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I am Lyosha [343]
3 years ago
7

A force of 100. newtons is used to move an object a distance of 15 meters with a power of 25 watts. Find the time it takes to do

the work.
Physics
1 answer:
Flauer [41]3 years ago
3 0
<h3>It takes 60 seconds to do the work</h3>

<em><u>Solution:</u></em>

Given that,

Force = 100 newtons

Distance = 15 meters

Power = 25 watts

To find: time it takes to do the work

<em><u>Find the work done:</u></em>

work = force \times displacement\\\\work = 100\ newtons \times 15\ meters\\\\work = 1500\ joule

<em><u>Find the time taken</u></em>

power = \frac{work}{time}\\\\25\ watts = \frac{1500\ joule}{time}\\\\time = \frac{1500\ joule}{25\ watt}\\\\time = 60\ second

Thus it takes 60 seconds to do the work

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A car is moving in the positive direction along a straight highway and accelerates at a constant rate while going from point A t
rusak2 [61]

Answer:

The time where the avergae speed equals the instaneous speed is T/2

Explanation:

The velocity of the car is:

v(t) = v0 + at

Where v0 is the initial speed and a is the constant acceleration.

Let's find the average speed. This is given integrating the velocity from 0 to T and dividing by T:

v_{ave} = \frac{1}{T}\int\limits^T_0 {v(t)} \, dt

v_ave = v0+a(T/2)

We can esaily note that when <u><em>t=T/2</em></u><u><em> </em></u>

v(T/2)=v_ave

Now we want to know where the car should be, the osition of the car is:

x(t) = x_A + v_0 t + \frac{1}{2}at^2

Where x_A is the position of point A. Therefore, the car will be at:

<u><em>x(T/2) = x_A + v_0 (T/2) + (1/8)aT^2</em></u>

8 0
3 years ago
What is needed to set a body in motion
kumpel [21]

Answer:

Newtons law

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According to this law, a body at rest tends to stay at rest, and a body in motion tends to stay in motion, unless acted on by a net external force.

3 0
3 years ago
In an inkjet printer, letters and images are created by squirting drops of ink horizontally at a sheet of paper from a rapidly m
Serga [27]

Answer:

q = 6.48 \times 10^{-14} C

Explanation:

Deflection in the drop is due to electric field force

so we will have

F = qE

acceleration of the drop is given as

a = \frac{qE}{m}

a = \frac{q(7.75 \times 10^4)}{1.00 \times 10^{-11}}

a = 7.75 \times 10^{15} q

now we know that time to cross the plates is given as

t = \frac{D}{v}

t = \frac{0.02}{18}

t = 1.11 \times 10^{-3} s

now the deflection is given as

d = \frac{1}{2}at^2

0.310 \times 10^{-3} = \frac{1}{2}(7.75 \times 10^{15} q)(1.11 \times 10^{-3})^2

0.310 \times 10^{-3} = 4.78 \times 10^9 q

q = 6.48 \times 10^{-14} C

5 0
3 years ago
two charges are placed at corners A and B of a square of side length. How much work is needed to move a charge from point C to D
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A study occasionally the effect of anxiety (low vs. high) and stress (low vs. moderate vs. high) on test.

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Learn more about anxiety here:
brainly.com/question/4913240

#SPJ4

6 0
2 years ago
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