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luda_lava [24]
3 years ago
6

How does the gravity in the space shuttle compare with the gravity on earth’s surface?

Physics
2 answers:
Zolol [24]3 years ago
7 0

These days they are exactly identical. All the space shuttles are currently in museums located on the surface of the Earth.

Years ago when the space shuttles were in orbit, the gravity inside a shuttle was about 85% of the gravity on the Earth's surface.

zheka24 [161]3 years ago
5 0
The gravity in the space shuttle is approximately equal to the gravity on the surface of the earth
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What is the difference between convex and concave mirrors and lens?
Sidana [21]
Concave makes things smaller and convex makes things bigger
5 0
3 years ago
A 1400 kg wrecking ball hangs from a 20-m-long cable. the ball is pulled back until the cable makes an angle of 30.0 ∘ with the
anastassius [24]

From the geometry of the problem, the 20 m-long cable creates the hypotenuse of a right triangle, with the extended of the other two sides of size 20 m * cos(30 deg), which is around 17.3 m. Therefore, the ball has increased by 20 m - 17.3 m = 2.7 m. 

The potential energy will have altered by m*g*h, which is 1400 kg * 9.8 m/s^2 * 1.6 m , or about 37044 joules.

5 0
3 years ago
Read 2 more answers
A skydiver of 75 kg mass has a terminal velocity of 60 m/s. At what speed is the resistive force on the skydiver half that when
ankoles [38]

Answer:

The speed of the resistive force is 42.426 m/s

Explanation:

Given;

mass of skydiver, m = 75 kg

terminal velocity, V_T = 60 \ m/s

The resistive force on the skydiver is known as drag force.

Drag force is directly proportional to square of terminal velocity.

F_D = kV_T^2

Where;

k is a constant

k = \frac{F_D_1}{V_{T1}^2} = \frac{F_D_2}{V_{T2}^2}

When the new drag force is half of the original drag force;

F_D_2 = \frac{F_D_1}{2} \\\\\frac{F_D_1}{V_{T1}^2} = \frac{F_D_2}{V_{T2}^2} \\\\\frac{F_D_1}{V_{T1}^2} = \frac{F_D_1}{2V_{T2}^2} \\\\\frac{1}{V_{T1}^2} = \frac{1}{2V_{T2}^2}\\\\2V_{T2}^2 = V_{T1}^2\\\\V_{T2}^2= \frac{V_{T1}^2}{2} \\\\V_{T2}= \sqrt{\frac{V_{T1}^2}{2} } \\\\V_{T2}=  \frac{V_{T1}}{\sqrt{2} } \\\\V_{T2}=  0.7071(V_{T1})\\\\V_{T2}= 0.7071(60 \ m/s)\\\\V_{T2}= 42.426 \ m/s

Therefore, the speed of the resistive force is 42.426 m/s

8 0
3 years ago
A 5.6 cm diameter parallel-plate capacitor has a 0.58 mm gap. What is the displacement current in the capacitor if the potential
BARSIC [14]

Answer:

1.88\cdot 10^{-5} A

Explanation:

The capacitance of a parallel plate capacitor is given by:

C=\frac{\epsilon_0 A}{d} (1)

where

\epsilon_0 is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

The charge stored on the capacitor is given by

Q=CV (2)

where C is the capacitance and V is the voltage across the capacitor.

The displacement current in the capacitor is given by

J=\frac{Q}{t} (3)

where t is the time elapsed

Substituting (1) and (2) into (3), we find an expression for the displacement current:

J=\frac{CV}{t}=\frac{\epsilon_0 A}{d} \frac{V}{t}

where we have

A=\pi (\frac{d}{2})^2=\pi (\frac{0.056 m}{2})^2=2.46\cdot 10^{-3} m^2

d = 0.58 mm = 5.8\cdot 10^{-4} m

\frac{V}{t}=500,000 V/s

Substituting into the equation, we find

J=\frac{(8.85\cdot 10^{-12} F/m)(2.46\cdot 10^{-3} m^2)}{5.8\cdot 10^{-4}m}(500,000 V/s)=1.88\cdot 10^{-5} A

6 0
3 years ago
Supposing you were in space..in a weightless environment, would it require a force to set an object in motion? Explain..
Dvinal [7]

Answer:

Yes, a force is require to set an object in motion.

Explanation:

  • In space, even if you feel weightless, you are subject to motion. If you are orbiting the Earth, you are under the constant influence of Earth having a free-fall acceleration equal to the centripetal acceleration.
  • To disturb this orbital motion, an external force is required.
  • According to Newton's laws of motion, a force is required to change the state of the rest of a body or to change the velocity or direction if it is moving with uniform velocity along a straight line.
  • Whenever there is a change in velocity or direction of a body there is a force acting on it.
8 0
3 years ago
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