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amid [387]
2 years ago
7

What does a meteoroid become when it enters Earth's atmosphere?

Physics
2 answers:
Reika [66]2 years ago
8 0

Answer:

A. A meteor

What is the difference between a meteor and a meteoroid?

Meteoroids are objects in space that range in size from dust grains to small asteroids. Think of them as “space rocks." When meteoroids enter Earth's atmosphere (or that of another planet, like Mars) at high speed and burn up, the fireballs or “shooting stars” are called meteors.

allochka39001 [22]2 years ago
7 0
A METEOR I LEARNED IT ON EDGE PLEASE GIVE BRAINLY
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Steel wire rope is used to lift a heavy object. We use a 3.1m steel wire that
kaheart [24]

Answer:

Young's modulus for the rope material is 20.8 MPa.    

 

Explanation:

The Young's modulus is given by:

E = \frac{FL_{0}}{A\Delta L}

Where:

F: is the force applied on the wire

L₀: is the initial length of the wire = 3.1 m

A: is the cross-section area of the wire

ΔL: is the change in the length = 0.17 m

The cross-section area of the wire is given by the area of a circle:

A = \pi r^{2} = \pi (\frac{0.006 m}{2})^{2} = 2.83 \cdot 10^{-5} m^{2}

Now we need to find the force applied on the wire. Since the wire is lifting an object, the force is equal to the tension of the wire as follows:

F = T_{w} = W_{o}

Where:

T_{w}: is the tension of the wire

W_{o}: is the weigh of the object = mg

m: is the mass of the object = 1700 kg

g: is the acceleration due to gravity = 9.81 m/s²

F = mg = 1700 kg*9.81 m/s^{2} = 16677 N

Hence, the Young's modulus is:

E = \frac{16677 N*0.006 m}{2.83 \cdot 10^{-5} m^{2}*0.17 m} = 20.8 MPa          

Therefore, Young's modulus for the rope material is 20.8 MPa.                

I hope it helps you!                                    

7 0
3 years ago
A 2.00-kg block is pushed against a spring with negligible mass and force constant k = 400 N/m, compressing it 0.220 m. When the
marin [14]

Answer:

0.82052 m

Explanation:

potential energy of spring = kinetic energy

=> 0.5kx^2 = 0.5mv^2

=> v=\sqrt{\frac{kx^2}{m} }

v=\sqrt{\frac{400\times 0.220^2}{2} }

v= 3.11127 m/s

angle = 37°

thus height = distance×sin(37) = D×0.60182

Also,

m×g×h = 0.5×m×v^2

=> 2×9.8×D×0.60182 = 0.5×2×3.11127×3.11127

=> D = 0.82052 m

4 0
3 years ago
Newton's Second Law
lara [203]
One of the last two answers
4 0
3 years ago
What do astronomers expect will eventually happen to the Milky Way galaxy?
Nataly [62]
<span> The Andromeda Milkyway is a galactic collision predicted to occur in 4 billion years. I believe thats the answer</span>
5 0
3 years ago
Read 2 more answers
The mass of a hypothetical planet is 1/100 that of Earth and its radius is 1/4 that of Earth. If a person weighs 600 N on Earth,
omeli [17]

To solve this problem we will apply the Newtonian concept of gravitational acceleration produced by a planet. This relationship is given by:

g = \frac{GM}{r^2}

Where,

G = Gravitational Universal Constant

M = Mass of Earth

r = Radius

The values given are based on the constants of the earth, so they can be expressed as

M_p = \frac{1}{100} M_e

r_p = \frac{1}{4} r_e

The relationship of gravity would then be given:

g_e = \frac{GM_e}{r_e^2}

The relationship with the new planet, from the gravity of the earth would be given

g_p = \frac{GM_p}{r_p^2}

g_p = \frac{G(1/100)M_e}{(1/4 r_e)^2}

g_p = \frac{GM_e 16}{100 r_e^2}

g_p = 0.16 \frac{GM_e}{r_e^2}

g_p = 0.16g_e

The relationship with the weight of the earth would be given as:

W_e = m*g_e = 600N

W_p = m*g_p = m(0.16g_p)

W_p = (m*g_p)(0.16)

W_p = 600*0.16

W_p = 96N

Therefore the weigh on this planet would be 96N

3 0
3 years ago
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