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2 years ago

4. How much work is done when you lift a 25 N box up to a 2 m shelf?

Physics

1 answer:

Semmy [17]2 years ago

6 0

Answer:

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question

force = 25 N

distance = 2 m

We have

workdone = 25 × 2 = 50

We have the final answer as

Hope this helps you

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^^ help :c

ipn [44]

I need help to I have a test and I need help to

6 0

3 years ago

The question states: two large, parallel conducting plates are 12cm

AnnZ [28]

Answer:

1. 24375 N/C

2. 2925 V

Explanation:

d = 12 cm = 0.12 m

F = 3.9 x 10^-15 N

q = 1.6 x 10^-19 C

1. The relation between the electric field and the charge is given by

F = q E

So, $E=\frac{F}{q}$

$E=\frac{3.9 \times 10^{-15}}{1.6 \times 10^{-19}}$

E = 24375 N/C

2. The potential difference and the electric field is related by the given relation.

V = E x d

where, V be the potential difference, E be the electric field strength and d be the distance between the electrodes.

By substituting the values, we get

V = 24375 x 0.12 = 2925 Volt

6 0

2 years ago

When a pendulum is at the position all the way to the left when it is swinging (at the top of the arc), what is true of the kine

JulsSmile [24]

Choice - B is the correct one.

At the top of the arc, at one end of the swing:
-- it's not going to get any higher, so the potential energy is maximum
-- it stops moving for an instant, so the kinetic energy is zero

At the bottom of the arc, in the center of the swing:
-- it's not going to get any lower, so the potential energy is minimum
-- it's not going to move any faster, so the kinetic energy is maximum

7 0

3 years ago

Real-world examples of power

Paha777 [63]

By definition, power is the amount of energy consumed (or produced) in a second. (or more precisely, it is the rate of change in energy).
so anything which uses energy in a known time period can be labeled with a power rating.

an example for power could be a nuclear plant; traditional nuclear plants produce somewhat close to 1 giga watts (which means 1 giga joules in a second)

3 0

2 years ago

A machine that operates a ride at the fair requires 2500 J to lift a 294 N child 5.0 m. What is the efficiency of this machine?

NeX [460]

Answer:

η = 58.8%

Explanation:

Work is defined as the force applied by the distance traveled by the body.

$W =F*d$

where:

W = work [J] (units of joules)

F = force = 294 [N]

d = distance = 5 [m]

$W = 294*5\\W = 1470 [J]\\$

Efficiency is defined as the energy required to perform an activity in relation to the energy actually added to perform some activity. This can be better understood by means of the following equation.

$efficiency = W_{done}/W_{required}\\efficiency = 1470/2500\\efficiency = 0.588 = 58.8%$

5 0

2 years ago