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3 years ago

Calculate the force between two objects that have masses of 70 Kilograms and 2,000 kilograms separated by a distance of 1 meter

Physics

1 answer:

Hunter-Best [27]3 years ago

5 0

9.3 x 10⁻⁶N

Explanation:

Given parameters:

Mass 1 = 70kg

Mass 2 = 2000kg

distance = 1m

Unknown:

force between them =

Solution:

The force between the two masses will be a gravitational force of attraction.

F = $\frac{G m_{1}m_{2} }{r^{2} }$

G is universal gravitation constant = 6.67430×10−¹¹ N⋅m²/kg²

r is the distance between the two masses

Substituting the parameters:

F = $\frac{6.67 x 10^{-11} x 70 x 2000}{1^{2} }$ = 9.3 x 10⁻⁶N

Learn more:

Universal gravitation constant brainly.com/question/1724648

#learnwithBrainly

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How much heat (in calories) are needed to raise the temperature of 60 g of water

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Well, each ml of water requires one calorie to go up 1 degree Celsius, so this liter of water takes 1000 calories to go up 1 degree Celsius.

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2 years ago

If the distance between the Earth and Moon were half what it is now, by what factor would the force of gravity between them be c

hichkok12 [17]

Answer:

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

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Dividing the equations

$\dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{\dfrac{1}{4}r^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=4$

The ratio is $\dfrac{F_n}{F_o}=4$

The new force would be 4 times the old force

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2 years ago

A 16.2 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The

S_A_V [24]

To solve this problem we will apply the concepts related to the balance of forces. We will decompose the forces in the vertical and horizontal sense, and at the same time, we will perform summation of torques to eliminate some variables and obtain a system of equations that allow us to obtain the angle.

The forces in the vertical direction would be,

$\sum F_x = 0$

$f-N_w = 0$

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The forces in the horizontal direction would be,

$\sum F_y = 0$

$N_f -W =0$

$N_f = W$

The sum of Torques at equilibrium,

$\sum \tau = 0$

$Wdcos\theta - N_wLsin\theta = 0$

$WdCos\theta = fLSin\theta$

$f = \frac{Wd}{Ltan\theta}$

The maximum friction force would be equivalent to the coefficient of friction by the person, but at the same time to the expression previously found, therefore

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