<h3>
Answer: Approximately 4.67 m/s^2</h3>
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Explanation:
Let's say you want to push the brick to the right. The free body diagram will have an arrow pointing right on the rectangle (the brick) and the arrow is labeled with 35 N.
Friction always counteracts whatever force you apply. The friction force arrow will point left and be labeled with 7 N.
The net horizontal force is therefore 35-7 = 28 N and the direction is to the right. The positive net force means you've overcome the force of friction and the brick is moving.
F = 28 is the net force
m = 6 is the mass
a = unknown acceleration
F = m*a .... newton's second law
28 = 6a
6a = 28
a = 28/6
a = 4.67
The acceleration of the brick is approximately 4.67 m/s^2
This means that for every second, the brick's velocity is increasing by about 4.67 m/s.
Answer:
1. False
2. True
3. True
Explanation:
1- False —> The relation between electric potential and electric field is given such that
Therefore, for a uniform E field, electric potential is linearly proportional to the distance.
2- True —> The electric field lines always cross the equipotential lines perpendicularly.
3- True —> In order to be a potential difference, one source of electric field is enough. The electric potential will decrease radially according to the following formula:
There is no test charge in the formula, only the source charge. Even when there is no test charge, the potential difference between points in space can exist.
Accelerating objects are objects which are changing their velocity - either the speed (i.e., magnitude of the velocity vector) or the direction. An object undergoing uniform circular motion is moving with a constant speed. Hope you find this helpful
Answer:
30 m
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Explanation:
<u>We are given:</u>
Initial velocity (u) = 10 m/s
Final velocity (v) = 20 m/s
Time interval (t) = 2 seconds
Distance travelled by the bus (s) = s meters
<u>Solving for the distance travelled:</u>
<u>Solving for the acceleration:</u>
v = u + at [<em>first equation of motion</em>]
20 = 10 + a(2) [<em>replacing the given values</em>]
2a = 10 [<em>subtracting 10 from both sides</em>]
a = 5 m/s² [<em>dividing both sides by 2</em>]
<u>Solving for the distance:</u>
s = ut + 1/2 (at²) [<em>second equation of motion</em>]
s = 10(2) + 1/2(5)(2)² [<em>replacing the given values</em>]
s = 20 + 10
s = 30 m
Therefore, the bus travelled 30 m in the given time interval