the velocity of a body of mass 60kg reaches 15m/s from 0m/s in 12 second. calculate the kinetic energy and power of the body.​

Answers

Answer 1

Answer:

KE=1/2m v^2

1/2*60*15*15

30*15*15

6750 joules

power=work/time


Related Questions

Earthquakes
Shaking of the ground
Seismographs
Scientific method

Answers

Answer:

C. seismographs

Explanation:

Theirs another name for seismographs but c is correct

A car traveling at 50 m/s comes to a stop in 5 seconds. What is the
acceleration of this?

Answers

Answer:

10

Explanation:

10 is the answer because you divide 50  and 5 and your answer is 10.

Hope this helps:)

Pls mark brainlist

Answer:

a = -10m/s²

Explanation:

Givens:

Vf = 0m/s

Vi = 50m/s

t = 5s

Use the following kinematic equation:

Vf = Vi+at

Rearrange

a = (Vf-Vi)/t

a = -10m/s²

Where are alkaline earth metals found on the periodic table?

Group 1
Group 2
Groups 3–12
Group 17

Answers

Answer:

B

Explanation:

Answer:

B

Explanation:

I had a feeling

Which source of energy causes most of the water evaporation on Earth’s surface?
A.
wind
B.
solar
C.
gravity
D.
electrical

Answers

Answer:

I think it will be b. solar

Hope I can help.

Answer:

Explanation:

its b

What year was the RoboSapien toy robot released?
a) 2007
b) 2004
c) 2020
d) dunno

Answers

option b )2004........

Charles, the 75 kg trampoline artist, lands on a trampoline with a speed of 9.0 m/s.
If the trampoline behaves like a spring with a spring constant of 52,000 N/m, what maximum distance will Charles push down the trampoline before bouncing back up? (Hint: at maximum compression, Charles is not moving.)

Answers

Answer:

The maximum distance Charles will push down the trampoline ≈ 0.342 m

Explanation:

The given parameters are;

The mass of the trampoline artist, m = 75 kg

The speed with which the trampoline artist lands, v = 9.0 m/s

The value of the spring constant of the trampoline, k = 52,000 N/m

Let x represents the maximum distance Charles will push down the trampoline

Therefore, we have;

Kinetic energy = 1/2·m·v²

The kinetic energy with which the trampoline artist lands  = 1/2 × 75 × 9.0² = 3037.5

The kinetic energy with which the trampoline artist lands  =  3037.5 J

The potential energy stored in a spring = 1/2·k·x² = The kinetic energy with which the trampoline artist lands

∴ 1/2 × 52,000 × x² = 3037.5

∴ x = √(3037.5/(1/2 × 52,000)) ≈ 0.342

The maximum distance Charles will push down the trampoline = x ≈ 0.342 m

A 2450 kg stunt airplane accelerates from 120 m/s to 162 m/s in 2.10s. If the airplane is putting out an average force of 5.8810x10^4 N during this time, what is the average friction force exerted on the airplane by the air?

Answers

Given :

A 2450 kg stunt airplane accelerates from 120 m/s to 162 m/s in 2.10 s.

If the airplane is putting out an average force of [tex]5.8810\times 10^4 \ N[/tex].

To Find :

The average friction force exerted on the airplane by the air.

Solution :

Acceleration is given by :

[tex]a = \dfrac{162-120}{2.10}\ m/s^2\\\\a = 20 \ m/s^2[/tex]

Now, force equation is given by :

[tex]F - F_{friction} = ma\\\\F_{friction} = F-ma\\\\F_{friction} = 58810 - (2450\times 20 )\\\\F_{friction} = 9810\ N[/tex]

Therefore, frictional force exerted in the airplane by the air is 9810 N.

As we know, the moon is a satellite of our earth, what is the
theoretical period of the moon? The average radius of the
moon's orbit is 3.84 108 m and the mass of the earth is 5.97 x
1024 kg (in hours, G = 6.67 x 10-9 N (m/kg) 3).

Answers

Answer:c

Explanation:c

This question involves the concepts of the time period, orbital radius, and gravitational constant.

The theoretical period of the moon is "658 hr".

The theoretical time period of the moon around the earth can be found using the following formula:

[tex]\frac{T^2}{R^3}=\frac{4\pi^2}{GM}[/tex]

where,

T = Time Period of Moon = ?

R = Orbital Radius = 3.84 x 10⁸ m

G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of Earth = 5.97 x 10²⁴ kg

Therefore,

[tex]\frac{T^2}{(3.84\ x\ 10^8\ m)^3}=\frac{4\pi^2}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(5.97\ x\ 10^{24}\ kg)}\\\\T^2=(9.91\ x\ 10^{-14}\ s^2/m^3)(56.62\ x\ 10^{24}\ m^3)\\\\T=\sqrt{561.34\ x\ 10^{10}\ s^2}[/tex]

T = 2.37 x 10⁶ s[tex](\frac{1\ h}{3600\ s})[/tex]

T = 658 hr

Learn more about the orbital time period here:

https://brainly.com/question/14494804?referrer=searchResults

The attached picture shows the derivation of the formula for orbital speed.

Which type of energies make up the mechanical energy of a roller coaster moving along a track?

Answers

Answer:

gravitational potential energy and kinetic energy

Explanation:

In its American colonies, Spain helped the Catholic Church meet its goal of
finding gold.
gaining territory.
converting people.
achieving glory.
This is from ed please help thank you

Answers

Answer: Converting people

Explanation: In Spanish colonies such as Florida, Spanish moved up into the other states to spread their christianity and were successful with doing so.

Converting people but I’m not sure

A girl pushes a wagon at constant velocity. If the
momentum of the wagon is 50 kg*m/s at a
velocity of 2 m/s, the mass of the wagon is what

Answers

Answer; 100 m/s
Explanation;
F= ma
50 kg x 2 m/s

What 2 factors affect the impulse on an object in a collision?

Answers

force an friction?

i looked it up an thats all i can find

How much power is used if a force of 35 newtons is used to push a box a distance of 10 meters in 5 seconds?w=350j

Answers

The answer your looking for is 70watts

Answer:

How much power is used if 350J of work is done when pushing a box for 5 seconds.

Explanation:

the answer is 70watts

What is the work energy transfer equation?

Answers

Answer:

The equation used to calculate the work done is: work done = force × distance. W = F × d. This is when: work done (W) is measured in joules (J)

Answer:

The equation used to calculate the work done is: work done = force × distance. W = F × d. This is when: work done (W) is measured in joules (J)

Explanation:

The net work done on a particle equals the change in the particle's kinetic energy:

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