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vekshin1
4 years ago
8

A box on a ramp is connected by a rope to a winch. The winch is turned so that the box moves down the ramp at a constant speed.

The box experiences kinetic friction with the ramp. Which forces on the box do zero work as the box moves down the ramp?
a. Weight (gravitational force)
b. Normal force
c. Kinetic friction force
d. Tension force
e. None
Physics
1 answer:
Marrrta [24]4 years ago
7 0

Answer:

Option B:

The normal force

Explanation:

The normal force does no work as the box slides down the ramp.

Work can only be done when the force succeeds in moving the object in the direction of the force.

All the other forces involved have a component that is moving the box in their direction.

However, the normal force does not, as it points downwards into the ramp. Since the normal force is pointing into the ramp, and the box is sliding down the ramp, we can say that no work is being done by the normal force because the box is not moving in its direction (which would have been the box moving into the ramp)

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Answer:

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distance travelled in reaction time = v*t

= 20 m/s * 0.70 s

= 14 m

So, when brake is applied, distance remaining= 110 m - 14 m = 96 m

Answer: 96 m

b)

vf = 0 m/s

d = 96 m

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use:

vf^2 = vi^2 + 2*a*d

0 = 20^2 + 2*a*96

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a = -2.08 m/s^2

Answer: -2.08 m/s^2

c)

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4 years ago
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Automobile A and B are initially 30 m apart travelling in adjacent highway lanes at speeds VA = 14.4 km/hr., VB 23.4 km/hr. at t
marshall27 [118]

Answer:

        x = 240 m

Explanation:

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           x = x₀ₐ+ v₀ₐ t + ½ aₐ t²

         

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for car B

          x = x_{ob} + v_{ob} t - ½ a_b t²

     

car B's starting position is 30 m

         x = 30 + v_{ob} t - ½ 0.4 t²

at the point where they meet, the position of the two vehicles is the same

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let's reduce the speeds to the SI system

        v₀ₐ = 14.4 km / h (1000 m / 1 km) (1h / 3600s) = 4 m / s

        v_{ob} = 23.4 km / h = 6.5 m / s

        4 t + 0.4 t² = 30 + 6.5 t - 0.2 t²

        0.2 t² - 2.5 t - 30 = 0

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we solve the quadratic equation

       t = \frac{12.5 \pm \sqrt{12.5^2 + 4 \ 150}  }{2}

       t = \frac{12.5 \  \pm 27.5}{2}

       t₁ = 20 s

       t₂ = -7.5 s

time must be a positive quantity so the correct result is t = 20 s

let's look for the distance

        x = 4 t + ½ 0.8 t²

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8 0
3 years ago
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nevsk [136]

Answer:

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y = 1 500 10⁻⁹ 1.5 / 1 10⁻³  

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y = 1 600 10⁻⁹ 1.5 / 1.10⁻³  

y = 9 10⁻⁴ m  

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violet, green, orange, red  

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d) To find the wavelength we must measure the distance y from the center of the slit for various interference orders and using the equation we can find the wavelength of rationality

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