Solution :
Energy of photon, E = 6.7 eV
E = [tex]$6.7 \times 1.602 \times 10^{-7}$[/tex] joule
Kinetic energy, [tex]$K.E. =\frac{1}{2} mv^2 = 1.602 \times 6.7 \times 10^{-7}$[/tex]
[tex]$v^2=\frac{2 \times 1.602 \times 6.7 \times 10^{-7}}{1.6726 \times 10^{-27}}$[/tex]
[tex]$=12.834 \times 10^{-20}$[/tex]
Kinetic energy at high speeds
[tex]$(r-1)\times mc^2 = 6.7 \ eV$[/tex]
[tex]$(r-1)=\frac{6.7 \times 1.602 \times 10^{-7}}{1.6726 \times 10^{-27} \times 9 \times 10^{16}}$[/tex]
r - 1 = 7130
r = 7130 + 1
r = 7131
[tex]$\frac{1}{\sqrt{1-\frac{v^2}{C^2}}}=7131$[/tex]
[tex]$1-\frac{v^2}{C^2} = \left(\frac{1}{7131}\right)^2$[/tex]
[tex]$v^2=C^2\left[1-\left(\frac{1}{7131}\right)^2\right]$[/tex]
[tex]$v=0.99999999017C$[/tex]
Δ = 1 - 0.99999999017
= 0.00000000933
Relative mass, [tex]$m_{rel}=r.m$[/tex]
[tex]$=7131 \times 1.6728 \times 10^{-27}$[/tex]
[tex]$=1.1927 \times 10^{-23}$[/tex] kg
A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The ball moves so fast that the string is always taut and perpendicular to the velocity of the ball. As the ball swings from its lowest point to its highest point Group of answer choices the work done on it by gravity is -118 J and the work done on it by the tension in the string is zero. the work done on it by gravity is -118 J and the work done on it by the tension in the string is 118 J. the work done on it by gravity and the work done on it by the tension in the string are both equal to -118 J. the work done on it by gravity is 118 J and the work done on it by the tension in the string is -118 J. the work done on it by gravity and the work done on it by the tension in the string are both equal to zero.
Answer:
The ball moves from lowest to highest point:
W = M g h = 3 * 9.8 * 4 = 118 J
This is work done "against" gravity so work done by gravity is -118 J
The tension of the string does no work because the tension does not
move thru any distance W = T * x = 0 because the length of the string is fixed.
Assume that the change in pressure of H2S is small enough to be neglected in the following problem.
H2S(g) ⇌ 2H2(g) + S2(g) Kp =2.2 x 10^-6
a. Calculate the equilibrium pressures of all species in an equilibrium mixture that results from the decomposition of H2S with an initial pressure of 0.824 atm.
b. Show that the change is small enough to be neglected.
Answer:
pH2S = 0.816 atm
pH2 = 0.0153 atm
pS2 = 0.00765 atm
Explanation:
Step 1: Data given
Kp = 2.2 *10 ^-6
Initial pressure H2S = 0.824 atm
Step 2: The equation
H2S(g) ⇌ 2H2(g) + S2(g)
Step 3: The initial pressure
pH2S = 0.824 atm
pH2 = 0 atm
pS2 = 0 atm
Step 4: The pressure at the equilibrium
pH2S = 0.824 - x atm
pH2 = 2x atm
pS2 = x atm
Kp = ((pH2)² * (pS2))/pH2S
2.2 * 10^-6 = ((2x)² * x) / (0.824 - x)
2.2*10^-6 = 4x³ / (0.824 - x)
4x³ = 2.2*10^-6 * (0.824 - x)
x = 0.00765
pH2S = 0.824 - 0.00765 = 0.816 atm
pH2 = 2* 0.00765 = 0.0153 atm
pS2 = 0.00765 atm
The change is small enough to be neglected because 0.00765 << 0.824
If we calculate the result with or without this change the result will not be very different
2.2*10^-6 = 4x³ / (0.824 )
x = 0.00768 atm
pH2S = 0.824 - 0.00768 = 0.816 atm
pH2 = 2* 0.00768 = 0.0154 atm
pS2 = 0.00768 atm
A 70-kg astronaut (including spacesuit and equipment) is floating at rest a distance of 13 m from the spaceship when she runs out of oxygen and fuel to power her back to the spaceship. She removes her oxygen tank (3.0 kg) and flings it away from the ship at a speed of 15 m/s relative to the ship.
