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2 years ago

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A 235 kg object and a 1.37×1012 kg are located 2.59×104 m away from each other. What is the force due to gravity between the two

objects?

2 answers:

V125BC [204]2 years ago

8 0

Answer:

the force due to gravity between the two objects is 3.2 x 10⁻⁵ N.

Explanation:

Given;

mass of the first object, m₁ = 235 kg

mass of the second object, m₂ = 1.37 x 10¹² kg

distance between the two object, r = 2.59 x 10⁴ m

The gravitational force between the two object is calculated as;

$F= \frac{Gm_1m_2}{r^2}$

where;

G is gravitational constant = 6.67 x 10⁻¹¹ Nm²/kg²

$F= \frac{(6.67\times 10^{-11})(235)(1.37\times 10^{12})}{(2.59\times 10^4)^2} \\\\F = 3.2 \times 10^{-5} \ N$

Therefore, the force due to gravity between the two objects is 3.2 x 10⁻⁵ N.

Alexus [3.1K]2 years ago

4 0

Answer:

F = 3.2 x 10⁻⁵ N

Explanation:

The gravitational force of attraction between the two objects is given by Newton's Gravitational law through the following formula:

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

where,

F = gravitational force = ?

G = Gravitational Constant = 6.67 x 10⁻¹¹ Nm²/kg²

m₁ = mass of object 1 = 235 kg

m₂ = mass of object 2 = 1.37 x 10¹² kg

r = distance between objects = 2.59 x 10⁴ m

Therefore,

$F = \frac{(6.67\ x\ 10^{-11}\ Nm^{2}/kg^{2})(235\ kg)(1.37\ x\ 10^{12}\ kg)}{(2.59\ x\ 10^{4}\ m)^{2}}$

<u>F = 3.2 x 10⁻⁵ N</u>

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1) Distance down the hill: 1752 ft (534 m)

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Explanation:

1)

The motion of the shell is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion of the shell is a uniformly accelerated motion, so the vertical position is given by the following equation:

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At the same time, the horizontal motion of the shell is a uniform motion, so the horizontal position of the shell at time t is given by the equation

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We can re-write this last equation as

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$xtan \alpha = x tan \theta - \frac{1}{2}g(\frac{x}{ucos \theta})^2\\x (tan \theta - tan \alpha)-\frac{g}{2u^2 cos^2 \theta} x^2=0\\x(tan \theta - tan \alpha-\frac{gx}{2u^2 cos^2 \theta})=0$

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and

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So, the distance d down the hill at which the shell strikes the hill is

$d=\frac{x}{cos \alpha}=\frac{1322}{cos(-41.0^{\circ})}=1752 ft=534 m$

2)

In order to find how long the mortar shell remain in the air, we can use the equation:

$t=\frac{x}{u cos \theta}$

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x = 1322 ft is the final position of the shell when it strikes the hill

$u=156 ft/s$ is the initial velocity of the shell

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Substituting these values into the equation, we find the time of flight of the shell:

$t=\frac{1322}{(156)(cos 49^{\circ})}=12.9 s$

3)

In order to find the final speed of the shell, we have to compute its horizontal and vertical velocity first.

The horizontal component of the velocity is constant and it is

$v_x = u cos \theta =(156)(cos 49^{\circ})=102.3 ft/s$

Instead, the vertical component of the velocity is given by

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And substituting at t = 12.9 s (time at which the shell strikes the hill),

$v_y=(156)(cos 49^{\circ})-(32)(12.9)=-310.4ft/s$

Therefore, the final speed of the shell is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(102.3)^2+(-310.4)^2}=326.8 ft/s=99.6 m/s$

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