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Verizon [17]
3 years ago
14

A 1.5-kg block slides at rest starts sliding down a snow-covered hill Point A, which has an altitude of 10 m. There is no fricti

on on hill. After leaving the hill at point B, it travels horizontally toward a massless spring with force constant of 200 N/m. While travelling, it encounters a 20-m patch of rough surface CD where the coefficient of kinetic friction is 0.15. (a) What is the speed of the block when it reaches point B
Physics
1 answer:
damaskus [11]3 years ago
7 0

Answer:

the speed of the block when it reaches point B is 14 m/s

Explanation:

Given that:

mass of the block slides = 1.5 - kg

height = 10 m

Force constant  = 200 N/m

distance of rough surface patch = 20 m

coefficient of kinetic friction = 0.15

In order to determine the speed of the block when it reaches point B.

We consider the equation for the energy conservation in the system which can be represented by:

\dfrac{1}{2}mv^2=mgh

\dfrac{1}{2}v^2=gh

v^2=2 \times g \times   h

v^2=2 \times 9.8 \times   10

v=\sqrt{2 \times 9.8 \times   10

v=\sqrt{196

v = 14 m/s

Thus; the speed of the block when it reaches point B is 14 m/s

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A box rests on top of a flat bed truck. The box has a mass of m = 16.0 kg. The coefficient of static friction between the box an
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Answer:

1) 1.31 m/s2

2) 20.92 N

3) 8.53 m/s2

4) 1.76 m/s2

5) -8.53 m/s2

Explanation:

1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

a = \frac{\Delta v}{\Delta t} = \frac{17 - 0}{13} = 1.31 m/s2

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

F_s = am = 1.31*16 = 20.92 N

3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.

F_{\mu_s} = \mu_sN = mg\mu_s = 16*9.81*0.87 = 136.6N

So the maximum acceleration on the block is

a_{max} = F_{\mu_s} / m = 136.6 / 16 = 8.53 m/s^2

4)As the box slides, it is now subjected to kinetic friction, which is

F_{\mu_s} = mg\mu_k = 16*9.81*0.69 = 108.3 N

So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is

28.25 / 16 = 1.76 m/s2

5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2

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3 years ago
A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Ne
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Answer:

the frequency of this mode of vibration is 138.87 Hz

Explanation:

Given;

length of the copper wire, L = 1 m

mass per unit length of the copper wire, μ = 0.0014 kg/m

tension on the wire, T = 27 N

number of segments, n = 2

The frequency of this mode of vibration is calculated as;

F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz

Therefore, the frequency of this mode of vibration is 138.87 Hz

7 0
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Answer:

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Explanation:

Because it exerts higher pressure as

P= F/ A

Larger the area less will be the pressure and less is the area larger will be the pressure.

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