Required:
a. At what speed relative to the ship does she recoil toward the spaceship?
b. How long must she hold her breath before reaching the ship?
Answer:
Explanation:
mass of the astronaut including the spacesuit, [tex]M=30[/tex]
distance of astronaut from the spaceship, d = 13 m
mass of the oxygen tank, m = 3 kg
Speed of tank with respect to spaceship, [tex]v=15~m/s[/tex]
a)
Using the conservation of linear momentum:
total momentum before collision = total momentum after collision
[tex]M.u=m.v+(M-m)v'[/tex]
[tex]0=3\times 15+(70-15)\times v'[/tex]
[tex]v'=0.82~m/s[/tex]
b)
She mush hold her breath until she reaches the spaceship, i.e.
[tex]t=d/v'[/tex]
[tex]t=13/0.82[/tex]
[tex]t=15.89~s[/tex]
A plane flying horizontally at an altitude of 1 mi and a speed of 480 mi/h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is 3 mi away from the station.
Answer:
First remember that the distance between two points (a, b) and (c, d) is given by the equation:
[tex]d = \sqrt{(a - c)^2 + (b - d)^2}[/tex]
Now let's define the position of the radar as:
(0mi, 0mi)
Then we can write the position of the plane as:
(480mi/h*t, 1mi)
where t is time in hours.
Then we can write the distance equation as:
[tex]d(t) = \sqrt{(480\frac{mi}{h}*t - 0mi)^2 + (1mi -0mi)^2 } \\\\d(t) = \sqrt{(480\frac{mi}{h}*t )^2 + (1mi)^2 }[/tex]
Now we want to get:
the rate at which the distance from the plane to the station is increasing when it is 3 mi away from the station.
So first we want to find the value of t such that:
d(3) = 3mi
We will look at the positive value of t, because at this point the plane is increasing its distance to the station.
[tex]3mi = \sqrt{(480\frac{mi}{h}*t )^2 + (1mi)^2 }\\\\(3mi)^2 = (480\frac{mi}{h}*t )^2 + (1mi)^2\\\\9mi^2 - 1mi^2 = (480\frac{mi}{h}*t )^2\\\\8mi^2 = (230,400 mi^2/h^2)*t^2\\\\\\\sqrt{\frac{8mi^2}{230,400 mi^2/h^2} } = t = 0.0059 h[/tex]
The rate of change when the plane is 3 mi away from the station is:
d'(0.0059h)
remember that:
d'(t) = dd(t)/dt
We can write:
d(t) = h( g(t) )
such that:
h(x) = √x
g(t) = (480mi/h*t)^2 + (1mi)^2
then:
d'(t) = h'(g(t))*g'(t)
This is:
[tex]d'(t) = \frac{dd(t)}{dt} = \frac{1}{2}*\frac{2*t*480mi/h}{\sqrt{(480mi/h*t)^2 + (1mi)^2} }[/tex]
The rate of change at t = 0.0059h is then:
[tex]d'(0.0059h) = \frac{1}{2}*\frac{2*0.0059h*(480mi/h)^2}{\sqrt{(480mi/h*0.0059h)^2 + (1mi)^2} } =452.6 mi/h^2[/tex]
An electron with an initial speed of 660,000 m/s is brought to rest by an electric field.
A) What was the potential difference that stopped the electron?
B) What was the initial kinetic energy of the electron, in electron volts?
