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4 years ago

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A Batsled of mass 10kg sits on a horizontal surface. Batman (30kg) runs with a velocity of 5m/s. He jumps onto the sled (assume

a completely inelastic collision). What is the kinetic energy of the system (Batman and Batsled) immediately after the collision?

1 answer:

horsena [70]4 years ago

5 0

Answer:

30kg

Explanation:

because thats the weight

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In an experiment, an object is released from rest from the top of a building. Its speed is measured as it reaches a point that i

NikAS [45]

Answer:

so the speed will increase by 1.44 times then the initial speed if the distance is increased to double

Explanation:

As we know that the air friction or resistance due to air is neglected then we can use the equation of kinematics here

$v_f^2 - v_i^2 = 2 a d$

since we released it from rest so we have

$v_i = 0$

so here we have

$v_f = \sqrt{2gd}$

now if the distance is double then we have

$v_f' = \sqrt{2g(2d)}$

now from above two equations we can say that

$v_f' = \sqrt2 v_f$

so the speed will increase by 1.44 times then the initial speed if the distance is increased to double

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3 years ago

An archer pulls the bowstring back for a distance of 0.470 m before releasing the arrow. The bow and string act like a spring wh

never [62]

Answer:

(a) 46.94 J.

(b) 55.95 m/s

Explanation:

(a)

Potential Energy: This is the energy of a body, due to its position. The S.I unit of potential energy is Joules (J).

The formula of potential energy in a stretched spring is

Ep = 1/2ke² .......................... Equation 1

Where Ep = potential energy of the spring, k = Force constant of the spring, e = extension or compression.

Given: k = 425 N/m, e = 0.47 m.

Substitute into equation 1

Ep = 1/2(425×0.47²)

Ep = 46.94 J.

(b)

at the instant When the arrow leaves the bow, the potential energy of the arrow is converted kinetic energy of the bow.

I.e,

Ep = 1/2mv² ............. Equation 2

Where m = mass of the arrow, v = velocity of the arrow.

make v the subject of the equation

v = √(2Ep/m)............. Equation 3

Given: Ep = 46.94 J, m = 0.03 Kg.

Substitute into equation 3

v = √(2×46.96/0.03)

v = √(93.92/0.03)

v = √(3130.67)

v = 55.95 m/s

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3 years ago

Explanation A 5000 kg rocket is at rest in deep space. The rocket burns fuel pushing 10kg of exhaust gases rearward at 4000 m/s.

katovenus [111]

Answer:

F = 4000 N

Explanation:

given,

mass of rocket (M)= 5000 Kg

10 Kg gas burns at speed (m)= 4000 m/s

time = 10 s

average force = ?

at the end the rocket is at rest

by conservation of momentum

M v + m v' = 0

5000 x v - 10 x 4000 = 0

5000 v = 40000

v = 8 m/s

speed of rocket = 8 m/s

now,

we know

change in momentum = F x Δ t

$F = \dfrac{m(v_i-v_f)}{\Delta t}$

$F = \dfrac{5000(8-0)}{10}$

F = 4000 N

Hence, the average force applied to the rocket is equal to F = 4000 N

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Answer:

True

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Answer:

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