (9.180 N)(3.010 m) cos(67.8°) = 10.97 J
W3 = F3d cosθ = (3.850 N)(3.010 m) cos(67.8°) = 4.58 J
Net work = W1 + W2 + W3 = 6.99 J + 10.97 J + 4.58 J = 22.54 J
Therefore, the net work done on the trunk by the three forces is 22.54 J.
ΔK = ½ mvf² - ½ mvi²
Since the trunk moves a distance of 3.010 m and is initially at rest, we can use the equation:
vf² = 2ad
where a is the acceleration of the trunk, which is given by:
a = ΣF / m
where ΣF is the net force on the trunk, which we can find using:
ΣF = F1 + F2 + F3
ΣF = (5.86 N + 9.180 N + 3.850 N) = 18.89 N
Therefore, the acceleration of the trunk is:
a = ΣF / m = 18.89 N / m
Since the trunk moves leftward, the acceleration is also leftward, so we can use a negative value for a.
Substituting the values for a and d, we get:
vf² = -2(-18.89 N / m)(3.010 m) = 113.51 (m/s)²
Taking the square root, we get:
vf = 10.65 m/s
Therefore, the change in kinetic energy of the trunk is:
ΔK = ½ mvf² - ½ mvi² = ½ m(10.65 m/s)²- 0 = ½ mvf²
Kinetic energy is a type of energy that an object possesses by virtue of its motion. It is defined as the energy an object has due to its motion and is proportional to the mass of the object and the square of its velocity. The formula for kinetic energy is KE = 1/2mv^2, where m is the mass of the object and v is its velocity.
Kinetic energy is a scalar quantity and has units of joules in the International System of Units (SI). It is a fundamental concept in physics and is used to describe many physical phenomena, including the motion of particles, the behavior of gases, and the motion of waves. In many cases, kinetic energy can be transformed into other forms of energy. For example, when a ball is thrown upwards, its kinetic energy is gradually converted into gravitational potential energy as it moves higher and higher.
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questionwhen you heat an air-filled balloon, what happens inside with regard to the movement of air molecules?
When you heat an air-filled balloon, the movement of air molecules inside the balloon increases, causing the air to expand and the balloon to inflate.
Heating the air inside the balloon increases the temperature of the air molecules, causing them to move more rapidly and collide with each other more frequently.
This increased movement and collision between molecules causes them to spread out and fill a larger volume, which leads to the expansion of the air inside the balloon.
As the air inside the balloon expands, it exerts a greater pressure on the walls of the balloon, causing it to inflate.
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Is an object moving with a constany speed around a circular path veloctiy? why? why not?
Answer: The motion of a body with constant speed in a circular path is said to be accelerated, because it is moving with uniform speed, but not with uniform velocity, as velocity is a vector quantity, it can be represented in magnitude as well the direction.
Explanation:
a particle's velocity is described by the function vx=kt2 , where vx is in m/s , t is in s , and k is a constant. the particle's position at t0=0s is x0 = -5.40 m . at t1 = 2.00 s , the particle is at x1 = 5.80 m .
A particle's velocity is described by the function vx=kt2 , where vx is in m/s , t is in s , and k is a constant. The particle's position at t0=0s is x0 = -5.40 m. At t1 = 2.00 s , the particle is at x1 = 5.80 m. The value of k is 2.80 m/s2.
The given equation describes the velocity of a particle in terms of a constant, k, and time, t. The velocity, vx, is given in m/s. The initial position of the particle at t0=0s is x0=-5.40 m, and at t1=2.00 s the particle is at x1=5.80 m. To find the value of the constant k, we can solve the equation for the change in velocity Δvx.
Δvx = vx1 – vx0 = k(t12 – t02)
Δvx = 5.80 – (-5.40) = 11.20 m/s
k = (11.20 m/s) / (2.002 s2) = 2.80 m/s2
Now that we have found the value of the constant k, we can use it to find the velocity of the particle at any time t. For example, at t2=4.00 s the velocity of the particle is vx2=11.20 m/s. This can be calculated using the equation vx2 = k(t22) = 2.80(4.002) = 11.20 m/s.
