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Stells [14]
2 years ago
10

A spacecraft drifts through space at a constant velocity. suddenly a gas leak in the side of the spacecraft causes it to constan

tly accelerate in a direction perpendicular to the initial velocity. the orientation of the spacecraft does not change, so the acceleration remains perpendicular to the original direction of the velocity. what is the shape of the path followed by the spacecraft?
Physics
1 answer:
Ivanshal [37]2 years ago
6 0

Here space craft is moving with constant velocity

Now due to gas leakage it will have acceleration perpendicular to velocity

it is given that acceleration is always perpendicular to velocity

So it will neither increase nor decrease the velocity

it will only change the direction of the motion

so it is the centripetal acceleration of the space craft

in this case we will say that path of the space craft will be circular path

<em>So it will perform Uniform Circular Motion and its path will be a circle</em>

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A 2 kg stone moves with a speed of 1 m/s. A second 2 kg stone is moving twice as fast. Compare their kinetic energies.
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D
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5 0
2 years ago
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A 10kg rocket is traveling at 80 m/s when the booster engine applies a constant forward force of 60 N for 3.0 seconds. What impu
Lina20 [59]

Answer:

Impulse = 90

Resulting Velocity = 89

Explanation:

Use F * change in time = m * change in velocity.

For the first part of the question, the left side of the equation is the impulse. Plug it in.

60 * (3.0 - 0) = 90.

For the second half. we use all parts of the equation. I'm gonna use vf for the final velocity.

60 * (3.0 - 0) = 10 * (vf - 80). Simplify.

90 = 10vf - 800. Simplify again.

890 = 10vf. Divide to simplify and get the answer.

The resulting velocity is 89.

4 0
2 years ago
The weight of Earth's atmosphere exerts an average pressure of 1.01 ✕ 105 Pa on the ground at sea level. Use the definition of p
zloy xaker [14]

Answer:

The weight of Earth's atmosphere exert is 516.6\times10^{17}\ N

Explanation:

Given that,

Average pressure P=1.01\times10^{5}\ Pa

Radius of earth R_{E}=6.38\times10^{6}\ m

Pressure :

Pressure is equal to the force upon area.

We need to calculate the weight of earth's atmosphere

Using formula of pressure

P=\dfrac{F}{A}  

F=PA

F=P\times 4\pi\times R_{E}^2

Where, P = pressure

A = area

Put the value into the formula

F=1.01\times10^{5}\times4\times\pi\times(6.38\times10^{6})^2

F=516.6\times10^{17}\ N

Hence, The weight of Earth's atmosphere exert is 516.6\times10^{17}\ N

8 0
2 years ago
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Consider a large tank holding 1000 L of pure water into which a brine solution of salt begins to owat a constant rate of 6L/min.
lbvjy [14]

Answer:

T = 693.147 minutes

Explanation:

The tank is being continuously stirred. So let the salt concentration of the tank at some time t be x in units of kg/L.

Therefore, the total salt in the tank at time t = 1000x kg

Brine water flows into the tank at a rate of 6 L/min which has a concentration of 0.1 kg/L

Hence, the amount of salt that is added to the tank per minute = (6\times0.1)kg/min=0.6kg/min

Also, there is a continuous outflow from the tank at a rate of 6 L/min.

Hence, amount of salt subtracted from the tank per minute = 6x kg/min

Now, the rate of change of salt concentration in the tank = \frac{dx}{dt}

So, the rate of change of salt in the tank can be given by the following equation,

1000\frac{dx}{dt} =0.6-6x

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3 0
2 years ago
Astronomers have discovered a new planet called "Xandar" beyond the orbit of Pluto (No, not really but I need a fake planet for
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Answer:

m = 1.82E+23 kg

Explanation:

G = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

mG = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

m = 1.82E+23 kg

3 0
2 years ago
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