Help me please.......

Objects with three very different sizes and weights are dropped off a tower on the moon. which factor of the objects will determ

vovikov84 [41]

4) none of these matter. with no air resistance, they all hit the ground at the same time.

Explanation:

The motion of the objects is a free fall motion, since the only force acting on them is gravity. Therefore, it is a uniformly accelerated motion with constant acceleration $g$ (acceleration of gravity) towards the ground, and so we can find the time of flight by using the suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the vertical displacement (equal to height of the tower)

u = 0 is the initial velocity

t is the time

$a=g$ is the acceleration

Re-arranging the equation.

$t=\sqrt{\frac{2s}{g}}$

As we can see, the time of flight depends only on the height of the tower (s) and the acceleration due to gravity (g): therefore, the three objects reach the ground at the same time.

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4 0

3 years ago

a 91.5 kg football player running east at 3.73 m/s tackles a 63.5 kg player running east at 3.09 m/s. what is their velocity aft

Lesechka [4]

The final velocity is 3.47 m/s east

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, the total momentum of the two players before and after the collision must be conserved:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v$

where:

$m_1 = 91.5 kg$ is the mass of the first player

$u_1 = 3.73 m/s$ is the initial velocity of the first player (we take east as positive direction)

$m_2 = 63.5 kg$ is the mass of the second player

$u_2 = 3.09 m/s$ is the initial velocity of the second player

$v$ is their final combined velocity after the collision

Re-arranging the equation and substituting the values, we find:

$v = \frac{m_1 u_1+m_2 u_2}{m_1+m_2}=\frac{(91.5)(3.73)+(63.5)(3.09)}{91.5+63.5}=3.47 m/s$

So, their velocity afterwards is 3.47 m/s east.

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6 0

3 years ago

Please solve the Problem.

prisoha [69]

The scalar reading during the process is 170. 5 Newton

<h3>How to determine the scalar reading</h3>

force = mass × acceleration

Given the mass = 55kg

In finding the acceleration, use the first equation of motion

v = u + at

u = 0

t = 2s

v = 6.5 m/s

Substitute the values

6.5 = 0 + 2a, make 'a' the subject

a = 6. 5÷ 2 = 3. 1 m/s²

Substitute the value of 'm' and 'a' in the original equation

F = 55× 3.1 = 170. 5 Newton

Hence, the scalar reading during the process is 170. 5 Newton

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7 0

2 years ago

A test of the prototype of a new automobile shows that the minimum distance for a controlled stop from 95 km/h to zero is 55 m.

iris [78.8K]

Answer:

-0.64525g

Explanation:

t = Time taken for the car to stop

u = Initial velocity = 95 km/h

v = Final velocity = 0 km/h

s = Displacement

a = Acceleration

Equation of motion

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-95^2}{2\times 0.055}\\\Rightarrow a=-82045.45\ km/h^2$

Converting to m/s²

$a=82045.45=\frac{82045.45\times 1000}{3600\times 3600}=-6.33\ m/s^2$

g = Acceleration due to gravity = 9.81 m/s²

Dividing both the accelerations, we get

$\frac{a}{g}=\frac{-6.33}{9.81}=-0.64525\\\Rightarrow a=-0.64525g$

Hence, acceleration of the car is -0.64525g

8 0

4 years ago

Two spherical asteroids have the same radius R. Asteroid 1 has mass M and asteroid 2 has mass 1.97·M. The two asteroids are rele

nekit [7.7K]

Answer:

$0.536\sqrt{\frac{GM}{R}}$

Explanation:

We are given that

Mass of one asteroid 1, $m_1=M$

Mass of asteroid 2, $m_2=1.97 M$

Initial distance between their centers,d=13.63 R

Radius of each asteroid=R

d'=R+R=2R

Initial velocity of both asteroids

$u=0$

We have to find the speed of second asteroid just before they collide.

According to law of conservation of momentum

$(m_1+m_2)u=m_1v_1+m_2v_2$

$(M+1.97 M)\times 0=Mv_1+1.97Mv_2$

$Mv_1=-1.97 Mv_2$

$v_1=-1.97v_2$

According to law of conservation of energy

$Gm_1m_2(\frac{1}{d'}-\frac{1}{d})=\frac{1}{2}m_1v^2_1+\frac{1}{2}m_2v^2_2$

$GM(1.97M)(\frac{1}{2R}-\frac{1}{13.63R})=\frac{1}{2}M(-1.97v_2)^2+\frac{1}{2}(1.97M)v^2_2$

$1.97M^2G(\frac{13.63-2}{27.26R})=\frac{1}{2}Mv^2_2(3.8809+1.97)$

$1.97MG(\frac{11.63}{27.26 R})=\frac{1}{2}(5.8509)v^2_2$

$v^2_2=\frac{1.97GM\times11.63\times 2}{27.26R\times 5.8509}$

$v_2=\sqrt{\frac{1.97GM\times11.63\times 2}{27.26R\times 5.8509}}$

$v_2=0.536\sqrt{\frac{GM}{R}}$

Hence, the speed of second asteroid = $0.536\sqrt{\frac{GM}{R}}$

8 0

3 years ago