From the velocity equation, we can also calculate the position of the particle at any time t. The position of the particle at t2=4.00 s is x2= 11.20(4.00) = 44.80 m. We can also calculate the position of the particle at any other time t, by simply substituting in the corresponding value of t into the equation.
In conclusion, the equation vx = kt2 describes the velocity of a particle in terms of a constant, k, and time, t. Using this equation, we can calculate the velocity and position of the particle at any given time.
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Complete Question:
A particle’s velocity is described by the function vx = [tex]kt^2m/s[/tex], where k is a constant and t is in s. The particle’s position at [tex]t_0[/tex] = 0s is [tex]x_0[/tex] = -5.40 m. At [tex]t_1[/tex] = 2.00 s, the particle is at [tex]x_1[/tex] = 5.80 m. Determine the value of the constant k. Be sure to include the proper units
A car travelling at 22.4 m/s skids to a stop in 2.55s. Determine the skidding distance of the car (assume uniform acceleration).
Answer:
Approximately [tex]28.6\; {\rm m}[/tex].
Explanation:
Let [tex]u[/tex] denote the initial velocity of the vehicle, and let [tex]v[/tex] denote the velocity of the vehicle after skidding. It is given that the initial velocity was [tex]u = 22.4\; {\rm m\cdot s^{-1}}[/tex]. Since the vehicle skidded to a stop, [tex]v = 0\; {\rm m\cdot s^{-1}}[/tex].
Let [tex]t[/tex] denote the duration of the skid. It is given that [tex]t = 2.55\; {\rm s}[/tex].
Under the assumption that acceleration is constant, SUVAT equations will apply.
Specifically, the SUVAT equation [tex]x &= (1/2)\, (u + v)\, t[/tex] will be satisfied. In this equation, the displacement of the vehicle is equal to average velocity times duration. This equation allows the displacement [tex]x[/tex] to be found from [tex]u[/tex], [tex]v[/tex], and [tex]t[/tex] without knowing the exact value of acceleration:
[tex]\begin{aligned}x &= \left(\frac{u + v}{2}\right)\, t \\ &= \left(\frac{22.4 + 0}{2}\; {\rm m\cdot s^{-1}}\right)\; (2.55\; {\rm s}) \\ &\approx 28.6\; {\rm m}\end{aligned}[/tex].
the maximum energy of photoelectrons from aluminium is 2.3 ev for radiation of 2000 a and 0.90 ev for radiation of 3130 a. use this data to calculate plancks constant and the work function of aluminium
The maximum energy of photoelectrons from aluminium is 2.3 eV for radiation of 2000 Å and 0.90 eV for radiation of 3130 Å.
To calculate Planck's constant and the work function of aluminium, we need to use the equation:
h = E2 - E1/ λ2 - λ1
Where h is Planck's constant, E1 and E2 are the maximum energy of photoelectrons for each wavelength, and λ1 and λ2 are the wavelengths.
Using the given data, we have:
h = (2.3 - 0.90) / (2000 - 3130)
Therefore, h = -1.4 eV / -930 Å, which simplifies to h = 0.0015 eVÅ.
The work function of aluminium is equal to the maximum energy of the photoelectrons for the longest wavelength, in this case, 0.90 eV. Therefore, the work function of aluminium is 0.90 eV.
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In outer space will a liquid in a beaker exert a pressure on the bottom or on the sides of a beaker?
Answer:
Yo dude, if you had a beaker of liquid in outer space, it wouldn't push down on the bottom or the sides of the beaker like it would on Earth. In space, there's no gravity to make the liquid settle down, so it forms a round shape because of surface tension. So basically, the liquid would just float around in a ball inside the beaker. If you moved the beaker around, the liquid would just roll around with it like a bouncy ball.
A small grinding wheel has a moment of inertia of 4. 0×10−5 kg⋅m2
k
g
⋅
m
2. What net torque must be applied to the wheel for its angular acceleration to be 150 rad/s2
r
a
d
/
s
2
?
A net torque of [tex]6.0×10^−3 N⋅m[/tex] is sufficient to produce the desired angular acceleration of [tex]150 rad/s^2[/tex].
The net torque required to produce an angular acceleration in a rotating object can be calculated using the formula: net torque = moment of inertia × angular acceleration In this case, the moment of inertia of the grinding wheel is given as 4.0×10^−5 kg⋅m^2 and the angular acceleration required is 150 rad/s^2.
