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SpyIntel [72]
3 years ago
8

A 50kg boy jumps off the back of a 4 kg skateboard moving forward. Find the skateboard’s velocity immediately after the boy jump

s, assuming that the skateboard’s initial velocity is 5 m/s and the boy’s velocity when jumping off the back is 2.5 m/s in the forward direction.
Physics
1 answer:
AleksAgata [21]3 years ago
6 0

The mass of the boy and skateboard is there to throw you off. X and Y velocities are independent of each other, meaning that the only things you need are the velocity of the skateboard and the velocity of the boy. The total velocity of the boy is 7.5 m/s because the boy would still be moving with the 5 m/s that the skateboard has if the boy were to just jump straight up. Since the boy jumps forward with a velocity of 2.5 m/s, you would add them together.

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What is the electric field strength inside the capacitor after insulating handles are used to pull the electrodes away from each
Bezzdna [24]
This question apparently comes after an EARLIER one,
where you were told either the voltage across the same
capacitor or the total charge stored in it.  You can't answer
THIS one without that information.
3 0
3 years ago
A 2.35-kg rock is released from rest at a height of 21.4 m. Ignore air resistance and determine (a) the kinetic energy at 21.4 m
kvasek [131]

Explanation:

Given that,

The mass of rock, m = 2.35-kg

It was released from rest at a height of 21.4 m.

(a) The kinetic energy is given by : E_k=\dfrac{1}{2}mv^2

As the rock was at rest initially, it means, its kinetic energy is equal to 0.

(b) The gravitational potential energy is given by : E_p=mgh

It can be calculated as :

E_p=2.35\times 9.8\times 21.4\\\\E_p=492.84\ J

(c) The mechanical energy is equal to the sum of kinetic and potential energy such that,

M = 0 J + 492.84 J

M = 492.84 J

Hence, this is the required solution.

6 0
3 years ago
While moving a box 10.0 m across the floor, Sam pushes with a force of 150 N to the right. The force friction acting on the box
svet-max [94.6K]

Answer:

Explanation:required formula is

W 1=F*S

W1=work done by Sam =?

F=force applied by sam=150N

S=displacement =10m

again

W2=F*S

W2=work done by friction =?

S=displacement =10m

F=friction =25N

W=W1-W2=net work done

please feel free to ask if you have any questions

4 0
3 years ago
A tow truck exerts a net force of 1000 N on a 600 kilogram car What’s the cars acceleration
Mice21 [21]

Answer:

a = 1000/600= 5/3 m/s^2

5 0
3 years ago
A ball is launched with initial speed v from the ground level up a frictionless hill. The hill becomes steeper as the ball slide
Triss [41]

Answer:

H(max) = (v²/2g)

Explanation:

The maximum height the ball will climb will be when there is no friction at all on the surface of the hill.

Normally, the conservation of kinetic energy (specifically, the work-energy theorem) states that, the change in kinetic energy of a body between two points is equal to the work done in moving the body between the two points.

With no frictional force to do work, all of the initial kinetic emergy is used to climb to the maximum height.

ΔK.E = W

ΔK.E = (final kinetic energy) - (initial kinetic energy)

Final kinetic energy = 0 J, (since the body comes to rest at the height reached)

Initial kinetic energy = (1/2)(m)(v²)

Workdone in moving the body up to the height is done by gravity

W = - mgH

ΔK.E = W

0 - (1/2)(m)(v²) = - mgH

mgH = mv²/2

gH = v²/2

H = v²/2g.

7 0
3 years ago
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