Answer:
Explanation:
The formula for angular velocity is
[tex]\omega=\frac{\theta}{t}[/tex] where omega is the angular velocity, theta is the change in the angular rotation, and t is the time in seconds. First and foremost, we have the angular rotation in minutes and the time in seconds, so that's a problem we have to amend. Let's change the angular rotation to rotations per second:
[tex]281.133\frac{r}{min}*\frac{1min}{60s}=4.68555\frac{r}{s}[/tex]
Now we're ready to set up the problem:
[tex]4.68555=\frac{\theta}{3.686}[/tex] and we multiply both sides by 3.686 to get the rotations per seconds:
θ = 17.27 rotations
A block of mass 0.260 kg is placed on top of a light, vertical spring of force constant 5 200 N/m and pushed downward so that the spring is compressed by 0.090 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise
After being released, the restoring force exerted by the spring performs
1/2 (5200 N/m) (0.090 m)² = 12.06 J
of work on the block. At the same time, the block's weight performs
- (0.260 kg) g (0.090 m) ≈ -0.229 J
of work. Then the total work done on the block is about
W ≈ 11.83 J
The block accelerates to a speed v such that, by the work-energy theorem,
W = ∆K ==> 11.83 J = 1/2 (0.260 kg) v ² ==> v ≈ 9.54 m/s
Past the equilibrium point, the spring no longer exerts a force on the block, and the only force acting on it is due to its weight, hence it has a downward acceleration of magnitude g. At its highest point, the block has zero velocity, so that
0² - v ² = -2gy
where y is the maximum height. Solving for y gives
y = v ²/(2g) ≈ 4.64 m
A cylindrical swimming pool has a radius 2m and depth 1.3m .it is completely filled with salt water of specific gravity 1.03.The atmospheric preassure is 1.013 x 10^5 Pa.
a.calculate the density of salt water.
Answer:
the density of the salt water is 1030 kg/m³
Explanation:
Given;
radius of the cylindrical pool, r = 2 m
depth of the pool, h = 1.3 m
specific gravity of the salt water, γ = 1.03
The atmospheric pressure, P₀ = 1.013 x 10⁵ Pa
Density of fresh water, [tex]\rho _w[/tex] = 1000 kg/m³
The density of the salt water is calculated as;
[tex]Specific \ gravity \ of \ salt\ water \ (\gamma _s_w) = \frac{density \ of \ salt \ water \ (\rho_{sw})}{density \ of \ fresh \ water \ (\rho_{w})} \\\\1.03 = \frac{\rho_{sw}}{1000 \ kg/m^3}\\\\\rho_{sw} = 1.03 \times 1000 \ kg/m^3\\\\\rho_{sw} = 1030 \ kg/m^3[/tex]
Therefore, the density of the salt water is 1030 kg/m³
A 0.2 oz. bullet leaves the muzzle of a rifle with a speed of 1420 ft/s. If the length of the barrel is 24 inches, what is the magnitude of the force acting on the bullet while it travels down the barrel
Answer:
196 lbf
Explanation:
v² = u² + 2as
1420² = 0² + 2a(24/12)
a = 1420²/4 = 504,100 ft/s²
F = ma = 0.2oz(1lb/16 oz)(1slug/32.2 lb)(504,100) = 195.6909...
In part B of the lab, when the current flows through the orange part of the wire from right to left, the wire deflects (or moves) ____. This is in accordance with the right-hand-rule.
This seems to be incomplete, as we do not have any information about the magnetic field surrounding the wire, but we can answer in a general way.
We know that for a wire of length L, with a current I, and in a magnetic field B, the force can be written as:
F = L*(IxB)
if we define the right as the positive x-axis, and knowing that the current flows to the right, we can write:
I = i*(1, 0, 0)
And the field will be some random vector that can't be parallel to the current because in that case, we do not have any force.
To find the direction of the force, which will tell us the direction in which the wire deflects or moves, first, we need to point with our thumb in the direction of the current, in this case, to the right.
Now, with the hand open, using the tip of our other fingers we point in the direction of the magnetic field.
