As the charges’ distance increase, there is a weaker force of attraction between them hence the electrostatic force decreases as distance increases. It increase by 4 (times 4) so the force will decrease by 4 making the answer
=A (400 divided by 4 = 100)
When the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².
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Frictional force between the block and the horizontal surface</h3>
The frictional force between the block and the horizontal surface is determined by applying Newton's law;
∑F = ma
F - Ff = ma
Ff = F - ma
Ff = 4 - 2(1.2)
Ff = 4 - 2.4
Ff = 1.6 N
When the applied force increases to 5 N, the magnitude of the block's acceleration is calculated as follows;
F - Ff = ma
5 - 1.6 = 2a
3.4 = 2a
a = 3.4/2
a = 1.7 m/s²
Thus, when the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².
Learn more about frictional force here: brainly.com/question/4618599
Http://earthguide.ucsd.edu/virtualmuseum/virtualmuseum/OriginofGulfStream.shtml this website might help u find ur answer
We are given:
v0 = initial velocity = 18 km/h
d = distance = 4 km
v = final velocity = 75 km/h
a =?
<span>
We can solve this problem by using the formula:</span>
v^2 = v0^2 + 2 a d
75^2 = 18^2 + 2 (a) * 4
5625 = 324 + 8a
<span>a = 662.625 km/h^2</span>