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katen-ka-za [31]
4 years ago
11

two objects were lifted by a machine. One object had a mass of 2 kilograms, was lifted at a speed of 2m/sec . The other had a ma

ss of 4 kilograms and was lifted at 3m/sec. Which object had more kinetic energy while it was being lifted? Show all calculations
Physics
1 answer:
Lubov Fominskaja [6]4 years ago
5 0

Object 2 has more kinetic energy

Explanation:

The kinetic energy of an object is the energy possessed by the object due to its motion, and it is given by

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

In this problem, for object 1:

m = 2 kg

v = 2 m/s

So its kinetic energy is

K_1 = \frac{1}{2}(2)(2)^2=4 J

For object 2,

m = 4 kg

v = 3 m/s

So its kinetic energy is

K_2 = \frac{1}{2}(4)(3)^2=18 J

Therefore, object 2 has more kinetic energy.

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

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Answer:

10 m/s

Explanation:

Use the kinetic energy formula:

KE=(1/2)mv^2

I always remember it as Kevin is half-mad, and very square.

25J = (1/2)*0.5kg*(v^2)

50J = 0.5kg*(v^2)

100J = v^2

v = 10 m/s

Check it:

KE = (1/2)*0.5*(10^2)

KE = 25J

yep, it's right!

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The maximum current output of a 60 ω circuit is 11 A. What is the rms voltage of the circuit?
bezimeni [28]

Answer:

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Explanation:

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8 0
4 years ago
A lump of clay whose rest mass is 4 kg is travelling at three-fifths the speed of light when it collides head-on with an identic
Harman [31]

Answer:

mass of the composite lump is 10 kg

Explanation:

given data

mass = 4 kg

to find out

mass of composite lump

solution

we know energy is conserved so

so m1 = m2 = m0 that is 4kg

and

E(1) release+ E(2) release = E(1,2) rest

so γ(1)m(1)c² + γ(2)m(2)c²  = Mc²    ..........................1

that why here

|v(1)| = |v(2)| = 3/5 c        ......................2

and

γ = 1 / √(1 − v²/c²)           .......................3

put here v = 3 and c is 5

γ = 1 /√(1 − 9/25)

γ =  5/4

so

γ(1) = γ(2) = γ = 5/4  

so from equation 1

γ(1)m(1)c² + γ(2)m(2)c²  = Mc²

M = 2γm0

M = 2(5/4 )(4)

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Given a graph of volocity v . Time what does a horizontal line represent
kifflom [539]

Answer:

Acceleration

Explanation:

For example, if the acceleration is zero, then the velocity-time graph is a horizontal line (i.e., the slope is zero). If the acceleration is positive, then the line is an upward sloping line (i.e., the slope is positive).

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