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3 years ago

5

Planet a has three times the mass of planet b and the same radius. from this information, what can we conclude about the acceler

ation due to gravity at the surface of planet a compared to that at the surface of planet b?

1 answer:

omeli [17]3 years ago

6 0

It would be 3 times bigger (formula is g= mass divided by squared radius)
Hence mass grows 3 times - g will grow 3 times as well

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A person suffering from anaemia gets tired after a short walk

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Force can change the ____________ and _______________ of an object

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Answer:

Force can change the speed and direction of an object.

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3 years ago

An Astronaut lands on an Earthlike planet and drops a small lead ball with a mass from the top of her spaceship. The point of re

bearhunter [10]

First we have to find out the gravity on that planet. We use Newton second equation of motion. It is given as,

s = ut +(gt^2)/2

Distance s = 25m

Time t = 5 s

Velocity u = 0

By putting these values,

25 = 1/2.g.(5)²

g = 2

So the gravity on that planet is 2. Lets find out the weight of the astronaut.

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2 years ago

An all-electric car (not a hybrid) is designed to run from a bank of 12.0 V batteries with total energy storage of 2.30 ✕ 107 J.

AnnyKZ [126]

Answer:

a) $I=733.33\ A$

b) $d=52272.7273\ m$

c) $d'=51948.0519\ m$

Explanation:

Given:

voltage of the battery, $V=12\ V$

energy storage capacity of the battery, $E=2.3\times 10^7\ J$

speed of the car, $v=20\ m.s^{-1}$

a)

power drawn by the car, $P=8.8\ kW$

<u>Now the Current delivered to the motor:</u>

we the relation between the power and electrical current,

$P=V.I$

$8800=12\times I$

$I=733.33\ A$

b)

<u>Distance travelled before battery is out of juice:</u>

we first find the time before the battery runs out,

$t=\frac{E}{P}$

$t=\frac{2.3\times 10^7}{8800}$

$t=2613.636\ s$

Now the distance:

$d=v.t$

$d=20\times 2613.636$

$d=52272.7273\ m$

c)

When the head light of 55 W power is kept on while moving then the power consumption of the car is:

$P'=P+55$

$P'=8800+55$

$P'=8855\ W$

<u>Now the time of operation of the car:</u>

$t'=\frac{E}{P'}$

$t'=\frac{2.3\times 10^7}{8855}$

$t'=2597.4026\ s$

<u>Now the distance travelled:</u>

$d'=v.t'$

$d'=20\times 2597.4025$

$d'=51948.0519\ m$

5 0

3 years ago