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SVETLANKA909090 [29]
2 years ago
9

In what unit do we measure the work done by a force

Physics
1 answer:
Gala2k [10]2 years ago
7 0

Explanation:

the work done by a force is measured in Newton(N) in SI unit and in CGS unit it is measured in dyne

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The two spheres pictured above have equal densities and are subject only to their mutual gravitational attraction. Which of the
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Answer:

Gravitational force

Explanation:

If two spheres have equal densities and they are subject only to their mutual gravitational attraction. We need to say that the quantities that must have the same magnitude for both spheres. So, the correct option is (E) i.e. gravitational force.

It is because of Newton's third law of motion. It states that the force due to object 1 to object 2 is same as force due to object 2 to object 1. The two forces act in opposite direction.  

Hence, the correct option is (E) "Gravitational force".                        

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Which element is most likely to carry electric current easily?
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#2 A car accelerates from rest at 4 m/s^2. What is the velocity of the car after 4 second?
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3 years ago
A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the
Margaret [11]

Hi there!

\boxed{\omega = 0.38 rad/sec}

We can use the conservation of angular momentum to solve.

\large\boxed{L_i = L_f}

Recall the equation for angular momentum:

L = I\omega

We can begin by writing out the scenario as a conservation of angular momentum:

I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)

I_m = moment of inertia of the merry-go-round (kgm²)

\omega_m = angular velocity of merry go round (rad/sec)

\omega_f = final angular velocity of COMBINED objects (rad/sec)

I_b = moment of inertia of boy (kgm²)

\omega_b= angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

I = MR^2

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

\omega = \frac{v}{r}

L_b = MR^2(\frac{v}{R}) = MRv

Plug in the given values:

L_b = (20)(3)(5) = 300 kgm^2/s

Now, we must solve for the boy's moment of inertia:

I = MR^2\\I = 20(3^2) = 180 kgm^2

Use the above equation for conservation of momentum:

600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}

8 0
2 years ago
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