For example, if the magnetic field is in the positive z-axis, we will point upwards.
Now the palm of our hand tells us in which direction the force is applied.
This is the right-hand rule.
For example, in the case that the current goes to the right and the magnetic field is upwards, we could see that the force is to the front.
A person pulls on a 9 kg crate against a 22 Newton frictional force, using a rope attached to the center of the crate. If the The crate began with a speed of 1.5 m/s and speeded up to 2.7 m/s while being pulled a horizontal distance of 2.0 meters. What is the work in J done by the force applied by the rope on the crate
Answer:
uninstell this apps nobody will give u ans its happening to me alsoi was in exam hall i thought this app will give answer but no
Use the values from PRACTICE IT to help you work this exercise. If the current in each wire is doubled, how far apart should the wires be placed if the magnitudes of the gravitational and magnetic forces on the upper wire are to be equal
What is the volume of a metal block 3cm long by 2cm wide by 4cm high? What would be the volume of a block twice as long, wide, and high?
Answer:
Volume of a metal block = 24 cm^3
Volume of a block twice as long, wide and high = 192 cm^3
Explanation:
Volume of a block is measured in l*w*h and in the first block, the sides are 3, 2 and 4 and 3*2*4 = 24
Second block, just double each of the lengths to get 6*4*8 = 192
Why are scientific models important?
Answer:
Scientific models are representations of objects, systems or events and are used as tools for understanding the natural world. Models use familiar objects to represent unfamiliar things. Models can help scientists communicate their ideas, understand processes, and make predictions.
what are the limitation of clinical thermometer
Answer:
Their main disadvantage is that they are fairly easy to break and if they do, it results in small splinters of glass and the release of mercury which is quite toxic if absorbed into the body.
A lightning bolt has a current of 56,000 A and lasts for 80 x 10-6 seconds (80 μs). How much charge (in Coulombs) has flowed in this bolt?
Answer:
A cloud can discharge as much as 20 coulombs in a lightning bolt.
You are to connect resistors R1 andR2, with R1 >R2, to a battery, first individually, then inseries, and then in parallel. Rank those arrangements according tothe amount of current through the battery, greatest first. (Useonly the symbols > or =, for exampleseries>R1=R2>parallel.)
Answer:
The current is more in the parallel combination than in the series combination.
Explanation:
two resistances, R1 and R2 are connected to a battery of voltage V.
When they are in series,
R = R1 + R2
In series combination, the current is same in both the resistors, and it is given by Ohm's law.
V = I (R1 + R2)
[tex]I = \frac{V}{R_1 + R_2}[/tex]..... (1)
When they are connected in parallel.
the voltage is same in each resistor.
The effective resistance is R.
[tex]R = \frac{R_1R_2}{R_1 + R_2}[/tex]
So, the current is
[tex]I = \frac{V(R_1+R_2)}{R_1 R_2}[/tex]..... (2)
So, the current is more is the parallel combination.
Develop a hypothesis regarding one factor you think might affect the period of a pendulum or an oscillating mass on a spring. Potential factors include the mass, the spring constant, and the length of the pendulum's string. Write down your hypothesis. 2. Design a controlled experiment to test your hypothesis. Take extreme care to keep all factors constant except the variable you are testing.
Answer:
A hypothesis for the period of a pendulum is:
"The period of the pendulum varies with its length"
Explanation:
A hypothesis for the period of a pendulum is:
"The period of the pendulum varies with its length"
To test this hypothesis we can carry out a measurement of a simple pendulum keeping the angle fixed, in general the angle used is about 5º since when placing this value in radiand and the sine of this angle they differ little <5%. therefore measured the time of some oscillations, for example about 10 oscillations, changing the length of the pendulum to test the hypothesis.
If the hypothesis and the model used is correct, the relationship to be tested is
T² =(4π² /g) L
by making a graph of the period squared against the length if obtaining, os a line, the hypothesis is tested.
