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ANEK [815]
3 years ago
13

8.) If a car moving at 50km/h skids 15m with locked brakes, how far does the same car moving at 100km/h

Physics
1 answer:
pantera1 [17]3 years ago
4 0

(8) A car starting with a speed <em>v</em> skids to a stop over a distance <em>d</em>, which means the brakes apply an acceleration <em>a</em> such that

0² - <em>v</em>² = 2 <em>a</em> <em>d</em> → <em>a</em> = - <em>v</em>² / (2<em>d</em>)

Then the car comes to rest over a distance of

<em>d</em> = - <em>v</em>² / (2<em>a</em>)

Doubling the starting speed gives

- (2<em>v</em>)² / (2<em>a</em>) = - 4<em>v</em>² / (2<em>a</em>) = 4<em>d</em>

so the distance traveled is quadrupled, and it would move a distance of 4 • 15 m = 60 m.

Alternatively, you can explicitly solve for the acceleration, then for the distance:

A car starting at 50 km/h ≈ 13.9 m/s skids to a stop in 15 m, so locked brakes apply an acceleration <em>a</em> such that

0² - (13.9 m/s)² = 2 <em>a</em> (15 m) → <em>a</em> ≈ -6.43 m/s²

So the same car starting at 100 km/h ≈ 27.8 m/s skids to stop over a distance <em>d</em> such that

0² - (27.8 m/s)² = 2 (-6.43 m/s²) <em>d</em> → <em>d</em> ≈ 60 m

(9) Pushing the lever down 1.2 m with a force of 50 N amounts to doing (1.2 m) (50 N) = 60 J of work. So the load on the other end receives 60 J of potential energy. If the acceleration due to gravity is taken to be approximately 10 m/s², then the load has a mass <em>m</em> such that

60 J = <em>m g h</em>

where <em>g</em> = 10 m/s² and <em>h</em> is the height it is lifted, 1.2 m. Solving for <em>m</em> gives

<em>m</em> = (60 J) / ((10 m/s²) (1.2 m)) = 5 kg

(10) Is this also multiple choice? I'm not completely sure, but something about the weight of the tractor seems excessive. It would help to see what the options might be.

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3 years ago
Jin knows that the initial internal energy of a closed system is 78 J and the final internal energy is 180 J. He also knows that
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As we know by the first law of thermodynamics

Q = \Delta U + W

here we know that

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now here we can say

\Delta U = 180 - 78 = 102 J

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now we can say that heat will be given as

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So best describe Jin's Error is

<em>B )For step 1, he should have subtracted 78 J from 180 J to find the change in internal energy. </em>

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As SCUBA divers go deeper underwater, the pressure from the weight of all the water above them increases tremendously which comp
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The problem is that a diver, experiences an increased pressure of water compresses nitrogen and more of it dissolves into the body. Just as there is a natural nitrogen saturation point at the surface, there are saturation points under water. Those depend on the depth, the type of body tissue involved, and also how long a diver is exposed to the extra pressure. The deeper a diver go, the more nitrogen the body absorbs.

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4 0
3 years ago
What are the things being compared in the following argument?
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C

Explanation:

3 0
2 years ago
A small 1.0 kg steel ball rolls west at 3.0 m/s collides with a large 3.0 kg ball at rest. After the collision, the small ball m
77julia77 [94]

Answer:

The direction of the momentum of the large ball after the collision with respect to east is 146.58°.

Explanation:

Given that,

Mass of large ball = 3.0 kg

Mass of steel ball = 1.0 kg

Velocity = 3.0 kg

After collision,

Velocity = 2.0 m/s

Using conservation of momentum

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

3.0\times0+1.0\times(3.0)(-i)=1.0\times2(-j)+3.0\times v_{2}

-3i+2j=3.0\times v_{2}

v_{2}=-i+0.66j

The direction of the momentum

tan\theta=\dfrac{0.66}{-1}

\theta=tan^{-1}\dfrac{0.66}{-1}

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The direction of the momentum with respect to east

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Hence, The direction of the momentum of the large ball after the collision with respect to east is 146.58°.

7 0
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