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SCORPION-xisa [38]
3 years ago
12

Meteoroids are small objects in space, usually made of rock or metal. These bodies often get close to a planet and fall through

its atmosphere. A burning streak of light, called a , is visible as it falls through the sky. Sometimes the object does not burn up completely and a part of it survives the fall to the planet’s surface. These rocky bodies, called , are studied by scientists to gain a better understanding of our solar system. NextReset
Physics
1 answer:
rewona [7]3 years ago
5 0
First chose is meteor and the second one is asteroids
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Please need help with this
NeTakaya

<u>Note that</u>:

The gravitational potential energy = mgh

where m: is the mass, g: the acceleration due to the gravity and h is the height from the earth surface

Then, we can increase the gravitational potential energy by increasing the mass or the height from the earth surface

<u>In our question</u>, we can increase the gravitational potential energy by

<u>A) Strap a boulder to the car so that it wights more.</u>

7 0
2 years ago
Is electricity a fuel
labwork [276]

YES, ELECTRICITY CONCERNS ENERGY WHICH IS USED AS A FUEL . IN MODERN DAY TECH, MOST MACHINES USE ELECTRICITY AS A FUEL SUCH AS THE ELECTRONIC TRAIN IN TOKYO, JAPAN.

8 0
2 years ago
Read 2 more answers
As a woman walks, her entire weight is momentarily placed on one heel of her high-heeled shoes. Calculate the pressure exerted o
STatiana [176]

Answer:

3335400 N/m² or 483.75889 lb/in²

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

A = Area = 1.5 cm²

m = Mass of woman = 51 kg

F = Force = mg

When we divide force by area we get pressure

P=\frac{F}{A}\\\Rightarrow P=\frac{mg}{A}\\\Rightarrow P=\frac{51\times 9.81}{1.5\times 10^{-4}}\\\Rightarrow P=3335400\ N/m^2

1\ N/m^2=\frac{1}{6894.757}\ lb/in^2

3335400\ N/m^2=3335400\times \frac{1}{6894.757}\ lb/in^2=483.75889\ lb/in^2

The pressure exerted on the floor is 3335400 N/m² or 483.75889 lb/in²

7 0
3 years ago
A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t
Mumz [18]

Answer:

v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}

Explanation:

We will apply the equations of kinematics to both stones separately.

First stone:

Let us denote the time spent after the second stone is thrown as 'T'.

y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t

Second stone:

y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}

6 0
3 years ago
Read 2 more answers
Two cars are traveling on a desert road. After 5.0 seconds they are side by side at the next telephone pole.The distance between
stira [4]
The question looks incomplete, but according to the information given above seem like they have <span>identical journeys.
</span>a. the displacement of car A - <span> 65.5 m 
</span>b. the displacement of car B  - <span>65.5 m
c. average velocir</span>y of A  65.5 / 5 = 13.1 m/s
d. the average velocity of car B  has the same. 
8 0
3 years ago
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