Answer:
A) ΔV = 1.237 V
B) K.E = 1.237 eV
Explanation:
B)
The initial kinetic energy of the electron is given by the following formula:
[tex]K.E = \frac{1}{2}mv^2\\\\[/tex]
where,
K.E = Kinetic Energy of electron = ?
m = mass of elctron = 9.1 x 10⁻³¹ kg
v = speed of electron = 660000 m/s
Therefore,
[tex]K.E = \frac{1}{2}(9.1\ x\ 10^{-31}\ kg)(660000\ m/s)^2[/tex]
K.E = 1.98 x 10⁻¹⁹ J
K.E = (1.98 x 10⁻¹⁹ J)([tex]\frac{1\ eV}{1.6\ x\ 10^{-19}\ J}[/tex])
K.E = 1.237 eV
A)
The energy applied by the potential difference must be equal to the kinetic energy of the electron, in order to stop it:
[tex]e\Delta V = K.E\\\\\Delta V = \frac{K.E}{e}[/tex]
where,
e = charge on electron = 1.6 x 10⁻¹⁹ C
Therefore,
[tex]\Delta V = \frac{1.98\ x\ 10^{-19}\ J}{1.6\ x\ 10^{-19}\ C}[/tex]
ΔV = 1.237 V
An object moves in a direction parallel to its length with a velocity that approaches the velocity of light. The length of this object, as measured by a stationary observer:________
a. approaches infinity.
b. approaches zero.
c. increases slightly.
d. does not change.
Answer:
b. approaches zero.
Explanation:
The phenomenon is known as length contraction.
Length contraction is a result of Einstein's special theory of relativity. This theory states that an observer in an inertial frame of reference will observe a decrease in the length of any moving object placed at another inertial frame of reference.
let the length of the train = L
Let the length observed when the train is in motion = L₀
Apply Einstein's special theory of relativity;
[tex]L_0 = L \times \sqrt{1 - \frac{v^2}{c^2} } \\\\where;\\\\v \ is \ the \ velocity \ of \ the \ train\\\\c \ is \ the \ speed \ of \ light\\\\[/tex]
from the equation above, when v = 0, the length observed is equal to the initial length of the train. (L₀ = L)
As the velocity of the train (v) approaches the speed of light (c), the length of the train observed (L₀) becomes smaller than the initial length of the train (L). (L₀ < L)
Eventually, when v equals c, we will have a square root of zero (0), and the length observed will become zero. (L₀ = 0)
Thus, the length of this object, as measured by a stationary observer approaches zero
Forced response is just the current or voltage that occures due to an input voltage or current.
a. True
b. False
Answer:
Option B
Explanation:
Generally
The forced response of a system is what the circuit does with the sources turned on, but with the initial conditions set to zero.
Therefore
Option B
A 2.5 kg rock is released from rest at the surface of a pond 1.8 m deep. As the rock falls, a constant upward force of 5.0 N is exerted on it by water resistance. Let y=0 be at the bottom of the pond.
Required:
a. Calculate the nonconservative work, Wnc, done by water resistance on the rock, the gravitational potential energy of the system, U, the kinetic energy of the rock, K, and the total mechanical energy of the system, E, when the depth of the rock below the water's surface is 0 m.
b. Calculate the nonconservative work, Wnc, done by water resistance on the rock, the gravitational potential energy of the system, U, the kinetic energy of the rock, K, and the total mechanical energy of the system, E, when the depth of the rock below the water's surface is 0.50 m.
c. Calculate the nonconservative work, Wnc, done by water resistance on the rock, the gravitational potential energy of the system, U, the kinetic energy of the rock, K, and the total mechanical energy of the system, E, when the depth of the rock below the water's surface is 1.0 m.
Answer:
a) Nonconservative Work
[tex]W_{disp} = 9\,J[/tex]
Final Gravitational Potential Energy
[tex]U_{f} = 0\,J[/tex]
Final Translational Energy
[tex]K_{f} = 35.131\,J[/tex]
b) Nonconservative Work
[tex]W_{disp} = 6.5\,J[/tex]
Final Gravitational Potential Energy
[tex]U_{f} = 12.259\,J[/tex]
Final Translational Energy
[tex]K_{f} = 25.373\,J[/tex]
c) Nonconservative Work
[tex]W_{disp} = 4\,J[/tex]
Final Gravitational Potential Energy
[tex]U_{f} = 24.518\,J[/tex]
Final Translational Energy
[tex]K_{f} = 15.614\,J[/tex]
Explanation:
The nonconservative work due to water resistance is defined by definition of work:
[tex]W_{disp} = F\cdot (y_{o}-y_{f})[/tex] (1)
Where:
[tex]W_{disp}[/tex] - Dissipate work, in joules.
