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lana66690 [7]
3 years ago
8

How does the gravitational force between two objects change if the mass of

Physics
1 answer:
GalinKa [24]3 years ago
3 0

Answer:

When one of the two objects mass is doubled the gravitational force of one of the objects becomes twice as strong.

Explanation:

Because jesus said so

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What of the following does NOT influence resistance? *
Marina86 [1]

Answer:

D. Number of components

Explanation:

4 0
2 years ago
Given: G = 6.672 × 10−11 N · m2 /kg2 Io, a satellite of Jupiter, has an orbital period of 1.24 days and an orbital radius of 4.1
Dahasolnce [82]

Answer:

Mass of Jupiter = 4.173×10^15kg

Explanation:

Using Kepler's 3rd law, it states that the orbital period T is related to the distance,r as:

T^2 = GM/4 pi × r^3

Where G = universal gravitational constant

r = radius

M = masd of jupiter

Rearranging the formular to make M the subject of formular

T^2 × 4 pi = G M × r^3

(T^2 × 4 pi) / (G× r^3) = M

(1.24^2 × 4 × 3.142) /(6.672×10^-11)(4.11×10^8)^3

M = 19.32 /6.672×10^-11)(4.11×10^8)^3

M = 19.32 / 4.63 ×10^15

M = 4.173×10^15kg

6 0
2 years ago
On a hot summer day in the state of Washington while kayaking, I saw several swimmers jump from a railroad bridge into the Snoho
Svetlanka [38]

Answer: part a: 19.62m

part b: 19.62 m/s

part a: 2.83 secs

Explanation:If the air resistance is ignored then the swimmer experience free fall under gravity hence

u=0

a=9.81 m/s2

t=2 secs

s=ut+0.5at^2

s=h

h=0*2+0.5*9.81*2^2\\h=19.62 meters

Part b

v=u+at\\v=0+9.81*2\\v=19.62m/s

Part c

now we have h=2*19.62=39.24

39.24=0+0.5*9.81*t^2\\t^2=8\\t=2.83 secs

4 0
3 years ago
If an object accelerates from rest, what will its velocity be after 1.3 s if it has a constant acceleration of 9.1 m/s^2?
HACTEHA [7]

\text{Given that,}\\\\\text{Initial velocity,} ~v_0 = 0~ \text{m~s}^{-1}\\\\\text{Time,  t = 1.3~sec}\\\\\text{Acceleration, a = 9.1 m s}^{-2}\\\\\\\\\text{Velocity,}\\\\v = v_0  +at\\\\\implies v = 0 + 9.1 \times 1.3 = 11.83~~ \text{m~s}^{-1}

5 0
2 years ago
2. An airplane traveling north at 220. meters per second encounters a 50.0-meters-per-second crosswind
Alex777 [14]

The resultant speed of the plane  is (3) 226 m/s

Why?

We can calculate the resultant speed of the plane by using the Pythagorean Theorem since both speeds are perpendicular (forming a right triangle).

So, calculating we have:

ResultantSpeed=\sqrt{VerticalSpeed^{2}+HorizontalSpeed^{2}}\\\\ResultantSpeed=\sqrt{(220\frac{m}{s})^{2}+50\frac{m}{s})^{2}

ResulntantSpeed=\sqrt{48400\frac{m^{2} }{s^{2} }+2500\frac{m^{2} }{s^{2} } } \\\\ResultantSpeed=\sqrt{50900\frac{m^{2} }{s^{2} }}=226\frac{m}{s}

Hence, we have that the resultant speed of the plane  is (3) 226 m/s

Have a nice day!

5 0
3 years ago
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