Therefore, the net torque required can be calculated as: net torque = [tex](4.0×10^−5 kg⋅m^2) × (150 rad/s^2) = 6.0×10^−3 N⋅m[/tex]To explain this result, we need to understand the relationship between torque and angular acceleration. Torque is the rotational equivalent of force and it is defined as the product of force and the perpendicular distance between the line of action of the force and the axis of rotation.
When a torque is applied to a rotating object, it produces an angular acceleration in the object, which is a measure of how quickly the object's rotational speed changes.
The moment of inertia of an object is a measure of its resistance to changes in its rotational motion. It depends on the object's mass distribution and the distance of each element of mass from the axis of rotation. Objects with larger moments of inertia require more torque to produce a given angular acceleration than objects with smaller moments of inertia.
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the car passes over the top of a vertical curve at a with a speed of 50 km/hr and then passes through the bottom of a dip at b. the radii of curvature of the road at a and b are both 70 m. find the speed of the car at b if the normal force between the road and the tires at b is twice that at a. the mass center of the car is 1.2 meter from the road.
The speed of the car at b if the normal force between the road and the tires at b is twice that at a is about 44.1 km/h.
What is Speed?Speed of the car at A = 50 km/h
Radius of curvature at A = 70 m
Radius of curvature at B = 70 m
Normal force between the road and the tires at B = 2 × Normal force between the road and the tires at A= 2N
Mass center of the car = 1.2 m
The speed of car at B be v km/h
From the conservation of energy at the point A and B, we get:
1/2 mv² + mgh = 1/2 m(50)² + mg(70 - r)
1/2 mv² + mg(70 + r) = 1/2 m(50²)
1/2 mv² = 1/2 m50² - mg(70 + r) …… equation (1)
From the conservation of energy at point B, we get:
1/2 mv² + mg(2r + 1.2) = 1/2 m(50)² + mg(70 - r)
2× Normal force between the road and the tires at A = Normal force between the road and the tires at B
Normal force between the road and the tires at B = 2 × Normal force between the road and the tires at A
Therefore, mg - 2 × N = mv²/rmg - N = mv²/2r
2mg - 4N = mv²/rmg - 2N = mv²/2r
Subtracting, we get:
N = mg/3
Normal force between the road and the tires at A = mg/3
Normal force between the road and the tires at B = 2mg/3
Normal force between the road and the tires at B = 2(mg/3) = mg/3
From the above equations, we get the value of v. Putting the values, we get:
1/2 mv² = 1/2 m(50)² - mg(70 + r) - mg(2r + 1.2) + mg(70 - r)1/2 v² = 1/2(50)² - g(70 + r) - g(2r + 1.2) + g(70 - r)v = 44.1 km/h
Therefore, the speed of the car at B is 44.1 km/h.
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of the three states of matter, which one has the most kinetic energy?
Of the three states of matter (solid, liquid, and gas), gas has the most kinetic energy. This is because the particles in a gas have the highest average speed compared to the particles in solids and liquids.
In a gas, the particles are in constant motion, colliding with each other and the walls of the container. This motion generates kinetic energy, which is proportional to the speed and mass of the particles. In contrast, solids have the lowest kinetic energy because their particles are tightly packed and have limited movement. The particles in a solid vibrate around a fixed position, and only experience small oscillations. Liquids have an intermediate amount of kinetic energy. The particles in a liquid are less tightly packed than in a solid, and can move more freely, resulting in more kinetic energy. However, liquids have more intermolecular forces between the particles compared to gases, which restricts their movement and reduces their average speed. Therefore, of the three states of matter, gases have the most kinetic energy, followed by liquids and then solids.
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A 2. 00-kg object is attached to an ideal massless horizontal spring of spring constant 100. 0 N/m and is at rest on a frictionless horizontal table. The spring is aligned along the x-axis and is fixed to a peg in the table. Suddenly this mass is struck by another 2. 00-kg object traveling along the x-axis at 3. 00 m/s, and the two masses stick together. What are the amplitude and period of the oscillations that result from this collision? 0. 300 m, 1. 26 s 0. 424 m, 5. 00 s 0. 424 m, 0. 889 s 0. 300 m, 0. 889 s 0. 424 m, 1. 26 s
The correct option is A, the amplitude and period of the oscillations that result from this collision are 0.300 m in 1.26s.
