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2 years ago

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

Physics

1 answer:

zepelin [54]2 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

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a. Find the magnitude of the gravitational force (in N) between a planet with mass 6.50 ✕ 1024 kg and its moon, with mass 2.45 ✕

Vanyuwa [196]

Answer:

a. $1.70\times 10^{20} N$ b. $6.94\times 10^{-3}m/s^2$ c. $2.62\times 10^{-5} m/s^2$

Explanation:

Use Gravitational Force:

$F= G\frac{Mm}{r^2}$

Use values given to have the results.

$g=\frac{F}{m}$

Use m as the mass of the Moon. Using values of a. and m you have the answer.

$g = \frac{F}{M}$

Use M mass of the planet. Use value of a and Mass of Planet.

3 0

3 years ago

Which theory proposes that the moon was a passing asteroid pulled into orbit by Earth's gravity?

kolezko [41]

The correct answer to the question above is the capture theory. The capture theory proposes that the moon was a passing asteroid pulled into the orbit by the Earth's gravity. It says that the moon was originally orbiting the sun, not the Earth.

3 0

3 years ago

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Y-axis of a velocity-time graph represents

skelet666 [1.2K]

Answer:

The Y-axis in a Velocity/Time graph is the Velocity.

Explanation:

The speed is the velocity. Like in math, speed(velocity)=distance over time. Where time is the X-axis.

3 0

2 years ago

A newly discovered planet is found to have a density if 2/3pe and a radius of 2RE, where PE and RED are the density and radius o

Liono4ka [1.6K]

Missing question in the text of the exercise. Found on internet:
"What is the acceleration due to gravity on the surface of the planet?"

Solution:
The gravitational acceleration at Earth's surface is given by:
$g= \frac{GM}{r^2}$
where
G is the gravitational constant
M is the Earth's mass
r is the Earth's radius

The Earth's mass can be rewritten also as the product between the Earth's density, $d$ , and its volume (the volume of a sphere of radius r):
$M=dV=d ( \frac{4}{3} \pi r^3)= \frac{4}{3} \pi d r^3$

Now let's call M' the mass of the new planet, r' its radius and d' its density. The acceleration due to gravity on the surface of the new planet is
$g' = \frac{GM'}{r'^2}$ (1)
so we need to find M' and r'.

The problem says the radius of the new planet is twice the Earth's radius:
$r'=2r$ (2)
and that its density is 2/3 of Earth's density:
$d'= \frac{2}{3} d$
so the mass M' of the new planet is, with respect to the Earth's mass:
$M' = d'V' = \frac{4}{3} \pi d' (r')^3 = \frac{4}{3} \pi ( \frac{2}{3}d) (2r)^3 = ( \frac{4}{3} \pi d r^3 )( \frac{16}{3}) = \frac{16}{3} M$ (3)

And if we substitute (2) and (3) into (1), we find the gravitational acceleration on the surface of the new planet:
$g'= \frac{G( \frac{16}{3}M) }{(2r)^2}= \frac{GM}{r^2} \frac{4}{3} = \frac{4}{3}g$
And since $g=9.81 m/s^2$ , we find
$g'= \frac{4}{3}(9.81 m/s^2)=13.1 m/s^2$

8 0

3 years ago

I need help with this question. Anyone?

cluponka [151]

The closest answer i would go with is b, I'm not completely sure because I'm not sure what the question actually means because there could be any meaning

5 0

3 years ago

Read 2 more answers