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Usimov [2.4K]
2 years ago
14

A meteroid is in a circular orbit 600 km above the surface of a distant planet. the planet has the same mass as Earth but has a

radius that is 90% of Earths (where Earths radius is approximately 6370 km).
The acceleration of the meteroid due to the gravitational force exerted by the planet is most nearly

A) 9 m/s^2 toward the center of the planet

B) 9 m/s^2 in the direction of the meteoroids motion

C) 10 m/s^2 toward the center of the planet

D) 10 m/s^2 in the direction of the meteoroids motion​
Physics
2 answers:
taurus [48]2 years ago
6 0
Well, the answer to this question is D) 10 m/s^2 in the direction of the meteoroids motion.
Glad I could help and have a fantastic day!
Also, can you please mark my answer as the brainliest answer?
Thanks.
Llana [10]2 years ago
3 0

Answer:

C)  10 m/s^2 toward the center of the planet

Explanation:

g = GM/r^2

The masses of earth and this planet are the same and the radius is only diffferent by 10% so as gravitational force acceleration is 9.8m/s^2 on earth, it can be estimated to be around 10m/s^2.

Using the third law, we know acceleration is toward the larger mass from attraction, thus having it point toward the center of the planet.

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Sphinxa [80]
<h2>Answer: Resonance </h2>

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4 0
2 years ago
What are groups 1,2 and 3 examples of on the periodic table
pishuonlain [190]
<span>The number of the group identifies the column of the standard periodic table in which the element appears.</span>
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4 0
3 years ago
A 150 kg uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. Calculate the magn
xenn [34]

Answer:

T = 2010 N

Explanation:

m = mass of the uniform beam = 150 kg

Force of gravity acting on the beam at its center is given as

W = mg

W = 150 x 9.8

W = 1470 N

T = Tension force in the wire

θ = angle made by the wire with the horizontal =  47° deg

L = length of the beam

From the figure,

AC = L

BC = L/2

From the figure, using equilibrium of torque about point C

T (AC) Sin47 = W (BC)

T L Sin47 = W (L/2)

T Sin47 = W/2

T Sin47 = 1470

T = 2010 N

6 0
3 years ago
A steady beam of alpha particles (q = + 2e, mass m = 6.68 × 10-27 kg) traveling with constant kinetic energy 22 MeV carries a cu
cluponka [151]

Answer:

Explanation:

q = 2e = 3.2 x 10^-19 C

mass, m = 6.68 x 10^-27 kg

Kinetic energy, K = 22 MeV

Current, i = 0.27 micro Ampere = 0.27 x 10^-6 A

(a) time, t = 2.8 s

Let N be the alpha particles strike the surface.

N x 2e = q

N x 3.2 x 10^-19 = i t

N x 3.2 x 10^-19 = 0.27 x 10^-6 x 2.8

N = 2.36 x 10^12

(b) Length, L = 16 cm = 0.16 m

Let N be the alpha particles

K = 0.5 x mv²

22 x 1.6 x 10^-13 = 0.5 x 6.68 x 10^-27 x v²

v² = 1.054 x 10^15

v = 3.25 x 10^7 m/s

So, N x 2e = i x t

N x 2e = i x L / v

N x 3.2 x 10^-19 = 2.7 x 10^-7 x 0.16 / (3.25 x 10^7)

N = 4153.85

(c) Us ethe conservation of energy

Kinetic energy = Potential energy

K = q x V

22 x 1.6 x 10^-13 = 2 x 1.5 x 10^-19 x V

V = 1.17 x 10^7 V

5 0
3 years ago
A situation that restricts something
Ostrovityanka [42]
Answer:

Laws

Explanation:
i don’t have one but i hope ur doing well during this quarantine
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