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Paraphin [41]
3 years ago
9

An object moves from point A to point B then to point C. It then moves back to point A. What is the displacement of the object?

Physics
1 answer:
lana [24]3 years ago
6 0
The displacement is 0
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PLEASE HELP ME
sashaice [31]

Given that,

Electrical energy of the motor = 2280 J

Kinetic energy = 1606 J

Wasted energy in thermal and sound energy = 550 J

To find,

The efficiency of the motor in percent.

Solution,

Efficiency is defined as the ratio of useful output energy to the total energy.

Total energy = 2280 J

Useful energy = 1606 J

\eta=\dfrac{1606}{2280}\times 100\\\\=70.43\%

So, the efficiency of the motor is 70.43%.

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Which resource would be the best choice to learn more information about studying martial arts?
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Explanation:

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What is the average velocity in the time interval 5 to 6<br> seconds?
Alexeev081 [22]

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The average velocity from 5 to 6 seconds is -27 m/s

6 0
3 years ago
A mysterious rocket-propelled object of mass 49.5 kg is initially at rest in the middle of the horizontal, frictionless surface
Romashka-Z-Leto [24]

The distance covered by the rocket in the first 5 seconds is 7.13 m

Explanation:

In order to solve the problem, we have to find the acceleration of the rocket. We can do it by using Newton's second law, which states that the net force on the rocket is equal to the product between its mass (m) and its acceleration (a):

F(t)=ma(t)

Where here we have:

F(t)=16.8t is the time-dependent force

m = 49.5 kg is the mass

Solving for a(t),

a(t)=\frac{F(t)}{m}=\frac{16.8t}{49.5}=0.34t[m/s^2]

This is the time-dependent acceleration of the rocket.

By integrating this expression, we  find the velocity of the rocket at time t:

v(t)=\int a(t)dt=\int (0.34t)dt=\frac{0.34t^2}{2}=0.17t^2

where we didn't add the constant term since the velocity at t = 0 is zero (the rocket starts from rest).

By integrating this expression, we then find an expression for the position of the rocket at time t:

x(t)=\int v(t) dt = \int (0.17t^2)dt=\frac{0.17t^3}{3}=0.057t^3

Therefore, the distance covered by the rocket in the first t = 5.00 s is:

x(5)=0.057(5.0)^3=7.13 m

Learn more aboiut force and acceleration:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

#LearnwithBrainly

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