The expression for Period of spring is,
[tex]T = 2\pi\sqrt{\frac{2m}{k} }[/tex]
Here, m is the mass of the spring and k is the spring constant
Substitute 2 kg
for m
and 100N/m
for k
in equation [tex]T = 2\pi\sqrt{\frac{2m}{k} }[/tex]
and solve for T .
[tex]T = 2\pi\sqrt{\frac{(2)2 kg}{100 N/m} }[/tex]
T = 1.26s
In physics, amplitude refers to the maximum displacement or distance moved by a wave from its equilibrium or mean position. It is a measure of the intensity or strength of a wave, and it is usually represented as the height of the crest or depth of the trough of the wave.
The amplitude of a wave can be measured in various units, depending on the type of wave and the context in which it is being studied. For example, the amplitude of a sound wave is measured in decibels (dB), while the amplitude of an electromagnetic wave is measured in volts per meter (V/m). Amplitude plays an important role in the behavior of waves. It determines the energy carried by the wave and affects other properties such as frequency, wavelength, and phase.
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Complete Question: -
A 2.00-kg object is attached to an ideal massless horizontal spring of spring constant 100.0 N/m and is at rest on a frictionless horizontal table. The spring is aligned along the x-axis and is fixed to a peg in the table. Suddenly this mass is struck by another 2.00-kg object traveling along the x-axis at 3.00 m/s, and the two masses stick together. What are the amplitude and period of the oscillations that result from this collision
A) 0.300 m, 1.26 s
B) 0.300 m, 0.889 s
C) 0.424 m, 0.889 s
D) 0.424 m, 1.26 s
E) 0.424 m, 5.00 s
2.2 VECTORS IN TWO 120 N bearing 70° and 160 N bearing 40°
Answer:
Explanation:
Assuming you want to find the resultant vector of the two given vectors:
We can use the graphical method or the component method to find the resultant vector. Here, I will demonstrate the component method:
Step 1: Convert the given vectors into their component form (i.e., horizontal and vertical components).
Vector 1: 120 N bearing 70°
Horizontal component = 120 cos(70°) ≈ 38.23 N
Vertical component = 120 sin(70°) ≈ 113.41 N
Vector 2: 160 N bearing 40°
Horizontal component = 160 cos(40°) ≈ 122.15 N
Vertical component = 160 sin(40°) ≈ 103.08 N
Step 2: Add the horizontal components and vertical components separately to get the components of the resultant vector.
Horizontal component of resultant vector = 38.23 N + 122.15 N ≈ 160.38 N
Vertical component of resultant vector = 113.41 N + 103.08 N ≈ 216.49 N
Step 3: Use the Pythagorean theorem to find the magnitude of the resultant vector.
Magnitude of resultant vector = √(160.38 N)^2 + (216.49 N)^2 ≈ 268.15 N
Step 4: Find the direction of the resultant vector.
Direction of resultant vector = tan^-1(216.49 N / 160.38 N) ≈ 53.12°
Therefore, the resultant vector of the two given vectors is approximately 268.15 N at a bearing of 53.12°.
Two 4.4 kg bodies, A and B, collide. The velocities before the collision are A = (28i + 27j) m/s and B = (9.8i + 1.8j) m/s. After the collision, 'A = (3.7i + 3.2j) m/s. What are (a) the x-component and (b) the y-component of the final velocity of B? (c) What is the change in the total kinetic energy (including sign)?
Answer:jfnvufhdfiprhfpiurgh8rhvjm vjfnb
Explanation:
you live on an island in the pacific. an earthquake of magnitude 8.5 off the coast of japan, 8000 km away, generates a tsunami with a wavelength of 200 km. the average water depth between your island and japan is 4900 m. if a tsunami warning is issued for your island, how many hours will you have before the waves arrive?
If a tsunami warning is issued for the island, they will have approximately 11.7 hours before the waves arrive.
What is Magnitude?
