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3 years ago

Beth moves a 15 N book 20 meters in 10 seconds. How much power was produced?

Physics

1 answer:

Mice21 [21]3 years ago

5 0

Answer:

30 Watts

Explanation:

Power = Work/Time

Work = Force*Distance

Power = Force * Distance / Time

Power = 15 N * 20 meters / 10 sec

Power = 30 Watts

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Starting at 1.0 m/s, a cheetah runs with a constant acceleration for 4.8 s reaching a speed of 28 m/s. What is the acceleration

LenaWriter [7]

Answer:

c. $5.6m/s^2$

Explanation:

Initial velocity of cheetah,u=1 m/s

Time taken by cheetah =4.8 s

Final velocity of cheetah,v=28 m/s

We have to find the acceleration of this cheetah.

We know that

Acceleration, $a=\frac{v-u}{t}$

Where v=Final velocity of object

u=Initial velocity of object

t=Time taken by object

Using the formula

Then, we get

Acceleration, a= $\frac{28-1}{4.8}=\frac{27}{4.8} m/s^2$

Acceleration= $a=5.6 m/s^2$

Hence, the acceleration of cheetah= $5.6m/s^2$

5 0

3 years ago

The intensity at distance from a spherically symmetric sound source is 100 W/m2. What is the intensity at five times this distan

ss7ja [257]

To solve this problem it is necessary to apply the concepts related to intensity as a function of power and area.

Intensity is defined to be the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity I is

$I = \frac{P}{A}$

The area of a sphere is given by

$A = 4\pi r^2$

So replacing we have to

$I = \frac{P}{4\pi r^2}$

Since the question tells us to find the proportion when

$r_1 = 5r_2 \rightarrow \frac{r_2}{r_1} = \frac{1}{5}$

So considering the two intensities we have to

$I_1 = \frac{P_1}{4\pi r_1^2}$

$I_2 = \frac{P_2}{4\pi r_2^2}$

The ratio between the two intensities would be

$\frac{I_1}{I_2} = \frac{ \frac{P_1}{4\pi r_1^2}}{\frac{P_2}{4\pi r_2^2}}$

The power does not change therefore it remains constant, which allows summarizing the expression to

$\frac{I_1}{I_2}=(\frac{r_2}{r_1})^2$

Re-arrange to find $I_2$

$I_2 = I_1 (\frac{r_1}{r_2})^2$

$I_2 = 100*(\frac{1}{5})^2$

$I_2 = 4W/m^2$

Therefore the intensity at five times this distance from the source is $4W/m^2$

3 0

3 years ago

the cross section area of a hole is 725cm^2. Given that the area of a circle is A=3.14r^2 , find the radius of the hole.

asambeis [7]

$The\ area\ of\ a\ circle\ = \pi r^2 \\ 725\ cm^2 = 3.14r^2 \\
230.57 cm^2 = r^2 \\ \sqrt{230.57\ cm^2} = r \\ \boxed{r = 15.18\ cm}$

7 0

2 years ago

Which statement provides the most complete description of an object's motion?

vivado [14]

Answer:

A. The bird watcher followed the south trail a distance of five kilometers in 45 minutes.

5 0

2 years ago

How does the orbital speed of an asteroid in a circular solar orbit with a radius of 4.0 AU compare to a circular solar orbit wi

Murrr4er [49]

Answer:

The circular solar orbital speed at 4.0AU is 1/4( one fourth) that at 1.0AU

Explanation:

am = mvr= angular momentum

am4= 4mvt

am1= mvp1

Vt=1/4vp

Vp=4vt

am1= 4mvt

am1=am4

The circular solar orbital speed at 4.0AU is 1/4 (one fourth) that at 1.0AU

6 0

3 years ago