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3 years ago

A machine does 1200 J of work in 1 min. What is the power developed

Physics

1 answer:

densk [106]3 years ago

8 0

Answer:

20 watts

Explanation:

Big brain mode activated:

Power=1200J/60sec

Power=20 watts

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Answer:

If the velocity and acceleration are in the same direction (both have the same sign - both positive or both negative) the object is speeding up. If the velocity and acceleration are in opposite directions (they have opposite signs), the object is slowing down.

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Answer:

Explanation:

The question does not specify any outside forces that could slow down the ball horizontally. There fore the ball does not accelerate or decelerate horizontally. Therefore, a = 0m/s2

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Why does a round pizza come in a square box? Random questions

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A box is pushed 40 m by a mover. The amount of work done was 2,240 j. How much force was exerted on the box

Georgia [21]

The force exerted on the box is 56 N

Explanation:

The work done by a force on an object is given by

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the direction of the force and of the displacement

For the box in this problem, we have:

W = 2240 J is the work done

d = 40 m is the displacement of the box

Assuming that the force is parallel to the displacement, $\theta=0$

Solving the equation for F, we find the force exerted on the box:

$F=\frac{W}{d cos \theta}=\frac{2240}{(40)(cos 0)}=56 N$

Learn more about work:

brainly.com/question/6763771

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3 years ago

A bike race against the clock takes place on a straight road. Yan drives at 37 km / h and he starts the course 30s before Christ

joja [24]

Given data:

Yan speed;

$u_1=37\text{ km/h}$

Christopher speed;

$u_2=38.9\text{ km/h}$

Christophe starts 30 s later than Yan. Therefore, Christophe takes 30 s less than Yan to reach the same distance.

Part (A)

The distance is given as,

$d=ut$

Let both Yan and Christophe meet at d distance from the start position. Therefore,

$u_1t=u_2(t-30)$

Substituting all known values,

$\begin{gathered} (37\text{ km/h})t=(38.9\text{ km/h})\times(t-30) \\ \frac{(37\text{ km/h})}{(38.9\text{ km/h})}=\frac{(t-30)}{t} \\ 0.95=1-\frac{30}{t} \\ \frac{30}{t}=1-0.95 \\ \frac{30}{t}=0.05 \\ t=\frac{30}{0.05} \\ t=600\text{ s} \end{gathered}$

Therefore, 600 s after Yan's departure Christophe will join him.

Part (B)

The distance is given as,

$d=u_1t$

Substituting all known values,

$\begin{gathered} d=(37\text{ km/h})\times(600\text{ s}) \\ =(37\text{ km/h})\times(600\text{ s})\times(\frac{1\text{ hr}}{3600\text{ s}}) \\ \approx6.17\text{ km} \end{gathered}$

Therefore, Christophe joins Yan after 6.17 km from the start.

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1 year ago