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arlik [135]
2 years ago
14

A girl rolls a ball up an incline and allows it to re- turn to her. For the angle and ball involved, the acceleration of the bal

l along the incline is constant at 0.25g, directed down the incline. If the ball is re- leased with a speed of 4 m /s, determine the distance s it moves up the incline before reversing its direc- tion and the total time t required for the ball to re- turn to the child’s hand.
Physics
1 answer:
zalisa [80]2 years ago
7 0

Answer:

3.28 m

3.28 s

Explanation:

We can adopt a system of reference with an axis along the incline, the origin being at the position of the girl and the positive X axis going up slope.

Then we know that the ball is subject to a constant acceleration of 0.25*g (2.45 m/s^2) pointing down slope. Since the acceleration is constant we can use the equation for constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

X0 = 0

V0 = 4 m/s

a = -2.45 m/s^2 (because the acceleration is down slope)

Then:

X(t) = 4*t - 1.22*t^2

And the equation for speed is:

V(t) = V0 + a * t

V(t) = 4 - 2.45 * t

If we equate this to zero we can find the moment where it stops and begins rolling down, that will be the highest point:

0 = 4 - 2.45 * t

4 = 2.45 * t

t = 1.63 s

Replacing that time on the position equation:

X(1.63) = 4 * 1.63 - 1.22 * 1.63^2 = 3.28 m

To find the time it will take to return we equate the position equation to zero:

0 = 4 * t - 1.22 * t^2

Since this is a quadratic equation it will have to answers, one will be the moment the ball was released (t = 0), the other will eb the moment when it returns:

0 = t * (4 - 1.22*t)

t1 = 0

0 = 4 - 1.22*t2

1.22 * t2 = 4

t2 = 3.28 s

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A magnetic force can act on an electron even when it A) is at rest B) moves parallel to magnetic field lines C) both of these D)
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This force is given below as

F=qvB*sinθ

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What defines the force exerted on the charge is the angle between the field and it velocity.

If magnetic field is parallel to velocity, then it means that θ=0° which means sin 0 = 0, which means

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A 4 kg toy car moves horizontally on a rough road with coefficient of kinetic friction 0.2. It accelerates from rest to 20 m/s i
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The total work done on the car is 784Joule.

<h3>What's the acceleration of the car?</h3>
  • As per Newton's equation of motion, V= U+at
  • U= initial velocity= 0 m/s

V= vinal velocity= 20m/s

t= time = 10s

a= acceleration

  • So, 20= 0+ 10a

=> a= 20/10= 2m/s²

<h3>What's the distance covered by the car in 10 seconds?</h3>
  • As per Newton's equation of motion,

V²-U² = 2aS

  • S= distance covered by the car
  • So, 20²-0=2×2×S=4S

=> 400= 4S

=> S= 400/4= 100m

<h3>What's the work done on the car due to frictional force?</h3>

Work done by frictional force= frictional force × distance

= (0.2×4×9.8)×100

= 784Joule

Thus, we can conclude that the work done on the car is 784Joule.

Learn more about the work done here:

brainly.com/question/25573309

#SPJ1

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