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2 years ago

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Floor lamps usually have a base with large inertia, while the long body and top have much less inertia. Part A If you want to sh

ow off by spinning a floor lamp like a baton, where should you grab the lamp?

1 answer:

Novay_Z [31]2 years ago

5 0

Answer:

When a an object is been rotated its resistance capacity to that rotational force is know as rotational inertia and this mathematically given as

$I = mr^2$

Where m is the mass

r is the rotation radius

For the spinning of the lamp as a baton to work the location of the center of mass of the floor lamp needs to be located

This is more likely to be located closer to base of the lamp as compared to the top, so success of spinning a floor lamp like a baton is highly likely if the lamp is grabbed closer to the base because that is where the position of its center of mass is likely to be.

Explanation:

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Alex is standing still and throws a football with a speed of 10 m/s to his friend, who is also standing still. The two friends a

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The question is incomplete. It comes with a set of answer choices.

These are the answer choices:

Alex observes it as 10 m/s, and his friend observes it as less than 10 m/s.

Alex observes it as less than 10 m/s, and his friend observes it as 10 m/s.

Both Alex and his friend observe it as 10 m/s.

Both Alex and his friend observe it as less than 10 m/s.

Answer: Both Alex and his friend observe it as 10 m/s.

Justification:

1) The speed is relative to the frame of reference.

2) It is said that the both Alex and his friend are standing still.

3) Then, the speed they both see is the same, 10 m/s, respect the Earth (where they are standing still).

Of course, Alex is watching the ball moving away and his friend is seing it approaching, but it is not relevant for the question, as it deals with the speed which is only about magnitude, not direction.

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1 year ago

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A sample of an ideal gas is in a tank of constant volume. The sample absorbs heat energy so that its temperature changes from 38

oksano4ka [1.4K]

Answer:

the ratio is $\frac{V_2}{V_1}=\sqrt{2}$

Explanation:

Given

$Initial Temperature T_1=387 KFinal Temperature T_2=774 K$

The RMS velocity of molecules in a gas is given by

$V_{rms}=\sqrt{\dfrac{3k_bT}{m}}$

where T=temperature

$k_b=constant$

For T = 387K

$V_1=\sqrt{\frac{3k_b\cdot 387}{m}}----1$

For T = 774

$V_2=\sqrt{\frac{3k_b\cdot 774}{m}}----(2)$

dividing eqn 1 and eqn 2

$\frac{V_2}{V_1}=\sqrt{\frac{774}{387}}$

$\frac{V_2}{V_1}=\sqrt{2}$

Thus,the ratio is $\frac{V_2}{V_1}=\sqrt{2}$

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2 years ago

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A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equa

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Answer:

I = 16 kg*m²

Explanation:

Newton's second law for rotation

τ = I * α Formula (1)

where:

τ : It is the moment applied to the body. (Nxm)

I : it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)

α : It is angular acceleration. (rad/s²)

Kinematics of the wheel

Equation of circular motion uniformly accelerated :

ωf = ω₀+ α*t Formula (2)

Where:

α : Angular acceleration (rad/s²)

ω₀ : Initial angular speed ( rad/s)

ωf : Final angular speed ( rad

t : time interval (rad)

Data

ω₀ = 0

ωf = 1.2 rad/s

t = 2 s

Angular acceleration of the wheel

We replace data in the formula (2):

ωf = ω₀+ α*t

1.2= 0+ α*(2)

α*(2) = 1.2

α = 1.2 / 2

α = 0.6 rad/s²

Magnitude of the net torque (τ )

τ = F *R

Where:

F = tangential force (N)

R = radio (m)

τ = 80 N *0.12 m

τ = 9.6 N *m

Rotational inertia of the wheel

We replace data in the formula (1):

τ = I * α

9.6 = I *(0.6 )

I = 9.6 / (0.6 )

I = 16 kg*m²

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1 year ago

An electric clock is hanging on a wall. As you are watching the second hand rotate, the clock's battery stops functioning, and t

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Answer:

B. W is positive and a is negative

Explanation:

As we know that the angular speed of the second clock is in positive direction so as it comes to halt from its initial direction of motion then we have

initial angular velocity is termed as positive angular velocity

$\omega = positive$

now it comes to stop so angular acceleration is taken in opposite to the direction of angular speed

so we will have

$\alpha = negative$

so here correct answer is

B. W is positive and a is negative

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1 year ago

It's your birthday, and to celebrate you're going to make your first bungee jump. You stand on a bridge 110 m above a raging riv

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Answer:

h=20.66m

Explanation:

First we need the speed when the cord starts stretching:

$V_2^2=V_o^2-2*g*\Delta h$

$V_2^2=-2*10*(-31)$

$V_2=24.9m/s$ This will be our initial speed for a balance of energy.

By conservation of energy:

$m*g*h+1/2*K*(h_o-l_o-h)^2-m*g*(h_o-l_o)-1/2*m*V_2^2=0$

Where

$h$ is your height at its maximum elongation

$h_o$ is the height of the bridge

$l_o$ is the length of the unstretched bungee cord

$800h+21*(79-h)^2-63200-24800.4=0$

$21h^2-2518h+43060.6=0$ Solving for h:

$h_1=20.66m$ and $h_2=99.24m$ Since 99m is higher than the initial height of 79m, we discard that value.

So, the final height above water is 20.66m

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2 years ago

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