Which is larger, the Sun's pull on Earth or Earth's pull on the Sun?

The masses of the Moon and Earth are 7.35 x 1022 kg and 5.97 x 1024 kg, respectively. The strength of the gravitational force be

bixtya [17]

Answer:

Distance between centre of Earth and centre of Moon is 3.85 x 10⁸ m

Explanation:

The attractive force experienced by two mass objects is known as Gravitational force.

The gravitational force is determine by the relation:

$F=\frac{Gm_{1} m_{2} }{d^{2} }$ ....(1)

According to the problem,

Mass of Moon, m₁ = 7.35 x 10²² kg

Mass of Earth, m₂ = 5.97 x 10²⁴ kg

Gravitational force experienced by them, F = 1.98 x 10²⁰ N

Universal gravitational constant, G = 6.67 x 10⁻¹¹ Nm²kg⁻²

Substitute these values in equation (1).

$1.98\times10^{20} =\frac{6.67\times10^{-11}\times7.35\times10^{22}\times5.97\times10^{24} }{d^{2} }$

$d^{2}=\frac{2.93\times10^{37}}{1.98\times10^{20}}$

$d=\sqrt{1.48\times10^{17}}$

d = 3.85 x 10⁸ m

3 0

3 years ago

the diameter and the length of a thin wire, approximately 1m in length, are measured as accurately as possible. what are the bes

slamgirl [31]

Answer:

1.) Micrometres screw gauge

2.) Tape rule.

Explanation:

Given that the diameter and the length of a thin wire, approximately 1m in length, are measured as accurately as possible.

what are the best instruments to use ?

To measure the diameter of a thin wire, the best instrument to use is known as micrometres screw gauge.

And to measure the length of a thin wire up to 1 m, the measuring device can be tape rule or long metre rule.

5 0

3 years ago

A 0.03 kg golf ball is hit off the tee at a speed of 34 m/s. The golf club was in contact with the ball for 0.003 s. What is the

Liula [17]

Answer:

The average force on ball by the golf club is 340 N.

Explanation:

Given that,

Mass of the golf ball, m = 0.03 kg

Initial speed of the ball, u = 0

Final speed of the ball, v = 34 m/s

Time of contact, $\Delta t=0.003\ s$

We need to find the average force on ball by the golf club. We know that the rate of change of momentum is equal to the net external force applied such that :

$F=\dfrac{\Delta p}{\Delta t}\\\\F=\dfrac{mv-mu}{\Delta t}\\\\F=\dfrac{mv}{\Delta t}\\\\F=\dfrac{0.03\ kg\times 34\ m/s}{0.003\ s}\\\\F=340\ N$

So, the average force on ball by the golf club is 340 N.

4 0

4 years ago

Which element of variation would not be affected by adding the data value 15 to the data set {3, 5, 6, 8, 9, 10, 13, 14}?

Anna11 [10]

The answer to this would be 15

4 0

3 years ago

Can you guys help to understand this graph I'm so confused why I getting wrong?

dedylja [7]

The one fact that needs to be mentioned but isn't given anywhere on or around the graph is: The distance, on the vertical axis, is the distance FROM home. So any point on the graph where the distance is zero ... the point is in the x-axis ... is a point AT home.

Segment D ...
Walking AWAY from home; distance increases as time increases.

Segment B ...
Not walking; distance doesn't change as time increases.

Segment C ...
Walking away from home, but slower than before; distance increases as time increases, but not as fast. Slope is less than segment-D.

Segment A ...
Going home; distance is DEcreasing as time increases. Walking pretty fast ... the slope of the line is steep.

4 0

3 years ago

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