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3 years ago

13

if you were floating in the solar system at equal distances between mars (small mass) and jupiter (large mass) which planet woul

d you be gravitational pulls to? Why

1 answer:

tester [92]3 years ago

3 0

You would gravitate towards Jupiter because if it’s large mass it has a stronger gravitational pull

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I need help pls now plleeeeeeeeaaassseeeee

GrogVix [38]

Answer:

$r = \frac{v}{i} = v = ri \\ i = \frac{v}{r}$

4 0

3 years ago

75kg man climbs a mountain 1000m high in 3hrs and uses 4100 joulse/min. (a) calculate the power consumption in watt, (b) what is

mr Goodwill [35]

Answer:

Explanation:

a) Power consumption is 4100 J/min / 60 s/min = 68.3 W(atts)

work done raised the potential energy

b) 75(9.8)(1000) / (3(3600)) = 68.055555... 68.1 W

c) efficiency is 68.1 / 68.3 = 0.99593... or nearly 100%

Not a very likely scenario.

3 0

3 years ago

A mid-ocean ridge under the ocean is MOST similar to a _____ on land.

Alexxx [7]

A mid ocean ridge is a under water mountain range, knowing this what would you say it is most similar to on land? hope this helped!

5 0

3 years ago

S A voltage ΔV is applied to a series configuration of n resistors, each of resistance R. The circuit components are reconnected

Flura [38]

The power of is series combination is Vn^2 times that of a parallel combination.

For series combination :

Req = R + R + R + ............... n times = nR

I = Δv/nr

Power = (Δv/nr)^2 × nr = Δv^2/nr

For parallel combination

1/req = 1/R + 1/R + 1/R +................(n times) = n/R

Req = R/n

Power = Δv/(R/n) = nΔv^2/R

Ratio = Δv^2/nr/n·Δv^2/R = 1/n^2

Hence, power of is series combination is Vn^2 times that of a parallel.

Learn more about parallel combination here:

brainly.com/question/12400458

#SPJ4

3 0

2 years ago

Calculate the orbital speed (in m/s) of a satellite that circles the Earth with a time period of 12.00 hours. The mass of the Ea

Hoochie [10]

Answer:

$v = 3869 m/s$

Explanation:

As we know that the orbital speed of the satellite is given as

$v = \sqrt{\frac{GM}{r}}$

also we know that

time period of the revolution is given as

$T = \frac{2\pi r}{v}$

now from above equation we know that

$T = \frac{2\pi (\frac{GM}{v^2})}{v}$

$T = \frac{2\pi GM}{v^3}$

so we will have

$v = (\frac{2\pi GM}{T})^{1/3}$

now plug in all data in this equation

$v = (\frac{2\pi (6.67 \times 10^{-11})(5.97 \times 10^{24})}{12 \times 3600})^{1/3}$

$v = 3869 m/s$

3 0

3 years ago