All categories

3 years ago

If 478 watts of power are used in 14 seconds,how much work was done

Physics

1 answer:

zepelin [54]3 years ago

6 0

Answer:

6692J

Explanation:

Power is defined as the rate at which work is being done.

So,

Power = $\frac{workdone}{time }$

Work done = Power x time

Given parameters:

Power = 478watts

Time = 14s

So;

Work done = 478 x 14 = 6692J

You might be interested in

Radon is a radioactive gas found below ground. If a sample of radon occupies 29 mL at 25 oC, what volume will it occupy at 500 K

Valentin [98]

Answer:

48.63 mL

Explanation:

Using Charles's law

$\frac {V_1}{T_1}=\frac {V_2}{T_2}$

Given ,

V₁ = 29 mL

V₂ = ?

T₁ = 25 °C

T₂ = 500 K

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (25 + 273.15) K = 298.15 K

Using above equation as:

$\frac{29\ mL}{298.15\ K
}=\frac{V_2}{500\ K}$

$V_2=\frac{29\cdot \:500}{298.15}\ mL$

<u>New volume = 48.63 mL</u>

8 0

4 years ago

If there's air in a ball and it bounces how come a room doesn't bounce

finlep [7]

Answer:

Funny answer but works!

Explanation:

Not enough air. Go outside, crack open the window and blow air into the room like you would a balloon then close the window quickly. If it doesn't start bouncing, try giving the house a push. Good luck.

4 0

3 years ago

. A telescope is constructed with two lenses separated by a distance of 25 cm. The focal length of the objective is 20 cm. The f

beks73 [17]

Answer:

the magnitude of the angular magnification of the telescope. is 4

Explanation:

Calculate the magnitude of the angular magnification of the telescope.

Given that,

distance = 25cm

focal length from the objective f₀ = 20cm

focal length from eye piece f₁ = 5cm

The angular magnification of the telescope is

$M = \frac{f_0}{f_1}$

Magnification = 20 / 5

magnification = 4

Hence, the magnitude of the angular magnification of the telescope. is 4

5 0

3 years ago

A motorcycle has a constant acceleration of 3.49 m/s2. Both the velocity and acceleration of the motorcycle point in the same di

Vilka [71]

Answer:

(a)2.865 s

(b)2.865 s

Explanation:

We are given that

Acceleration,a= $3.49 m/s^2$

a.Initial speed,u=29 m/s

Final speed,v=39 m/s

We know that

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

Using the formula

$t=\frac{39-29}{3.49}=2.865 s$

b.Initial speed,u=59 m/s

Final speed,v=69 m/s

Again using the formula

$t=\frac{69-59}{3.49}=2.865 s$

7 0

3 years ago

1150 g - 0.652 kg = ______________g?

beks73 [17]

Answer is 498 grams

6 0

3 years ago

Read 2 more answers