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ser-zykov [4K]
3 years ago
9

Determine the thrust produced if 1.5 x 10^3 kg of gas exits the combustion chamber each second, with a speed of 4.00 x 10^3 m/s.

Physics
1 answer:
ozzi3 years ago
3 0

Answer:

The thrust is 6\times 10^6\ N

Explanation:

Given that,

Mass of gas, m=1.5\times 10^3\ kg

The rate at which the gas is expelling, \dfrac{dv}{dt}=4\times 10^{3}\ m/s

We need to find the thrust produced by the gas.

We know that force is equal to the rate of change of momentum. So,

F=\dfrac{p}{t}

Also, p = mv

F=\dfrac{mv}{t}

So,

F=1.5\times 10^3\times 4\times 10^3\\\\F=6\times 10^6\ N

So, the thrust is 6\times 10^6\ N

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Pretend you (80 kg) are making repairs on the outside of the International Space Station. You are floating 20 meters away from t
djyliett [7]

Answer:

32 seconds

Explanation:

m1 = 80 kg

m2 = 10 kg

v2 = 5m/s

According to the property of conservation of momentum, assuming that both you and the bag are stationary before the safety rope comes lose:

m_{1} v_{1} =m_{2} v_{2} \\80v_{1} =10*5 \\v_{1} = 0.625\ m/s

Since the space station is 20 meters away, the time taken to reach it is given by:

t = \frac{20}{0.625}\\t=32\ s

It takes you 32 seconds to reach the station.

7 0
3 years ago
an object with a mass of 5kg is moving with an initial velocity of 20m/s.the object accelerates at a rate of 15m/s/s for 8s.what
Alinara [238K]

It doesn't matter what the object's initial velocity is, or how long
the acceleration lasts.  All that matters is the object's mass and
acceleration.

Force = (mass) x (acceleration) =

                (5kg) x (15 m/s²) =

                         75 kg-m/s² = <em>75 newtons .</em>


5 0
3 years ago
Besides gravity, what factor keeps the moon and Earth in orbit?
Gelneren [198K]

Answer:

(D) Inertia

Explanation:

Inertia words with Gravity to keep the Moon, Earth and Sun All in Orbit!

                                                      <em>-Aslina</em>

7 0
3 years ago
A block of mass 14.9 kg is pulled to the right by an applied force of 39.4 N. If it moves with constant velocity, how much frict
lakkis [162]

The frictional force is 39.4 N

Explanation:

We can solve this problem by applying Newton's 2nd law of motion: in fact, the net force acting on the block is equal to the product between its mass and its acceleration. So we can write

\sum F = ma

where

\sum F is the net force

m is the mass

a is the acceleration

Here we know that the box is moving with constant velocity, so its acceleration is zero:

a=0

This means that the net force is also zero:

\sum F=0

The net force on the block is given by the applied force, forward, and the frictional force, backward:

\sum F = F_a-F_f=0

where

F_a=39.4 N is the applied force

F_f is the frictional force

Therefore, solving for F_f,

F_f=F_a=39.4 N

Learn more about friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

8 0
3 years ago
A stone is dropped into water from a bridge 52 m above the water's
dusya [7]

Answer:

Its final velocity and how much time it takes to reach the water

Explanation:

The motion of the stone is a uniformly accelerated motion, so we can use the following suvat equation to determine its final velocity:

v^2-u^2=2as

where

v is the final velocity

u = 0 is the initial velocity

a=g=9.8 m/s^2 is the acceleration of gravity

s = 52 m is the distance covered during the fall

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0^2+2(9.8)(52)}=31.9 m/s

We can also find how much time it takes to reach the water, using the equation

v=u+at

where

v = 31.9 m/s is the final velocity

u = 0 is the initial velocity

a=g=9.8 m/s^2

t is the time

And solving for t,

t=\frac{v-u}{a}=\frac{31.9-0}{9.8}=3.26 s

3 0
3 years ago
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