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wlad13 [49]
2 years ago
15

When the body said to be in equilibrium​

Physics
2 answers:
Schach [20]2 years ago
7 0

Answer:

Oh! I just learned abt this

Explanation:

the body is said to be in equilibrium if it is resting or moving with uniform velocity.

zheka24 [161]2 years ago
5 0

Answer: The body is said to be in balance if it is at rest or moving at a uniform speed. In other words, if the body's linear and angular acceleration is zero, the body is said to be in balance. 1

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How high up is a 5kg object that has 300j of energy
Eddi Din [679]

-- If the object is moving with speed of 10.954 meters per second, then
it has 300J of kinetic energy no matter where it may be located.

-- If the object is 6.118 meters above somewhere, then it has 300J of
gravitational potential energy relative to that place. 


7 0
3 years ago
If there are 50 people per square kilometer in a city, and the area of the city is 1.5 × 10 square kilometers. What is the total
qaws [65]

Answer:

750 people

Explanation:

From the question,

Number of people in the city = population density×Area of the city

N = D×A.......................... Equagtion 1

Where N = Number of people in the city, D = population density, A = Area of the city.

Given: D = 50 people per square kilometer, A = 1.5×10 square kilometer.

Substitute into equation 1

N = 50(1.5×10)

N = 750 people.

Hence the total number of people in the city is 750 people.

6 0
3 years ago
I'm confusion, can I please get help???
kotykmax [81]

Answer:

Anything below 7.0 is acidic, so the range would be 0 to 7.

Neutral is simply 7, in the middle of the scale.

Lastly, anything above 7.0 is basic or alkaline, so that would be 7 to 14.

Good luck, I hope this helps

7 0
3 years ago
8. two +1 C charges are separated by 3000m. What is the magnitude of the electric force between them?
Sidana [21]

Answer:

1000 N

Explanation:

The magnitude of the electrostatic force between two charged object is given by

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb constant

q1, q2 is the magnitude of the two charges

r is the distance between the two objects

Moreover, the force is:

- Attractive if the two forces have opposite sign

- Repulsive if the two forces have same sign

In this problem:

q_1=q_2=+1C are the two charges

r = 3000 m is their separation

Therefore, the electric force between the charges is:

F=(9\cdot 10^9)\frac{(1)(1)}{3000^2}=1000 N

8 0
3 years ago
What should we do to be a computer engineer ‍ ‍?​
ycow [4]

Answer:

to be an eginere u would have to go to college and study hard

Explanation:

6 0
3 years ago
Read 2 more answers
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