A bus starts from rest and accelerates at 1.5m/s squared until it reaches a velocity of 9m/s .the bus continues at this velocity and then deccelerate at -2m/s squared until it comes to stop 400m from it's starting point. how much time did the bus takes to cover the 400m?
Answer:
23s
Explanation:
s=ut+1/2at^2
the distance (s) is 400, initial velocity (u) is 0, acceleration (a) is 1.5 therefore
400=0t+1/2(1.5)t^2
400/0.75=0.75t^2/0.75
t^2=√533.33
t=23s
I hope this helps and sorry if it's wrong
How much charge is stored on two parallel-plate capacitors by the 12V battery if one is filled with air and the other is filled with a dielectric (k=3.00)
The question is incomplete, the complete question is;
how much charge is stored on the parallel-plate capacitors by the 12.0 V battery? One is filled with air, and the other is filled with a dielectric for which k = 3.00; both capacitors have a plate area of 5.00×10 −3 m 2 and a plate separation of 2.00 mm.
The capacitance of the capacitor is the quantity of charge stored by the capacitor.
Given that;
C1= εo k * A/d
εo = permittivity of free space
C1 = 8.85 x 10-12 farad per meter *1 * 5.00×10 −3 m 2/2 * 10^-3
= 2.21 * 10^-11 F
C2 = 8.85 x 10-12 * 3 * 5.00×10 −3 m 2/2 * 10^-3
= 6.63 * 10^-11 F
q1 = C1V1 = 2.21 * 10^-11 C * 12 V
= 2.65 * 10^-10 C
q2 = C2V2 = 6.63 * 10^-11 F * 12 V
= 7.96 * 10^-10 C
qtotal = 2.65 * 10^-10 C + 7.96 * 10^-10 C
qtotal = 1.061 * 10^-9C
https://brainly.com/question/17176550
Astronaut Jill leaves Earth in a spaceship and is now traveling at a speed of 0.280c relative to an observer on Earth. When Jill left Earth, the spaceship was equipped with all kinds of scientific instruments, including a meter stick. Now that Jill is underway, how long does she measure the meter stick to be?
A) 0.280 m
B) 1.00 m
C) 0.960 m
D) 1.28 m
E) 1.04 m
(B) 1.00 m
Explanation:
Since the meter stick is traveling with Jill, it will have the same speed as she does so relative to Jill, the meter stick is stationary so its length remains 1.00 m as measured by her.
When astronaut Jill leaves Earth in a spaceship and is now traveling at a speed of 0.280c and measure the meter stick to be 1 meter. Hence, option B is correct.
What is length contraction?Length contraction is defined as the phenomenon of the moving object being shorter than its appropriate length, measured in the object's rest frame.
When the object travels with the speed of light, the length of the object gets more contracted than its original length, relative to the observer. It is also known as the Lorentz-Fitgerald contraction.
Length contraction, L = L₀√(1-v²/c²), where L is the original length, L₀ is the contracted length. c² is known as the velocity of light. v² is the velocity of the speed of the object.
From the given,
speed of the spaceship = 0.280c (c is the speed of the light)
Length contraction, L = L₀ √(1-v²/c²)
The stick also travels in the spaceship. Hence, the length of the meter stick does not change. It remains at its original length of one meter. Thus, the ideal solution is option B.
To learn more about length contraction:
https://brainly.com/question/10272679
#SPJ2
Which was a major effect of Pope Leo III crowning Charlemagne emperor of the Romans ?
Answer:
The crowning of Charlemagne by Pope Leo III was significant in a number of ways. For Charlemagne, it was necessary because it encouraged to give him higher reliability. It gave him the rank of a dictator, giving him the only ruler in Europe west of the Byzantine emperor in Constantinople.
A ball of mass m is dropped from a height h above the ground. neglecting air resistance then determine the speed of the ball when it is at a height y above the ground and determine the speed of the ball at y if at the instant of release it already has an initial upward speed vi at the initial altitude h.
