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lutik1710 [3]
3 years ago
7

A 48.5 kg student runs down the sidewalk and jumps with a horizontal speed of 4.25 m/s onto a stationary skateboard. The student

and skateboard move down the sidewalk with a speed of 4.05 m/s. a) Find the mass of the skateboard. Answer in units of kg.
Physics
1 answer:
storchak [24]3 years ago
7 0

<u>Answer:</u>

2.39 kg

<u>Explanation:</u>

There is conservation of momentum here in this problem so we will use the following problem:

m_1u_1+m_2u_2=(m_1+m_2)v

where the mass of the student m_1 is 48.5 kg,

the mass of the skateboard m_2 is m_2 kg,

the initial speed of the student u_1 is 4.25 m/s; and

the speed of the student and skateboard  v is 4.05 m/s.

So substituting the given values in the above formula to get:

(48.5*4.25) + (m_2 * 0) = (48.5 + m_2 ) * 4.05

206.125=196.425+4.05m_2

206.125 - 196.425 = 4.05m_2

m_2=\frac{9.7}{4.05}

m_2=2.39

Therefore, the mass of the skateboard is 2.39 kg.

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Answer:

conservation of momentum, general law of physics according to which the quantity called momentum that characterizes motion never changes in an isolated collection of objects; that is, the to

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3 years ago
What best describes an impulse acting on an object? the net force on an object divided by the time of impact the velocity of an
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Explanation:

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2 years ago
A plane rises from​ take-off and flies at an angle of 5 degrees5° with the horizontal runway. When it has gained 800800 ​feet, f
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Answer:

distance=9188149.567feet

Explanation:

Given Data

Angle α=5°

height h=800800 feet

To find

Distance r

Solution  

As we Know that

Sin\alpha =(\frac{Perpendicular}{hypotenuse} )\\Sin\alpha =\frac{h}{r}\\ r=\frac{h}{Sin\alpha}\\ r=\frac{800800feet}{Sin(5^{o} )}\\ r=9188149.567feet

4 0
3 years ago
Now imagine a person dragging a 50 kg box along the ground with a rope, as
ANTONII [103]

Answer:

The coefficient of static friction between the box and floor is, μ = 0.061

Explanation:

Given data,

The mass of the box, m = 50 kg

The force exerted by the person, F = 50 N

The time period of motion, t = 10 s

The frictional force acting on the box, f = 30 N

The normal force on the box, η = mg

                                                     = 50 x 9.8

                                                     = 490 N

The coefficient of friction,

                            μ = f/ η

                               = 30 / 490

                               = 0.061

Hence, the coefficient of static friction between the box and floor is, μ = 0.061

7 0
4 years ago
Part A If the velocity of a pitched ball has a magnitude of 47.5 m/s and the batted ball's velocity is 51.5 m/s in the opposite
nalin [4]

Explanation:

Let us assume that the mass of a pitched ball is 0.145 kg.

Initial velocity of the pitched ball, u = 47.5 m/s

Final speed of the ball, v = -51.5 m/s (in opposite direction)

We need to find the magnitude of the change in momentum of the ball and the impulse applied to it by the bat. The change in momentum of the ball is given by :

\Delta p=m(v-u)\\\\\Delta p=0.145\times ((-51.5)-47.5)\\\\\Delta p=-14.355\ kg-m/s

So, the magnitude of the change in momentum of the ball is 14.355 kg-m/s.

Let the the ball remains in contact with the bat for 2.00 ms. The impulse is given by :

J=\dfrac{\Delta p}{t}\\\\J=\dfrac{14.355}{2\times 10^{-3}}\\\\J=7177.5\ kg-m/s

Hence, this is the required solution.

7 0
4 years ago
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