Magnitude is a measure of the strength or intensity of a physical quantity or phenomenon, such as an earthquake or a sound wave. It is often expressed using a numerical scale, with higher values indicating greater strength or intensity. In the case of earthquakes, magnitude is typically measured using the Richter scale or the moment magnitude scale, which take into account the amplitude of seismic waves and the energy released by the earthquake.
To calculate the time it takes for a tsunami to travel from Japan to the island, we can use the following formula:
t = (2 * pi * d) / g * ln(1 + sqrt(h/d))
where t is the time it takes for the tsunami to travel, d is the average water depth, h is the wave height, and g is the acceleration due to gravity (9.8 m/s^2).
Magnitude of the earthquake: 8.5
Wavelength of the tsunami: 200 km = 200,000 m
Average water depth: 4,900 m
To calculate the wave height, we can use the following formula:
h = (M / 5) * (D / 10)^1/2
where M is the magnitude of the earthquake and D is the distance between the earthquake epicenter and the observation point (in this case, the island). Note that this formula is an approximation and may not be accurate for all cases.
Using the given values, we get:
D = 8,000 km = 8,000,000 m
h = (8.5 / 5) * ((8,000,000 / 10)^1/2) = 2,738.6 m
Substituting these values into the formula for t, we get:
t = (2 * pi * 4,900) / 9.8 * ln(1 + sqrt(2,738.6/4,900)) = 11.7 hours
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on a sunny day, a rooftop solar panel delivers 55 w of power to the house at an emf of 17 v . part a how much current flows through the panel? express your answer with the appropriate units.
On a sunny day, a rooftop solar panel delivers 55 w of power to the house at an emf of 17 v . the current flows through the panel is 3.235 A.
The amount of current flowing through the panel can be calculated using Ohm's Law (I = p/v). When working with the formula, Power = Voltage × Current (P = V × I), one can determine the current that flows through the panel by rearranging the formula to:
Current = Power/Voltage (I = P/V).
The calculation of the current (I) is given as follows: I = P/V = 55 W / 17 V = 3.235 A.
Therefore, 3.235 A of current flows through the solar panel on a sunny day.
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A small mass rests on a horizontal platform which vibrates vertically in simple harmonicmotion with period 0.50 s.(a) Find the maximum amplitude of the motion which will allow the mass to stay in contactwith the platform throughout the motion.The maximum acceleration that will allow the object to remain in contact with theplatform at all times is when amax = g = 9:81 m/s.But amax = !222A = (2¼=T )A ) 9:81 = (2¼=0:5)2A = 158A ) A = 0:062 m
The maximum amplitude of the motion which will allow the mass to stay in contact with the platform is 0.062 m.
This can be calculated by using the equation amax = (2π/T)2A, where A is the maximum amplitude, and T is the period of the motion. In this case, T is 0.50 s, and g (the acceleration due to gravity) is 9.81 m/s2, so we can calculate A:
A = (2π/T)2g = (2π/0.50)2 × 9.81 = 158 × 9.81 = 1543.38
Therefore, A = 1543.38/158 = 9.81 m/s2 = 0.062 m.
Alternatively: given,T = 0.50 s,The acceleration due to gravity, g = 9.81 m/s²Maximum acceleration, amax = g = 9.81 m/s². The maximum acceleration that will allow the object to remain in contact with the platform at all times is when amax = !222A = (2π/T )A ) 9.81 ...(1)From the equation (1), we get 158 A = 9.81 (2π/0.50)A = (9.81 (2π/0.50))/158 = 0.062 m. Therefore, the maximum amplitude of the motion which will allow the mass to stay in contact with the platform throughout the motion is 0.062 m.