Answer:
Explanation:
kinematic equation (g will have a negative value if we assume UP is positive)
v² = u² + 2as
a) v = √(0² + 2(g)(y - h))
b) v = √(vi² + 2(g)(y - h))
A 700N marine in basic training climbs a 10m vertical rope at constant speed in 8sec. what is power put
Answer:
875 Watts
Explanation:
P = W/t = mgh/t = 700(10)/8 = 875 Watts
A uniform disk turns at 3.6 rev/s around a frictionless spindle. A non rotating rod, of the same mass as the disk and length equal to the disk's diameter, is dropped onto the freely spinning disk . They then both turn around the spindle with their centers superposed.
What is the angular frequency in rev/s of the combination?
please express answer in proper significant figures and rounding.
Answer:
ω₁ = 2.2 rev/s
Explanation:
Conservation of angular momentum
moment of inertia uniform disk is ½mR²
moment of inertia uniform rod about an end mL²/3
We can think of our rod as two rods of mass m/2 and length R
L = ½mR²ω₀
L = (½mR² + 2(m/2)R²/3)ω₁
ω₁ = ω₀(½mR² / (½mR² + mR²/3))
ω₁ = ω₀(½ / (½ + 1/3))
ω₁ = 0.6ω₀
ω₁ = 2.16
Two motors in a factory are running at slightly different rates. One runs at 825.0 rpm and the other at 786.0 rpm. You hear the sound intensity increase and then decrease periodically due to wave interference. How long does it take between successive instances of the sound intensity increasing
Answer:
[tex]T=1.54s[/tex]
Explanation:
From the question we are told that:
Speed of Motor 1 [tex]\omega_1=825rpm=>2 \pi 13.75[/tex]
Speed of Motor 2 [tex]\omega_1=786rpm=>2 \pi 13.1[/tex]
Therefore
Frequency of Motor 1 [tex]f_1=13.75[/tex]
Frequency of Motor 2 [tex]f_2= 13.1[/tex]
Generally the equation for Time Elapsed is mathematically given by
[tex]T=\frac{1}{df}[/tex]
Where
[tex]df=f_1-f_2[/tex]
[tex]df=13.75-13.1[/tex]
[tex]df=0.65Hz[/tex]
Therefore
[tex]T=\frac{1}{65}[/tex]
[tex]T=1.54s[/tex]
A 207-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.750 rev/s in 2.00 s
Answer:
366 N
Explanation:
τ = Iα
FR = ½mR²α
F = ½mR(Δω/t)
F = ½(207)(1.50)(0.75)(2π) /2.00
F = 365.79919...
a. Do the waves made by the two faucets travel faster than the waves made by just one faucet?
b. How do you know this? Describe how the two-faucet wave pattern compares with the one-faucet pattern.
c. Describe what happens to the two-faucet wave pattern as the separation of the faucets is increased.
Answer:
asdasd dsa dasdasd sadas dasd asdasd asd asd dsa asdd 223 aasd ada dasd sa dasd dsaa sd adsd asasd
Explanation:
Which of the following behaviors would best describe someone who is listening and paying attention? a) Leaning toward the speaker O b) Interrupting the speaker to share their opinion c) Avoiding eye contact d) Asking questions to make sure they understand what's being said
Answer:
D
Explanation:
Anyone could be leaning forward toward the speaker but be distracted and I believe if you're paying attention to the speaker, you would ask questions to make sure you're understanding what they are speaking
Answer:
A
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Help! xoxo thank you
Like charges repel and opposite charges attract.
