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Darina [25.2K]
3 years ago
10

You are explaining to friends why astronauts feel weightless orbiting in the space shuttle, and they respond that they thought g

ravity was just a lot weaker up there. Convince them and yourself that it isn't so by calculating how much weaker gravity is 400 km above the Earth's surface.
Physics
1 answer:
bezimeni [28]3 years ago
6 0

Answer:

It's only 1.11 m/s2 weaker at 400 km above surface of Earth

Explanation:

Let Earth radius be 6371 km, or 6371000 m. At 400km above the Earth surface would be 6371 + 400 = 6771 km, or 6771000 m

We can use Newton's gravitational law to calculate difference in gravitational acceleration between point A (Earth surface) and point B (400km above Earth surface):

g = G\frac{M}{r^2}

where G is the gravitational constant, M is the mass of Earth and r is the distance form the center of Earth to the object

\frac{g_B}{g_A} = \frac{GM/r^2_B}{GM/r^2_A}

\frac{g_B}{g_A} = \left(\frac{r_A}{r_B}\right)^2

\frac{g_B}{g_A} = \left(\frac{6371000}{6771000}\right)^2

\frac{g_B}{g_A} = 0.94^2 = 0.885

g_B = 0.885 g_A

So the gravitational acceleration at 400km above surface is only 0.885 the gravitational energy at the surface, or 0.885*9.81 = 8.7 m/s2, a difference of (9.81 - 8.7) = 1.11 m/s2.

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Assume a satellite shines an unpolarized light on a telescope. The intensity of the light as it reaches the telescope is 1.1*10-
n200080 [17]

Answer:

4.125\times 10^{-11}\ W/m^2

Explanation:

I_0 = Intensity of unpolarized light = 1.1\times 10^{-10}\ W/m^2

\theta = Angle of the filter = 30^{\circ}

Intensity of light is given by

I=\dfrac{I_0}{2}cos^2\theta\\\Rightarrow I=\dfrac{1.1\times 10^{-10}}{2}cos^230\\\Rightarrow I=4.125\times 10^{-11}\ W/m^2

The intensity of light detected by the camera is 4.125\times 10^{-11}\ W/m^2

7 0
4 years ago
(01.06 MC)
Cerrena [4.2K]

Answer:

c

Explanation:

7 0
3 years ago
Helpp pls
Kipish [7]

Answer:

The intensity of the electric field is

|E|=10654.37 \:N/C

Explanation:

The electric field equation is given by:

|E|=k\frac{q}{d^{2}}

Where:

  • k is the Coulomb constant
  • q is the charge at 0.4100 m from the balloon
  • d is the distance from the charge to the balloon

As we need to find the electric field at the location of the balloon, we just need the charge equal to 1.99*10⁻⁷ C.

Then, let's use the equation written above.

|E|=(9*10^{9})\frac{1.99*10^{-7}}{0.41^{2}}

|E|=10654.37 \:N/C

I hope it helps you!

5 0
3 years ago
At a race car driving event, a staff member notices that the skid marks left by the race car are 9.06 m long. The very experienc
vaieri [72.5K]

Answer:

27.1 m/s

Explanation:

Given that at a race car driving event, a staff member notices that the skid marks left by the race car are 9.06 m long. The very experienced staff member knows that the deceleration of a car when skidding is -40.52 m/s2.

Using third equation of motion,

V^2 = U^2 + 2aS

Since the car is decelerating, the final velocity V = 0

Substitute all the parameter into the equation above,

0 = U^2 - 2 * 40.52 * 9.06

U^2 = 734.22

U = \sqrt{734.22}

U = 27.096

U = 27.1 m/s  approximately

Therefore, the staff member can estimate for the original speed of the race car to be 27.1 m/s if it came to a stop during the skid

5 0
3 years ago
Which of the following would add more resistance? Check all that apply.
Blababa [14]

Answer:C. E.

Explanation:

4 0
3 years ago
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