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two blocks with masses 4m and 7m are on a collision course with the same initial speeds vi. the block with mass 4m is traveling to the left, and the 7m block is traveling to the right. they undergo a head-on elastic collision and each bounces back, retracing its original path. find the final speeds of the particles. (enter your answers in terms of
The final speeds of the particles expressed in terms of the initial velocity are |v1'| = |v1| = 27/8|vi| and |v2'| = |v2| = 27/14|vi|
The conservation of momentum can be applied. The total momentum of the system before the collision is:
P before = m1v1 + m2v2
where m1 and v1 are the mass and velocity of the 4m block and m2 and v2 are the mass and velocity of the 7m block. Since the two blocks have the same initial speed, the momentum before the collision is:
P before = (4m)(-vi) + (7m)(vi)
P before = 3mvi
After the collision, the two blocks bounce back, so their final velocities are:
v1' = -v1
v2' = -v2
where v1 and v2 are the velocities of the blocks after the collision. Using the conservation of momentum again, the total momentum of the system after the collision is:
Pafter = m1v1' + m2v2'
Pafter = -4mv1 - 7mv2
Pafter = -4m(-v1) - 7m(-v2)
Pafter = 4mv1 + 7mv2
Since the collision is elastic, the total kinetic energy of the system is conserved. Therefore, the kinetic energy before the collision is equal to the kinetic energy after the collision:
Kbefore = Kafter
where Kbefore is the kinetic energy of the system before the collision and Kafter is the kinetic energy of the system after the collision. The kinetic energy can be expressed as:
K = 1/2mv²
Therefore, the total kinetic energy of the system before the collision is:
Kbefore = 1/2(4m)(vi)² + 1/2(7m)(vi)²
Kbefore = 27/2m(vi)²
The total kinetic energy of the system after the collision is:
Kafter = 1/2(4m)(-v1)² + 1/2(7m)(-v2)²
Kafter = 1/2(4m)(v1)² + 1/2(7m)(v2)²
Using the conservation of kinetic energy, Kbefore = Kafter:
27/2m(vi)² = 1/2(4m)(v1)² + 1/2(7m)(v2)²
Simplifying, the final velocities can be expressed in terms of the initial velocity:
v1 = 27/8vi
v2 = 27/14vi
Therefore, the final speeds of the particles are: |v1'| = |v1| = 27/8|vi| and |v2'| = |v2| = 27/14|vi|
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Assume that a drop of mercury is an isolated sphere. What is the capacitance in picofarads of a drop that results when two drops each of radius R = 5.61 mm merge?
The formula C=4R, where is the permittivity of open space, may be used to determine the capacitance of a merged mercury drop, assuming it is an isolated sphere. The capacitance is around 1.68 pF with R = 5.61 mm.
The formula C=4R, where R is the drop's radius and is the permittivity of free space, may be used to determine the capacitance of a merged mercury drop. As the capacitance of an isolated sphere is exactly proportional to its radius, the capacitance produced by the merger of two drops with similar radii is equal to the total of the capacitances of the individual drops. Given that the radius of the combined drop in this instance is R = 5.61 mm, the capacitance can be estimated using the formula C = 4(8.85 x 10-12 F/m) (5.61 x 10-3 m)2, yielding a capacitance of around 1.68 pF.
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Calculate the translational speed of a cylinder when it reaches the foot of an incline 7.20 m high. Assume it starts from rest and rolls without slipping.
Express your answer using three significant figures and include the appropriate units. Thank you!!
The translational speed of the cylinder when it reaches the foot of the incline is approximately 9.43 m/s.
We can use conservation of energy to solve this problem. The initial energy of the cylinder is all potential energy, and the final energy is all kinetic energy. The potential energy at the bottom of the incline is zero.
The potential energy of the cylinder at the top of the incline is given by:
PE = mgh
where m is the mass of the cylinder, g is the acceleration due to gravity, and h is the height of the incline. Substituting the given values, we get:
PE = (mass of cylinder) x (acceleration due to gravity) x (height of incline) = mgh
The kinetic energy of the cylinder at the bottom of the incline is given by:
KE = (1/2)mv^2
where v is the translational speed of the cylinder at the bottom of the incline.
According to the conservation of energy, the initial potential energy is equal to the final kinetic energy, so we can set these two expressions equal to each other:
mgh = (1/2)mv^2
We can cancel the mass of the cylinder from both sides, and solve for v:
v = sqrt(2gh)
Substituting the given values, we get:
v = sqrt(2 x 9.81 m/s^2 x 7.20 m) ≈ 9.43 m/s
Therefore, the translational speed of the cylinder when it reaches the foot of the incline is approximately 9.43 m/s.
