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ycow [4]
3 years ago
10

The diagrams show objects’ gravitational pull toward each other. Which statement describes the relationship between diagram X an

d Y?
A. Gravity attracts only larger objects toward one another.
B. Gravity attracts larger objects only if they are close to one another.
C.If the masses of the objects increase, then the force between them also increases.
D.If distance between the objects increases, then the amount of force also increases.
Physics
2 answers:
creativ13 [48]3 years ago
8 0

' C ' is the only correct statement on the list.  We don't know anything about diagram-x or diagram-y because we can't see them.

Nookie1986 [14]3 years ago
7 0

The answer is c on edugentiy hope this helped

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It will be traveling in the reverse direction it was originally going at 15.2 m/s
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3 years ago
Please help I literally don’t understand
Veseljchak [2.6K]

Answer:

A= 2

B=3

C=4

D=5

E=7

F=8

H=12

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Carol has several test tubes of different liquids. If Carol removes heat from the liquid substances, which of the following is m
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d

Explanation:

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An egg rolls off a kitchen counter and breaks as it hits the floor. The counter is 1.0 m high, the mass of the egg is about 50 g
crimeas [40]
Use equations of motion to find the velocity just before it hits the floor: 
<span>Vf^2 = Vi^2 + 2gx </span>
<span>Final velocity = 4.42m/s </span>

<span>Impulse is change in momentum so: </span>
<span>m(Vf - Vi) = 0.05(0 - 4.42) </span>
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3 0
3 years ago
A 25,000 kg traveling east collides with a 2,000 kg truck standing still on the tracks. After the collision the train and truck
Elis [28]

Answer:

24.084 m/s

Explanation:

From the law of conservation of linear momentum

Total momentum before collision equals to the total momentum after collision

Since momentum=mv where m is mass and v is velocity

M_{truck}V_{truck}=V_{common}*(M_{truck} +M_{standing}) where M_{truck} is the mass of the truck, V_{truck} is velocity of the truck, V_{common} is the common velocity of moving and standing truck after collision and M_{standing} is the mass of the standing truck

Making V_{truck} the subject we obtain

V_{truck}=\frac { V_{common}*(M_{truck} +M_{standing})}{M_{truck}}

Substituting M_{truck} as 25000 Kg, V_{common} as 22.3 m/s, M_{standing} as 2000 Kg we obtain

V_{truck}=\frac { 22.3 m/s *(25000 Kg +2000 Kg)}{25000}= 24.084 m/s

Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive

The truck was moving at 24.084 m/s

3 0
3 years ago
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