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3 years ago

What's the relationship between work done and force applied

1 answer:

3 0

When a force acts on a body along some path, the work done is W=F*s, where W is the work done, F is the force that is doing the work on the body and s is the path. The force doing the work has to be in the same direction, or parallel, as the path. This is called positive work. If the force and the path are anti-parallel, the work is negative. So the relationship between work and force is W=F*s.

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Using numbers to describe kinematic quantities is called what description?

MissTica

sorry dont know this

4 0

3 years ago

How fast can Usain Bolt run if it takes him 9.9 s to run 100m?

Arte-miy333 [17]

Answer:

27.8 mph

Explanation:

May I have brainliest please? :)

8 0

3 years ago

Read 2 more answers

A car is traveling at a speed of 54 km/h. Breaks are applied so as to produce a uniform acceleration of -0.5 m/s2. Find how far

Lena [83]

Explanation:

u=54 km/h

54*5/18=15 m/s

v=0m/s

t=?

acceleration=-0.5m/s^2

we know that a=v-u/t

so,

t=v-u/a

t=15-0/0.5

=15/0.5

=30

therefore, the time is 30 second

Hope this answer helps you..

8 0

3 years ago

A quick USB charger claims its output current is 1.97Amp. We know that the standard USB output voltage is 5V. What is the output

Arte-miy333 [17]

Answer:

Output power of the charger is 9.85 watts.

Explanation:

It is given that,

Output current of the USB charger, I = 1.97 A

The standard USB output voltage, V = 5 V

We need to find the output power of the charger. It can be determined using the following formula as :

P = V × I

$P=5\ V\times 1.97\ A$

P = 9.85 watts

The output power of the charger is 9.85 watts. Hence, this is the required solution.

3 0

2 years ago

A 5.0-μC charge is placed at the 0 cm mark of a meter stick and a -4.0 μC charge is placed at the 50 cm mark. At what point on a

Maksim231197 [3]

Answer:

The distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

Explanation:

Given that

q₁ = 5 μ C

q₂ = - 4 μ C

The distance between charges = 50 cm

d= 50 cm

Lets take at distance x from the charge μ C ,the electrical field is zero.

That is why the distance from the charge - 4 μ C = 50 - x cm

We know that ,electric field is given as

$E=K\dfrac{q}{r^2}$

$K\dfrac{5\ \mu}{x^2}=K\dfrac{4\mu }{(50-x)^2}\\\\\dfrac{5}{x^2}=\dfrac{4 }{(50-x)^2}\\\\\\5(50-x)^2=4x^2\\(50-x)^2=0.8x^2\\\\50-x =0.89x\\\ x=\dfrac{50}{1.89}\ cm\\\\\\x=26.45\ cm\\$

Therefore the distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

3 0

3 years ago