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3 years ago

15

A space shuttle burns fuel at the rate of 13,000kg in each second. Find the force exerted by the fuel on the shuttle if in 2s th

e shuttle experiences a change in momentum of 325,000kgm/s.

1 answer:

oksian1 [2.3K]3 years ago

4 0

Answer:

162,500

Explanation:

P=F(t)

32500=F(2)

162,500

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Jenny pushes a 40 N crate down the hall 2m. How much work did she do?

Shalnov [3]

Work done = force x distance = 40 x 2 = 80 Joules.

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3 years ago

An object weighing 49 N is dropped from a height of 30 m. It is found to be moving with a velocity of 24m/s just before it hits

faltersainse [42]

The frictional force will be 0.22N.

<h3>What is Frictional force?</h3>

Frictional force is the force generated between two surfaces that are in contact and slide against each other.

Given,

Weight=4N

mass =4.9/9.8=0.5kg

Hieght =30m

velocity=24m/s

Acceration , v²-u²=2as

24²/2×30 =a , u is zero

a= 1.5m/s²

By Using conservation of energy ,

30F+1/2mv²=mgh

30F=1150-144

F= 6/30

F=0.2N

The force will be 0.2N

to learn more about Friction clickhttps://brainly.com/question/17608236

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2 years ago

Suppose astronomers find an earthlike planet that is twice the size of Earth (that is, its radius is twice the radius of Earth).

JulijaS [17]

Answer:

4 times the mass of Earth

Explanation:

$M_1$ = Mass of Earth

$M_2$ = Mass of the other planet

r = Radius of Earth

2r = Radius of the other planet

m = Mass of object

The force of gravity on an object on Earth is

$F=\frac{GM_1m}{r^2}$

The force of gravity on an object on the other planet is

$F=\frac{GM_2m}{(2r)^2}$

As the forces are equal

$\frac{GM_1m}{r^2}=\frac{GM_2m}{(2r)^2}\\\Rightarrow M_1=\frac{M_2}{4}\\\Rightarrow M_2=4M_1$

So, the other planet would have 4 times the mass of Earth

6 0

3 years ago

Is our body in thermal equilibrium with the environment ?

LenKa [72]

Answer:

The human body runs at a constant 37 ºC but the air around you at room temperature is about 20-25 ºC which means heat is constantly leaving your body to warm your surroundings and maintain thermal equilibrium. You don't lose much energy doing this however as air heats reasonably quickly

Explanation:

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3 years ago

Consider identical spherical conducting space ships in deep space where gravitational fields from other bodies are negligible co

Illusion [34]

Answer:

q = 8.61 10⁻¹¹ m

charge does not depend on the distance between the two ships.

it is a very small charge value so it should be easy to create in each one

Explanation:

In this exercise we have two forces in balance: the electric force and the gravitational force

F_e -F_g = 0

F_e = F_g

Since the gravitational force is always attractive, the electric force must be repulsive, which implies that the electric charge in the two ships must be of the same sign.

Let's write Coulomb's law and gravitational attraction

$k \frac{q_1q_2}{r^2} = G \frac{m_1m_2}{r^2}$

In the exercise, indicate that the two ships are identical, therefore the masses of the ships are the same and we will place the same charge on each one.

k q² = G m²

q = $\sqrt{ \frac{G}{k} }$ m

we substitute

q = $\sqrt{ \frac{ 6.67 \ 10^{-11}}{8.99 \ 10^{9}} }$ m

q = $\sqrt{0.7419 \ 10^{-20}}$ m

q = 0.861 10⁻¹⁰ m

q = 8.61 10⁻¹¹ m

This amount of charge does not depend on the distance between the two ships.

It is also proportional to the mass of the ships with the proportionality factor found.

Suppose the ships have a mass of m = 1000 kg, let's find the cargo

q = 8.61 10⁻¹¹ 10³

q = 8.61 10⁻⁸ C

this is a very small charge value so it should be easy to create in each one

6 0

3 years ago