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3 years ago

What term generally describes small chunks of rocks and debris in space? PLEAS HURRY!

Physics

2 answers:

yan [13]3 years ago

4 0

B or meteors so yeanfjfjf

Brilliant_brown [7]3 years ago

4 0

Answer:

B) meteoroids. The key word here is "generally." Not when it enters the atmosphere or hits the surface of the Earth.

Explanation:

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As a substance melts, heat energy is used to break the connections between molecules, and temperature __________ until all melti

mario62 [17]

Temperature stays the same

4 0

3 years ago

Radon-220 is a radioactive gas. It escapes from underground rocks and causes a large part of the natural background radiation in

Sergeu [11.5K]

Answer: 134 neutrons are there in the nucleus of Radon-220

Explanation:

Atomic number : It is defined as the number of electrons or number of protons present in a neutral atom.

Thus, number of protons = atomic number = 86

Mass number is the number of the entities present in the nucleus which is the equal to the sum of the number of protons and electrons.

Mass number = Number of protons + Number of neutrons

220 = 86 + Number of neutrons

Number of neutrons = 134

5 0

3 years ago

does anyone know how to do this? the scenario is ""dominique reads that race cars have wide tires because the increased area of

Usimov [2.4K]

Answer:

Stupid

Explanation:

Because there is never a answer when we are trying to find one

7 0

3 years ago

A man pushed the box with a force of 20 N. He doubled his force to 40 N. The box did not

Alex787 [66]

Answer:

Explanation:

if the man doubles his force to 40 and the box was yet to move that means acceleration also doubled so your answer would be A

8 0

2 years ago

f an arrow is shot upward on the moon with velocity of 35 m/s, its height (in meters) after t seconds is given by h(t)=35t−0.83t

inysia [295]

Answer:

The velocity of the arrow after 3 seconds is 30.02 m/s.

Explanation:

It is given that,

An arrow is shot upward on the moon with velocity of 35 m/s, its height after t seconds is given by the equation:

$h(t)=35t-0.83t^2$

We know that the rate of change of displacement is equal to the velocity of an object.

$v(t)=\dfrac{dh(t)}{dt}\\\\v(t)=\dfrac{d(35t-0.83t^2)}{dt}\\\\v(t)=35-1.66t$

Velocity of the arrow after 3 seconds will be :

$v(t)=35-1.66t\\\\v(t)=35-1.66(3)\\\\v(t)=30.02\ m/s$

So, the velocity of the arrow after 3 seconds is 30.02 m/s. Hence, this is the required solution.

7 0

3 years ago