[tex]F[/tex] - Resistance force, in newtons.
[tex]y_{o}[/tex] - Initial height, in meters.
[tex]y_{f}[/tex] - Final height, in meters.
The final gravitational potential energy ([tex]U_{f}[/tex]), in joules, is calculated by means of the definition of gravitational potential energy:
[tex]U_{f} = m\cdot g\cdot y_{f}[/tex] (2)
Where:
[tex]m[/tex] - Mass of the rock, in kilograms.
[tex]g[/tex] - Gravitational acceleration, in meters per square second.
The final translational kinetic energy ([tex]K_{f}[/tex]), in joules, is obtained by means of the Principle of Energy Conservation, Work-Energy Theorem and definitions of gravitational potential energy and translational kinetic energy:
[tex]m\cdot g\cdot y_{o} = U_{f} + K_{f} + W_{disp}[/tex] (3)
[tex]K_{f} = m\cdot g\cdot y_{o} - U_{f} - W_{disp}[/tex]
Lastly, the mechanical energy of the system ([tex]E[/tex]), in joules, is the sum of final gravitational potential energy, translational kinetic energy and dissipated work due to water resistance:
[tex]E = U_{f} + K_{f} + W_{disp}[/tex] (4)
Now we proceed to solve the exercise in each case:
a) Nonconservative Work ([tex]F = 5\,N[/tex], [tex]y_{o} = 1.8\,m[/tex], [tex]y_{f} = 0\,m[/tex])
[tex]W_{disp} = (5\,N)\cdot (1.8\,m - 0\,m)[/tex]
[tex]W_{disp} = 9\,J[/tex]
Final Gravitational Potential Energy ([tex]m = 2.5\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{f} = 0\,m[/tex])
[tex]U_{f} = (2.5\,kg) \cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (0\,m)[/tex]
[tex]U_{f} = 0\,J[/tex]
Final Translational Energy ([tex]m = 2.5\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{o} = 1.8\,m[/tex], [tex]U_{f} = 0\,J[/tex], [tex]W_{disp} = 9\,J[/tex])
[tex]K_{f} = (2.5\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.8\,m) -0\,J-9\,J[/tex]
[tex]K_{f} = 35.131\,J[/tex]
b) Nonconservative Work ([tex]F = 5\,N[/tex], [tex]y_{o} = 1.8\,m[/tex], [tex]y_{f} = 0.50\,m[/tex])
[tex]W_{disp} = (5\,N)\cdot (1.8\,m - 0.5\,m)[/tex]
[tex]W_{disp} = 6.5\,J[/tex]
Final Gravitational Potential Energy ([tex]m = 2.5\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{f} = 0.5\,m[/tex])
[tex]U_{f} = (2.5\,kg) \cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (0.5\,m)[/tex]
[tex]U_{f} = 12.259\,J[/tex]
Final Translational Energy ([tex]m = 2.5\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{o} = 1.8\,m[/tex], [tex]U_{f} = 12.259\,J[/tex], [tex]W_{disp} = 6.5\,J[/tex])
[tex]K_{f} = (2.5\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.8\,m) -12.259\,J-6.5\,J[/tex]
[tex]K_{f} = 25.373\,J[/tex]
c) Nonconservative Work ([tex]F = 5\,N[/tex], [tex]y_{o} = 1.8\,m[/tex], [tex]y_{f} = 1\,m[/tex])
[tex]W_{disp} = (5\,N)\cdot (1.8\,m - 1\,m)[/tex]
[tex]W_{disp} = 4\,J[/tex]
Final Gravitational Potential Energy ([tex]m = 2.5\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{f} = 1\,m[/tex])
[tex]U_{f} = (2.5\,kg) \cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (1\,m)[/tex]
[tex]U_{f} = 24.518\,J[/tex]
Final Translational Energy ([tex]m = 2.5\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{o} = 1.8\,m[/tex], [tex]U_{f} = 24.518\,J[/tex], [tex]W_{disp} = 4\,J[/tex])
[tex]K_{f} = (2.5\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.8\,m) -24.518\,J-4\,J[/tex]
[tex]K_{f} = 15.614\,J[/tex]
You are cooking breakfast for yourself and a friend using a 1,140-W waffle iron and a 510-W coffeepot. Usually, you operate these appliances from a 110-V outlet for 0.500 h each day. (a) At 12 cents per kWh, how much do you spend to cook breakfast during a 30.0 day period
Answer:
The cost is 297 cents.
