3 years ago

A machine runs for 50 seconds with a steady power output of 100 watts. How many joules of work does the

Physics

1 answer:

liberstina [14]3 years ago

6 0

Answer:

The answer to your question is when time = 50 s, work = 5000 J

when time = 90 s, work = 9000 J

Explanation:

Data

time = 50 s or 90 s

Power = 100 watts

Power is defined as the rate of work done per unit of time.

Power = Work / time

-Solve for Work

Work = Power x time

-Substitution

Work = 100 x 50

-Result

Work = 5000

2.-When time = 90 s

Work = 100 x 90

-Result

Work = 9000 watts

You might be interested in

Which of the following was not a famous basketball player for the NBA? Select one: a. Wilt Chamberlain b. Lady Bird Johnson c. M

DiKsa [7]

Answer:

B Lady Bird Johnson

3 0

3 years ago

A man inside the water seea a fish in water. will the fish appear to man in it's actual position? will the man appear to fish in

weqwewe [10]

Answer:

Explanation:

due to the refraction of light,meaning bending of light when it passes from diffrent mediums with diffrent densities

(probably mot the complete ans but smtg like this)

4 0

3 years ago

Read 2 more answers

Describe how transverse waves can be produced by a rope. Then describe hiw pieces of the rope move as waves pass.

Oduvanchick [21]

We did this experiment before, when the rope moves, it represents the waves passing through in from the level of intensity. I hope this is a good answer.

3 0

3 years ago

a u tube contains a liquid of an unknown density an oil of density is poured into the right arm of the tube until the oil column

scoray [572]

Answer:

The answer is " $1155\ \frac{kg}{m^3}$ "

Explanation:

Please find the complete question in the attached file.

$p = p_0 + ?gh$

pi = pressure only at two liquids' devices

PA = pressure atmosphere.

1 = oil density

2 = uncertain fluid density

$h_1 = 11 \ cm\\\\h_2= 3 \ cm$

The pressures would be proportional to the quantity $11-3 = 8$ cm from below the surface at the interface between both the oil and the liquid.

$\to p_A + ?2g(h_1 - h_2) = p_A + ? 1gh_1\\\\\to ?2 = \frac{?1h_1}{(h_1 - h_2)} \\\\$

$= \frac{840 \frac{kg}{m^3}}{\frac{11}{8}} \\\\= 1155\ \frac{kg}{m^3}$

8 0

3 years ago

Igneous rocks would most likely be found near:

algol [13]

Erosion and hot whether

6 0

3 years ago

Read 2 more answers