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2 years ago

15

Calculate the force of gravity on the 0.60-kg mass if it were 1.3×107 m above Earth's surface (that is, if it were three Earth r

adii from Earth's center).

1 answer:

KIM [24]2 years ago

5 0

The force of gravity between two objects is given by:
$F=G \frac{m_1 m_2}{r^2}$
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is their separation

In this problem, the mass of the object is $m_1=0.60 kg$ , while the Earth's mass is $m_2=5.97 \cdot 10^{24} kg$ . Their separation is $r=1.3 \cdot 10^7 m$ , therefore the gravitational force exerted on the object is
$F=(6.67 \cdot 10^{-11}m^3 kg^{-1} s^{-2}) \frac{(0.60 kg)(5.97 \cdot 10^{24} kg)}{(1.3 \cdot 10^7 m)^2}=1.4 N$

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natulia [17]

Answer:

$\% Error = 2.6\%$

Explanation:

Given

$x: 1.54, 1.53, 1.44, 1.54, 1.56, 1.45$

Required

Determine the percentage error

First, we calculate the mean

$\bar x = \frac{\sum x}{n}$

This gives:

$\bar x = \frac{1.54+ 1.53+ 1.44+ 1.54+ 1.56+ 1.45}{6}$

$\bar x = \frac{9.06}{6}$

$\bar x = 1.51$

Next, calculate the mean absolute error (E)

$|E| = \sqrt{\frac{1}{6}\sum(x - \bar x)^2}$

This gives:

$|E| = \sqrt{\frac{1}{6}*[(1.54 - 1.51)^2 +(1.53- 1.51)^2 +.... +(1.45- 1.51)^2]}$

$|E| = \sqrt{\frac{1}{6}*0.0132}$

$|E| = \sqrt{0.0022}$

$|E| = 0.04$

Next, calculate the relative error (R)

$R = \frac{|E|}{\bar x}$

$R = \frac{0.04}{1.51}$

$R = 0.026$

Lastly, the percentage error is calculated as:

$\% Error = R * 100\%$

$\% Error = 0.026 * 100\%$

$\% Error = 2.6\%$

4 0

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Explanation:

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After the collision the bullet and the block have a total mass of 18.41 kg and they move at a speed of 8.8 m/s. The kinetic energy after the collision is

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Before the collision only the bullet has kinetic energy.

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Alex777 [14]

Answer:

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Rama09 [41]

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