Cho lực F ⃗=6x^3 i ⃗-4yj ⃗ tác dụng lên vật làm vật chuyển động từ A(-2,5) đến B(4,7). Vậy công của lực là:
The work done by [tex]\vec F[/tex] along the given path C from A to B is given by the line integral,
[tex]\displaystyle \int_C \mathbf F\cdot\mathrm d\mathbf r[/tex]
I assume the path itself is a line segment, which can be parameterized by
[tex]\vec r(t) = (1-t)(-2\,\vec\imath + 5\,\vec\jmath) + t(4\,\vec\imath+7\,\vec\jmath) \\\\ \vec r(t) = (6t-2)\,\vec\imath+(2t+5)\,\vec\jmath \\\\ \vec r(t) = x(t)\,\vec\imath + y(t)\,\vec\jmath[/tex]
with 0 ≤ t ≤ 1. Then the work performed by F along C is
[tex]\displaystyle \int_0^1 \left(6x(t)^3\,\vec\imath-4y(t)\,\vec\jmath\right)\cdot\frac{\mathrm d}{\mathrm dt}\left[x(t)\,\vec\imath + y(t)\,\vec\jmath\right]\,\mathrm dt \\\\ = \int_0^1 (288(3t-1)^3-8(2t+5)) \,\mathrm dt = \boxed{312}[/tex]
what are the two main types of sound like soundwave
Answer:
acoustic energy and mechanical energy
Explanation:
each type of sounds has to be tackled in their own way.
basic source of magnetism is a) charged particles alone b)Movement of charged particles c) Magnetic dipoles d)magnetic domains
Answer:
C . Magnetic dipoles is the correct
Answer:
b). movement of charged particles.
Explanation:
These charges create the nagnetic dipoles.
Calculate the kinetic energy of a mass 2kg moving with a velocity of 0.1m/s
ANSWER-:
1/2 mv²
K.E = 1/2 mv²
K.E = 0.01 J.
Hence, the kinetic energy of a body is 0.01 Joule.
!! HOPE ITS HELP U !!
Light of the same wavelength passes through two diffraction gratings. One grating has 4000 lines/cm, and the other one has 6000 lines/cm. Which grating will spread the light through a larger angle in the first-order pattern
Answer:
6000 lines/cm
Explanation:
From the question we are told that:
Grating 1=4000 lines/cm
Grating 2=6000 lines/cm
Generally The Spread of fringes is Larger when the Grating are closer to each other
Therefore
Grating 2 will spread the the light through a larger angle in the first-order pattern because its the closest with 6000 lines/cm
An electron is released from rest at a distance of 9.00 cm from a fixed proton. How fast will the electron be moving when it is 3.00 cm from the proton
Answer:
the speed of the electron at the given position is 106.2 m/s
Explanation:
Given;
initial position of the electron, r = 9 cm = 0.09 m
final position of the electron, r₂ = 3 cm = 0.03 m
let the speed of the electron at the given position = v
The initial potential energy of the electron is calculated as;
[tex]U_i = Fr = \frac{kq^2}{r^2} \times r = \frac{kq^2}{r} \\\\U_i = \frac{(9\times 10^9)(1.602\times 10^{-19})^2}{0.09} \\\\U_i = 2.566 \times 10^{-27} \ J[/tex]
When the electron is 3 cm from the proton, the final potential energy of the electron is calculated as;
[tex]U_f = \frac{kq^2}{r_2} \\\\U_f = [\frac{(9\times 10^9)\times (1.602 \times 10^{-19})^2}{0.03} ]\\\\U_f = 7.669 \times 10^{-27} \ J \\\\\Delta U = U_f -U_i\\\\\Delta U = (7.699\times 10^{-27} \ J ) - (2.566 \times 10^{-27} \ J)\\\\\Delta U = 5.133 \times 10^{-27} \ J[/tex]
Apply the principle of conservation of energy;
ΔK.E = ΔU
[tex]K.E_f -K.E_i = \Delta U\\\\initial \ velocity \ of \ the \ electron = 0\\\\K.E_f - 0 = \Delta U\\\\K.E_f = \Delta U\\\\\frac{1}{2} mv^2 = \Delta U\\\\where;\\\\m \ is \ the \ mass \ of\ the \ electron = 9.1 1 \times 10^{-31} \ kg\\\\v^2 = \frac{ 2 \Delta U}{m} \\\\v = \sqrt{\frac{ 2 \Delta U}{m}} \\\\v = \sqrt{\frac{ 2 (5.133\times 10^{-27})}{9.11\times 10^{-31}}}\\\\v = \sqrt{11268.935} \\\\v = 106.2 \ m/s[/tex]
Therefore, the speed of the electron at the given position is 106.2 m/s
