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givi [52]
3 years ago
15

2. A cinder block is sitting on a platform 20 m high. It weighs 16kg. The block has

Physics
1 answer:
Cerrena [4.2K]3 years ago
7 0

Answer:

3136 Joules

Explanation:

Applying,

P.E = mgh.............. Equation 1

Where P.E = potential energy, m = mass of the cinder block, h = height of the platform, g = acceleration due to gravity.

From the question,

Given: m = 16 kg, h = 20 m

Constant: g = 9.8 m/s²

Substitute these values into equation 1

P.E = 16(20)(9.8)

P.E = 3136 Joules

Hence the potential energy of the cinder block is 3136 Joules

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Answer:

C. An external downward field is created or an external downward field is removed

Explanation:

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Therefore because of the external field, the field out of page & flux would be reducing or the external upward field is eliminated

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A block is at rest on a plank whose angle can be varied. The angle is gradually increased from 0 deg. At 31.8°, the block starts
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Answer:

\mu_s = 0.62

\mu_k = 0.415

The motion of the block is downwards with acceleration 1.7 m/s^2.

Explanation:

First, we will calculate the acceleration using the kinematics equations. We will denote the direction along the incline as x-direction.

x - x_0 = v_0t + \frac{1}{2}at^2\\3.4 = 0 + \frac{1}{2}a(2)^2\\a = 1.7~m/s^2

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F = ma\\F = 1.7m

Now, let’s investigate the free-body diagram of the block.

Along the x-direction, there are two forces: The x-component of the block’s weight and the kinetic friction force. Therefore,

F = mg\sin(\theta) - \mu_k mg\cos(\theta)\\1.7m = mg\sin(31.8) - \mu_k mg\cos(31.8)\\1.7 = (9.8)\sin(\theta) - mu_k(9.8)\cos(\theta)\\mu_k = 0.415

As for the static friction, we will consider the angle 31.8, but just before the block starts the move.

mg\sin(31.8) = \mu_s mg\cos(31.8)\\\mu_s = tan(31.8) = 0.62

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