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2 years ago

15

How would the gravitational force between the earth and the moon change if the moon had half the mass?

1 answer:

Vikki [24]2 years ago

5 0

The moon's gravity, combined with the waltz of Earth and the moon around their center of mass, forces the oceans into an oval shape, with two simultaneous high tides. ... If the moon were half its mass, then the ocean tides would have been correspondingly smaller and imparted less energy to it.

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A football punker attempts to kick the football so that it lands on the ground 67.0 m from where it is kicked and stays in the a

Flauer [41]

To solve this problem we will apply the linear motion kinematic equations. We will find the two components of velocity and finally by geometric and vector relations we will find both the angle and the magnitude of the vector. In the case of horizontal speed we have to

$v_x = \frac{x}{t}$

$v_x = \frac{67}{4.5}$

$v_x = 14.89m/s$

The vertical component of velocity is

$-h = v_y t -\frac{1}{2} gt^2$

Here,

h = Height

g = Gravitational acceleration

t = Time

$v_y$ = Vertical component of velocity

$-1.23 = v_y(4.5)-\frac{1}{2} (9.8)(4.5)^2$

$-1.23= 4.5v_y - 99.225$

$v_y = 21.77m/s$

The direction of the velocity will be given by the tangent of the components, then

$tan\theta = \frac{v_y}{v_x}$

$\theta = tan^{-1} (\frac{21.77}{14.89})$

$\theta = 55.59\°$

The magnitude is given vectorially as,

$|V| = \sqrt{v_x^2+v_y^2}$

$|V| = \sqrt{14.89^2 +21.77^2}$

$|V| = 26.37m/s$

Therefore the angle is 55.59° and the velocity is 26.37m/s

6 0

2 years ago

An object is thrown directly downward from the top of a very tall building. The speed of the object just as it is released is 17

Softa [21]

Answer:

distance cover is = 102.53 m

Explanation:

Given data:

speed of object is 17.1 m/s

$t_1 = 3.32 sec$

$t_2 = 5.08 sec$

from equation of motion we know that

$d_1 = vt_1 + \frac{1}{2} gt_1^2$

where d_1 is distance covered in time t1

so $d_1 = 17.1 \times 3.32 + \frac{1}{2} 9.8 \times 3.32^2$ =

$d_1 = 110.78 m$

$d_2 = vt_2 + \frac{1}{2} gt_2^2$

where d_2 is distance covered in time t2

$d_2 = 17.1 \times 5.08 + \frac{1}{2}\times9.8 \times 5.08^2$

$d_2 = 213.31 m$

distance cover is = 213.31 - 110.78 = 102.53 m

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Vitek1552 [10]

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ludmilkaskok [199]

Answer:

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3 years ago

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