Explanation:
Power of iron, P = 1140 W
Power of coffee pot, P' = 510 W
Voltage, V = 110 V
Time, t = 0.5 h each day
Cost = 12 cents per kWh
(a) Total energy
E = P x t + P' x t
E = 1140 x 0.5 x 60 x 60 + 510 x 0.5 x 60 x 60
E = 2052000 + 918000 = 2970000 J
1 kWh = 3.6 x 10^6 J
E = 0.825 kWh
For 30 days
E' = 0.825 x 30 = 24.75 kWh
So, the cost is
= 12 x 24.75 = 297 cents
Use the following information to answer the next question.
Environmental Concerns
1. release of greenhouse gases
2. release of gases that cause acid rain
3. release of excess heat
4. depletion of solar energy
5. depletion of geothermal energy
6. flooding of land
Which of the above environmental concerns are associated with the production of electricity?
Select one:
O A. 2, 3, and 4
O B. 1, 2, 3, and 6
O C. 1, 2, 3, 5 and 6
O D. 1, 3, and 5
Answer:
1.Emitted primarily through the burning of fossil fuels (oil, natural gas, and coal), solid waste, and trees and wood products. Changes in land use also play a role. Deforestation and soil degradation add carbon dioxide to the atmosphere, while forest regrowth takes it out of the atmosphere.
2.Acid rain is caused by a chemical reaction that begins when compounds like sulfur dioxide and nitrogen oxides are released into the air. These substances can rise very high into the atmosphere, where they mix and react with water, oxygen, and other chemicals to form more acidic pollutants, known as acid rain.
3.Untreated, heat exhaustion can lead to heatstroke, a life-threatening condition that occurs when your core body temperature reaches 104 F (40 C) or higher. Heatstroke requires immediate medical attention to prevent permanent damage to your brain and other vital organs that can result in death.
4.The loss of solar energy in passing through the atmospheric layers is called the atmospheric deflection. ... The longer the path traversed, the greater the amount of radiant energy depleted. Various processes whereby heat energy is lost through the atmosphere are known as scattering, diffusion, absorption, and reflection.
5.Geothermal energy is renewable because the Earth has retained a huge amount of the heat energy that was generated during formation of the planet. In addition, heat is continuously produced by decay of radioactive elements within the Earth. The amount of heat within the Earth, and the amount that is lost though natural processes (e.g. volcanic activity, conduction/radiation to the atmosphere), are much, much more than the amount of heat lost through geothermal energy production. At any one geothermal field, however, the temperature of the geothermal reservoir or the fluid levels/fluid pressure in the reservoir may decrease over time as fluids are produced and energy is extracted. Produced fluids can be re-injected to maintain pressures, although this may further cool down the reservoir if care is not taken. Over time, it is commonly necessary to drill additional wells in order to maintain energy production as temperatures and/or reservoir fluid pressures decline.
6.Floods, Floodplains, and Flood-Prone Areas. ... Flooding is a result of heavy or continuous rainfall exceeding the absorptive capacity of soil and the flow capacity of rivers, streams, and coastal areas. This causes a watercourse to overflow its banks onto adjacent lands.
The environmental concerns associated with the production of electricity are the release of greenhouse gases, the release of gases that cause acid rain, the release of excess heat, the flooding of land, and the depletion of geothermal energy so, option C is correct.
What is electricity?The presence or movement of charged particles is electricity. The movement of electrons through a circuit is known as an electric current. The accumulation of electrons on an insulator causes static electricity.
Mostly released when solid trash, trees, and wood products are burned, along with fossil fuels (coal, gas, and oil).
