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sweet [91]
3 years ago
14

50 points!

Physics
1 answer:
Tanya [424]3 years ago
5 0
From Kepler's 3rd law, time period of satellite orbiting around planet in a circular orbit is given by,
T² = 4π²r³/GM 
M= mass of the Earth = 6×10²⁴ Kg. 
∴ T² = 4×3.14²×(2.56×10⁷)³/6.67×10⁻¹¹×6×10²⁴ 
   T² = 1653339719
∴ T = 40661.28 seconds.
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Essay about why people should not join a gang 300 word​
Salsk061 [2.6K]

Answer:

I don't know if it's fair for me to write an entire essay for you

but if you would like I can list some reasons you can incorporate into an essay.

-------------------------------------------------------------------------------------------------------

-If you are joining a gang you are most likely going to have a greater chance of being targeted, even if it seems like you would be protected.

-Most of the time when someone joins a gang a child is being separated from a parent and will face most aggression.

-This can also cause academic failure in students, from lack of education and probably a pretty bad background.

-For teens when joining a gang they will do things that aren't suited for their age (I would list things but I'm not trying to get myself reported if you know what I mean)

-You will also have a great risk of being imprisoned and facing bad consequences, and these consequences can even be having the death penalty.

-------------------------------------------------------------------------------------

I hope this is good enough, sorry I didn't give you a full essay. I think this should help with that though.

6 0
2 years ago
PLEASE PROVIDE AN EXPLANATION<br><br> THANK YOU!
Rus_ich [418]

Answer:

(a) 0.993 s

(b) 14.0 N/m

(c) -3.02 m/s

(d) -6.01 m/s²

Explanation:

(a) The block's position can be modeled as a cosine wave:

x(t) = A cos(ωt)

where A is the amplitude (in this case, 50 cm) and ω is the angular frequency.

At t = 0.200 s, x(t) = 15.0 cm.

15.0 cm = 50.0 cm cos((0.200 s) ω)

0.3 = cos((0.2 s) ω)

1.266 rad = (0.2 s) ω

ω = 6.33 rad/s

The period is:

T = (2π rad) (1 s / 6.33 rad)

T = 0.993

(b) For a spring-mass system, ω = √(k/m).  The mass of the block is 0.350 kg, so:

ω = √(k/m)

6.33 rad/s = √(k / 0.350 kg)

6.33 rad/s = √(k / 0.350 kg)

40.1 rad/s² = k / 0.350 kg

k = 14.0 N/m

(c) Energy is conserved:

EE₀ = EE + KE

½ kx₀² = ½ kx² + ½ mv²

kx₀² = kx² + mv²

(14.0 N/m) (0.50 m)² = (14.0 N/m) (0.15 m)² + (0.35 kg) v²

v = -3.02 m/s

Alternatively, we can take the derivative of our position equation:

v(t) = -Aω sin(ωt)

v = -(0.50 m) (6.33 rad/s) sin((6.33 rad/s) (0.2 s))

v = -3.02 m/s

(d) Sum of forces on the block:

∑F = ma

-kx = ma

a = -kx / m

a = -(14.0 N/m) (0.15 m) / (0.350 kg)

a = -6.01 m/s²

Alternatively, we can take the derivative of our velocity equation:

a(t) = -Aω² cos(ωt)

a = -(0.50 m) (6.33 rad/s)² cos((6.33 rad/s) (0.2 s))

a = -6.01 m/s²

6 0
3 years ago
Read 2 more answers
How many times higher could an astronaut jump on the Moon than on Earth if his takeoff speed is the same in both locations (grav
JulsSmile [24]

Answer:

maximum height on moon is 6 times more than the maximum height on Earth

Explanation:

Let the Astronaut has its maximum speed by which he can jump is "v"

now for the maximum height that it can jump is given as

v_f^2 - v_i^2 = 2 aH

now from above equation we will have

0 - v^2 = 2(-g)H

now we have

H = \frac{v^2}{2g}

now if Astronaut jump on the surface of moon with same speed

then we know that the acceleration of gravity on surface of moon is 1/6 times the gravity on earth

so at surface of moon we have

0 - v^2 = 2(-g/6) H

now we have

H = \frac{6v^2}{2g}

so maximum height on moon is 6 times more than the maximum height on Earth

8 0
3 years ago
What is the gravitational force between two identical 5000 kg asteroids whose centers of mass are separated by 100 m?
Alenkinab [10]

Answer: 1.67 x 10^-7N

Explanation:

5 0
3 years ago
A bar magnet is held in place while another bar magnet is placed near it. The second bar magnet spins around and
Vinil7 [7]

Answer:

o

Explanation:

the increase energy stored in thw system is proportional to the decrease in kinetic energy

4 0
3 years ago
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