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sp2606 [1]
4 years ago
6

A 1,800 kg car is parked on a road that has an elevation level of 7°. Suppose the coefficient of static friction is .65. What is

the force of static friction?

Physics
1 answer:
zmey [24]4 years ago
8 0

m = mass of the car = 1800 kg

g = acceleration due to gravity = 9.8 m/s²

F_{g} = force of gravity on the car

force of gravity on the car is given as

F_{g} = mg

inserting the values

F_{g} = (1800) (9.8) = 17640 N

f_{s} = static frictional force acting on the car

from the force diagram of the car  , the static frictional force is opposite to the parallel component of force of gravity on the car . hence static frictional force mus balance the component of force of gravity parallel to incline surface.

hence

f_{s} = F_{g} Sin7

f_{s} = (17640) Sin7

f_{s} = 2149.8 N


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3 years ago
A 65-kg ice skater stands facing a wall with his arms bent and then pushes away from the wall by straightening his arms. At the
Marrrta [24]

Our values can be defined like this,

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F = \frac {mv ^ 2} {2d}

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B) The work done by the wall is zero since there was no displacement of the wall, that is d = 0.

6 0
3 years ago
At which of the following temperature and pressure levels would a gas be most likely to follow the ideal gas law? A. 0 K and 100
bulgar [2K]
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3) Particles are in continuous, random motion.
4) Collisions between particles are elastic, meaning basically that when they collide, they don't lose any energy.
5) The average kinetic energy is the same for all gasses at a given temperature, regardless of the identity of the gas.

It's generally true that gasses are mostly empty space and their particles occupy very little volume. Gasses are usually far enough apart that they exhibit very little attractive or repulsive forces. When energetic, the gas particles are also in fairly continuous motion, and without other forces, the motion is basically random. Collisions absorb very little energy, and the average KE is pretty close.

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A gas with a low pressure and a high temperature will be spread out and therefore exhibit ideal properties.

So, in analyzing the four choices given, we look for low P and high T.

A is at absolute zero, which is pretty much impossible, and definitely does not describe a gas. We rule this out immediately.

B and D are at the same temperature (273 K, or 0 °C), but C is at 100 K, or -173 K. This is very cold, so we rule that out.

We move on to comparing the pressures of B and D. Remember, a low pressure means the particles are more spread out. B has P = 1 Pa, but D has 100 kPa. We need the same units to confirm. Based on our metric prefixes, we know that kPa is kilopascals, and is thus 1000 pascals. So, the pressure of D is five orders of magnitude greater! Thus, the answer is B.
6 0
4 years ago
As part
umka21 [38]

Answer:

Part a)

a = 3.68 m/s^2

Part b)

a = 11.8 m/s^2

Explanation:

Part a)

For force conditions of two blocks we will have

m_1g - T = m_1 a

T - m_2g = m_2 a

now from above equations we have

(m_1 - m_2) g = (m_1 + m_2) a

a = \frac{m_1 - m_2}{m_1 + m_2} g

now we know that

m_1 = \frac{908}{9.8} = 92.65 kg

m_2 = \frac{412}{9.8} = 42 kg

now from above equation we have

a = \frac{92.65 - 42}{92.65 + 42}(9.8)

a = 3.68 m/s^2

Part b)

When heavier block is removed and F = 908 N is applied at the end of the string then we have

F - mg = ma

908 - 412 = 42 a

a = 11.8 m/s^2

8 0
3 years ago
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Nana76 [90]

Answer: v = 880m/s

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