Land use changes also have an impact. Carbon dioxide is released into the atmosphere by deforestation and soil erosion, while it is removed from it by forest regeneration.
To know more about electricity:
https://brainly.com/question/29812640
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what is the final velocity if you have an initial velocity of 5 m/s with an acceleration of 3 m/s^2 over a 4 second interval
Answer:
initial velocity (u)=5m/s
final velocity (v)=?
acceleration (a)=3m/s^2
time (t)=4s
now,
acceleration (a)=v-u/t
3=v-5/4
3×4=v-5
12=v-5
12+5=v
17=v
v=17
A current of 1 mA flows through a copper wire. How many electrons will pass a point in each second?
Answer:
A current of 1ma flows through a copper wire, how many electron will pass a given point in one second? 1 Coulomb = 6.24 x 10^18 electrons (or protons)/1Sec which is also equal to 1 Amp/1 Sec. 1mA is 1/1000th of 1A so only 1/1000th of 6.24 x 10^18 electrons will pass a given point in 1 Sec.
Given the triangle shown below, what is the cosine of the angle 0 ?
triangle trig image 1
O Vx/V
O Vx/Vy
O Vy/Vx
O Vy/
Answer:
?
Explanation:
i am sorry but there is no triangle
Explanation: edmentum sample answer
a cleaner pushes a 4.50 kg laundry cart in such a way that the net external force on it is 60.0 N. Calculate the magnitude of it's accerlation
Answer:
Explanation:
F = ma so filling in:
60.0 = 4.50a and
a = 13.3 m/s/s
What bet force is required to stop a 2250 kg car if the decelerates at a rate of -4.3 m/s^2 please answer fast
Answer:
Force = Mass × Acceleration
[tex]{ \tt{force = 2250 \times 4.3}} \\ = { \tt{9675 \: newtons}}[/tex]
the current through a wire is measured as the potential difference is varied what is the wire resistance
Answer:
Resistance, R = 0.02 Ohms
Explanation:
Ohm's law states that at constant temperature, the current flowing in an electrical circuit is directly proportional to the voltage applied across the two points and inversely proportional to the resistance in the electrical circuit.
Mathematically, Ohm's law is given by the formula;
V = IR
Where;
V is the voltage or potential difference.
I is the current.
R is the resistance.
From the attachment, we would pick the following values on the graph of current against voltage;
Voltage, V = 0.5 V
Current = 25 A
To find resistance;
R = V/I
R = 0.5/2.5
Resistance, R = 0.02 Ohms
Note:
Resistance (R) is the inverse of slope i.e change in current with respect to change in voltage.
Uranus and Neptune may have a compressed liquid water ocean beneath their atmospheres. What three pieces of evidence support this conclusion?
Answer:
Distance from sun, orbit and rotation. Presence of interior oceans, and elements that forms compressed water.
Explanation:
Both the planets are Jovian planets and have layers formed by ice such as that of Uranus does not have any surface. Such as the planet is only rotating fluids. While 80% of the mass of Neptune is made up of fluids or icy water and also consists of ammonia and methane.Two resistors with resistance values of 4.5 Ω and 2.3 Ω are connected in series or parallel across a 30 V potential difference to a light bulb. a. Calculate the current delivered through the light bulb in the two cases. b. Draw the circuit connection that will achieve the brightest light bulb
Answer:
plz mark as a pure substance abuse and function of the equation for sulphur dioxide emissions and function of the equation for sulphur dioxide emissions and function of the equation for sulphur dioxide emissions and function of the equation for sulphur dioxide emissions and function of the equation forfriction between two flat surfaces can be divided into two categories. what are the two most common kinds of friction?
Answer:
kinetic and static
Explanation:
hope it helps! ^w^
A fan that is rotating at 960 rev/s is turned off. It makes 1500 revolutions before it comes to a stop. a) What was its angular acceleration(assuming it was constant)
Answer:
α = 1930.2 rad/s²
Explanation:
The angular acceleration can be found by using the third equation of motion:
[tex]2\alpha \theta=\omega_f^2-\omega_i^2[/tex]
where,
α = angular acceleration = ?
