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2 years ago

3. a) If the distance between two objects increases by a factor of 5 (5),

Physics

1 answer:

irina [24]2 years ago

3 0

Answer:

Answer for A

Explanation:

F1=GmM/r1^2

If r2 becomes r2=5r

F2=GmM/(25r^2)

Multiply with 25 gives to maintain the same force

I.e.,25F2=F1

F2=G(25m)M/25r^2=F1

By the factor 25 would change to increase to same.

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The period of time required for the moon to complete a cycle of phases is called the ________ month.

eduard

Synodic month, also known as a lunar month.

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3 years ago

The force of Earth's gravity on a capsule in space will lessen as it moves farther away. If the capsule moves to twice its dista

Bess [88]

Answer: One quarter of the force

Explanation:

According to Newton's law of Gravitation, the force $F$ exerted between two bodies of masses $m1$ and $m2$ and separated by a distance $r$ is equal to the product of their masses and inversely proportional to the square of the distance:

$F=G\frac{(m1)(m2)}{r^2}$ (1)

Where $G$ is the gravitational constant

This means that the gravity force decreases when the distance between these two bodies increases.

In this context, if the distance between the capsule and the Earth increases twice, the new distance will be $2r$ .

Substituting this distance in (1):

$F=G\frac{(m1)(m2)}{(2r)^2}$ (2)

$F=G\frac{(m1)(m2)}{4r^2}$

<u>Finally:</u>

$F=\frac{1}{4}G\frac{(m1)(m2)}{r^2}$ >>>This means the force toward Earth becomes one quarter "weaker"

3 0

2 years ago

A bird is about 6.26.2 in. long, with a thin, dark bill and a wide, white wing stripe. If the bird can fly 9292 mi with the w

Trava [24]

Answer:

$209$ mph

Explanation:

$V$ = Speed of bird in still air

$v$ = Speed of wind = 44 mph

Consider the motion of the bird with the wind

$D_{1}$ = distance traveled with the wind = 9292 mi

$t_{1}$ = time taken to travel the distance with wind

Time taken to travel the distance with wind is given as

$t_{1} = \frac{D_{1}}{V + v}$

$t_{1} = \frac{9292}{V + 44}$ eq-1

Consider the motion of the bird with the wind

$D_{2}$ = distance traveled against the wind = 6060 mi

$t_{2}$ = time taken to travel the distance against wind

Time taken to travel the distance against wind is given as

$t_{2} = \frac{D_{2}}{V + v}$

$t_{2} = \frac{6060}{V - 44}$ eq-2

As per the question,

Time taken with the wind = Time taken against the wind

$t_{1} = t_{2}$

$\frac{9292}{V + 44} = \frac{6060}{V - 44}$

$(9292) (V - 44) = (6060) (V + 44)$

$9292V - 408848 = 6060V + 266640$

$3232V = 675488$

$V = 209$ mph

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2 years ago

Electrons are important to electric current because they are able to

Agata [3.3K]

Electrons are important to the electric current because they are able to move from one atom to another. When an atom loses an electron, it becomes positively charged and when an atom gains an electron, it becomes negatively charged.

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3 years ago

Which of these choices is the smallest matter

Ksenya-84 [330]

What are the choices?

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2 years ago