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In which of the following cases does a car have a negative velocity and a positive acceleration? A car that is traveling in the ................. (A) -x direction at a constant 10 m/s. (B) - direction increasing in speed. (C) +x direction increasing in speed. (D) - direction decreasing in speed. (E) +x direction decreasing in speed.
In the case where the car is traveling in the -x direction and decreasing in speed, it has a negative velocity and a positive acceleration. Therefore, option D is the correct answer. In this case, the car is traveling in the - direction and decreasing in speed. Therefore, it has a negative velocity and a positive acceleration.
Let's discuss the given options one by one:
(A) In this case, the car is traveling in the -x direction at a constant speed. Therefore, it has a negative velocity and zero acceleration. This option is incorrect.
(B) In this case, the car is traveling in the - direction and increasing its speed. Therefore, it has a negative velocity and a positive acceleration. However, the given direction is not specified, and thus this option is not accurate.
(C) In this case, the car is traveling in the +x direction and increasing in speed. Therefore, it has a positive velocity and a positive acceleration. This option is incorrect.
(D) In this case, the car is traveling in the - direction and decreasing in speed. Therefore, it has a negative velocity and a positive acceleration. This option is correct.
(E) In this case, the car is traveling in the +x direction and decreasing in speed. Therefore, it has a positive velocity and a negative acceleration. This option is incorrect.
Therefore, Option D ( - direction decreasing in speed) is correct.
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a cliff diver drops from rest to the water below. how many seconds does it take for the driver to go from 0 mi/h to 60 mi/h? (for comparison, it takes about 3.5 s to 4.0 s for a powerful car to go from 0 to 60 mi/h.)
Assuming that the only force acting on the diver is gravity and neglecting air resistance, we can use the kinematic equations of motion to determine that it takes 2.7 s for the diver to reach a speed of 60 mi/h (or 88 ft/s).
Since the diver starts from rest, we can use the kinematic equation:
[tex]$$v_f = v_i + at$$[/tex]
where [tex]$v_i$[/tex] is the initial velocity (0 mi/h), [tex]$v_f$[/tex] is the final velocity (60 mi/h or 88 ft/s), [tex]$a$[/tex] is the acceleration due to gravity [tex](32.2 ft/s$^2$)[/tex], and [tex]$t$[/tex] is the time it takes to reach the final velocity.
Converting the final velocity to feet per second, we get:
[tex]$$v_f = 60\ \text{mi/h} \times \frac{5280\ \text{ft/mi}}{3600\ \text{s/h}} = 88\ \text{ft/s}$$[/tex]
Substituting the given values, we get:
[tex]$$88\ \text{ft/s} = 0\ \text{ft/s} + (32.2\ \text{ft/s}^2)t$$[/tex]
Solving for [tex]$t$[/tex], we get:
[tex]t = \frac{88\ \text{ft/s}}{32.2\ \text{ft/s}^2}[/tex]
Therefore, it takes approximately 2.73 seconds for the diver to go from 0 mi/h to 60 mi/h.
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This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Pin A, which is attached to link AB, is constrained to move in the circular slot CD. At t=0, the pin starts from rest and moves so that its speed increases at a constant rate of 1.2 in/s2 D 3.5 in. А B Determine the magnitude of its total acceleration when t= 0. The magnitude of its total acceleration is in/s2
The magnitude of the total acceleration of the pin when t=0 is 1.2 in/s^2.
To explain further, the acceleration of the pin is the sum of two components: tangential acceleration and centripetal acceleration. The tangential acceleration is responsible for increasing the speed of the pin, and its magnitude is constant at 1.2 in/s^2.
The centripetal acceleration is due to the circular motion of the pin in the slot CD and is directed towards the center of the circle.
To find the magnitude of the total acceleration at t=0, we need to first find the magnitude of the tangential acceleration and the centripetal acceleration separately. We know that the tangential acceleration is 1.2 in/s^2, and we can use the formula for centripetal acceleration, a_c = v^2/r, where v is the velocity of the pin and r is the radius of the circle. At t=0, the velocity of the pin is zero, and the radius of the circle is 3.5 inches.
Therefore, the centripetal acceleration is also zero.
Since the centripetal acceleration is zero, the magnitude of the total acceleration is equal to the magnitude of the tangential acceleration, which is 1.2 in/s^2.
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