θ = angular displacement = (1500 rev)(2π rad/1 rev) = 9424.78 rad
ωf = final angular speed = 0 rad/s
ωi = initial angular speed = (960 rev/s)(2π rad/1 rev) = 6031.87 rad/s
Therefore,
[tex]2\alpha(9424.78\ rad) = (0\ rad/s)^2-(6031.87\ rad/s)^2\\\\\alpha = -\frac{(6031.87\ rad/s)^2}{(2)(9424.78\ rad)}[/tex]
α = - 1930.2 rad/s²
negative sign shows deceleration
A boy of mass 50 kg on a motor bike is moveny coith 20m/see what is hio k.E
Mộtchấtđiểmtăngtốctừ v0 =20m/sđạtđến v=80m/strênđườngthẳngvớigia
tốc a = 45 m/s2. Khoảng thời gian thay đổi tốc độ đó là bao nhiêu?
Answer:
Waktu, t = 1.33 detik
Explanation:
Mengingat data berikut;
Kecepatan awal, Vo = 20 m/s
Kecepatan akhir, Vi = 80 m/s
Percepatan, a = 45 m/s²
Untuk menemukan interval perubahan kecepatan;
Dalam latihan ini, Anda diminta menghitung waktu untuk perubahan kecepatan.
Oleh karena itu, kita akan menggunakan persamaan gerak pertama;
Vi = Vo + at
Mengganti ke dalam rumus, kita memiliki;
80 = 20 + 45t
80 - 20 = 45t
45t = 60
Waktu, t = 60/45
Waktu, t = 1.33 detik
Modern wind turbines are larger than they appear, and despite their apparently lazy motion, the speed of the blades tips can be quite high-many times higher than the wind speed. A turbine has blades 53m long that spin at 14 rpm.
Required:
a. At the tip of a blade, what is the speed?
b. At the tip of a blade, what is the centripetal acceleration?
Explanation:
Given that,
The length of the blades, l = 53 m
The angular velocity = 14 rpm = 1.466 rad/s
(a) The speed at the tip of a blade.
[tex]v=r\omega\\\\=53\times 1.466\\\\=77.69\ m/s[/tex]
(b) The centripetal acceleration at the tip of the blade is :
[tex]a=\dfrac{v^2}{r}\\\\a=\dfrac{77.69^2}{53}\\\\a=113.88\ m/s^2[/tex]
Hence, this is the required solution.
You are a detective investigating why someone was hit on the head by a falling flowerpot. One piece of evidence is a smartphone video taken in a 4th-floor apartment, which happened to capture the flowerpot as it fell past a window. In a span of 8 frames (captured at 30 frames per second), the flowerpot falls 0.84 of the height of the window. You visit the apartment and measure the window to be 1.27 m tall. What can you conclude? Under what assumptions? Give as much detail as you can.
Answer:
Ok, first, suppose that a given object is dropped or thrown down.
Then the acceleration of the object is the gravitational acceleration, given by:
a(t) = -9.8m/s^2
The velocity of the object can be integrated from that, it gives:
v(t) = (-9.8m/s^2)*t + V0
where V0 is the initial speed of the object, in the case that it is dropped, V0 is equal to 0m/s
The position equation can be found integrating again:
p(t) = (1/2)*(-9.8m/s^2)*t^2 + V0*t + p0
Where p0 is the initial position.
Now let's see our problem.
We know that in 8 frames, the flowerpot falls 0.84 of the height of the window, which is 1.27m
Then the distance that it falls is:
D = 0.84*1.27m = 1.07m
Now we also know that the camera captures at 30 fps
Then we have the relation:
30 frames = 1 second
1 = (1 second)/(30 frames)
With this, we can rewrite:
8 frames = 8 frames* (1 second)/(30 frames) = 0.267 seconds.
So now we know the distance that the pot fall, and the time in which it did fall.
Remember that the position equation for a falling object, is:
p(t) = (1/2)*(-9.8m/s^2)*t^2 + V0*t + p0
let's define p0 = 0m
Then the position equation of the pot is given by:
p(t) = (1/2)*(-9.8m/s^2)*t^2 + V0*t
The distance that the pot would travel between t = 0s and t = t' is:
distance = p(t') - p(0s)
So, knowing that between t= 0s and t = 0.267s, the pot did travel -1.07m (the negative sign is because it traveled downwards), we can find the initial speed of the pot.
-1.07m = p(0.267s) - p(0s)
-1.07m = (1/2)*(-9.8m/s^2)*(0.267s)^2 + V0*0.267s
[-1.07m + (1/2)*(9.8m/s^2)*(0.267s)^2]/0.267s = V0 = -2.7 m/s
This means that the pot was thrown downwards with an initial speed of 2.7 m/s
So we can conclude that the pot did not fall down on its own, someone threw it intentionally downwards.
This is the only thing that we can conclude with the given information.
Consider the video tutorial you just watched. Suppose we repeat the experiment, but this time place the divider closer to one side of the tube than to the other. How will the speed of the air on the wide and narrow sides of the divider compare
Answer:
The answer is "The air will move faster on the narrow side".
Explanation:
The air on the top slows down in hypertensive. This is why light travels quicker on top. This results in air deflection downwards, required for its energy conservation to generate lift, and that is why air has to be moved quicker on the narrow side by the very same airflow per unit time as it departs.
B.F.Skinner emphesized the importance of-----?
Answer:
BFSkinner enfatizó la importancia de creía en la importancia de desarrollar la psicología experimental y dejar atrás el psicoanálisis y las teorías acerca de la mente basadas en el simple sentido común.
Explanation:
HELP NEEDED FAST (last cram sessions before finals)
BRAINLIEST!
Three resistors are connected in series across a 75-V potential difference. R, is 170 and R2 is 190. The potential difference across R3 is 21 V. Find the current in the circuit.
Explanation:
The sum of the voltages of the components connected in a series circuit is equal to the voltage across the battery.
[tex]V_T = V_1 + V_2 +V_3[/tex]
From Ohm's law ([tex]V=IR[/tex]) and in a series circuit, the amount of current flowing through the components is the same for all. So we can write [tex]V_T[/tex] as
[tex]V_T= 75\:\text{V} = I(170)+I(190) + 21\:\text{V}[/tex]
[tex]I(170+190)=54\:\text{V}[/tex]
[tex]I= \dfrac{54\:\text{V}}{360\:\text{ohms}}=0.15\:\text{A}[/tex]
In the graph below, why does the graph stop increasing after 30 seconds?
A. The hydrogen gas is absorbing heat to undergo a phase change.
B. A catalyst needs to be added to increase the amount of hydrogen produced.
C. No more hydrogen can be produced because all of the reactants have become products at this point.
D. It has reached the maximum amount of product it can make at this temperature. The temperature would need to increase to produce more.
Answer:
The answer is "Option C".
Explanation:
It's evident from the figure below that after thirty minutes, not no more hydrogen can be created because all of the reactants have converted into products.
hydrogen gas created in cm cubes per period x = 20 seconds, y = 45 centimeters squared, and so on.
A reaction's terminus (the graph's flat line) indicates that no further products are being created during the reaction.
A baseball pitcher brings his arm forward during a pitch, rotating the forearm about the elbow. If the velocity of the ball in the pitcher's hand is 34.0 m/s and the ball is 0.310 m from the elbow joint, what is the angular velocity (in rad/s) of the forearm
Answer:
[tex]\omega=109.67\ rad/s[/tex]
Explanation:
Given that,
The speed of the ball, u = 34 m/s
The ball is 0.310 m from the elbow joint.
We need to find the angular velocity (in rad/s) of the forearm.
We know that,
[tex]v=r\omega\\\\\omega=\dfrac{v}{r}\\\\\omega=\dfrac{34}{0.31}\\\\\omega=109.67\ rad/s[/tex]
So, the required angular velocity of the forearm is 109.67 rad/s.
How far away should a Cliff be from a source of sound to give an echo in 5.3 second?(given speed of sound at 0°c=331 m/s
Answer:
Total distance traveled by the sound wave = 2d . Thus we can conclude that the cliff should be at a distance of 877.15 metres from the source of sound to meet the required conditions
Explanation:
this is what i found on the web
