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3 years ago

On the surface of the earth the weight of an object is 200 lb. Determine the height of the

Physics

1 answer:

siniylev [52]3 years ago

7 0

Answer:

The height of the object is 5007.4 miles.

Explanation:

Given that,

Weight of object = 200 lb

We need to calculate the value of $Gmm_{e}$

Using formula of gravitational force

$F=\dfrac{Gmm_{e}}{r^2}$

Put the value into the formula

$200=\dfrac{Gmm_{e}}{(3958.756)^2}$

$200\times(3958.756)^2=Gmm_{e}$

$Gmm_{e}=3.134\times10^{9}$

We need to calculate the height of the object

Using formula of gravitational force

$F=\dfrac{Gmm_{e}}{r^2}$

Put the value into the formula

$125=\dfrac{200\times(3958.756)^2}{r^2}$

$r^2=\dfrac{200\times(3958.756)^2}{125}$

$r^2=25074798.5$

$r=\sqrt{25074798.5}$

$r=5007.4\ miles$

Hence. The height of the object is 5007.4 miles.

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A 0.0250-kg bullet is accelerated from rest to a speed of 550 m/s in a 3.00-kg rifle. The pain of the rifle’s kick is much worse

kondaur [170]

Answer:

a) 4.583 m/s

b) 31.505 J

c) 0.491 m/s

d) 3.375 J

p_player = (110 kg)(8 m/s) = 880 kg m/s

p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s

Explanation:

HI!

We can calculate the recoil velocity by conservation of momentum, remember that p=mv.

The momentum of the bullet is:

p_b = (0.0250 kg)*(550 m/s )

The momentum of the rifle is:

p_r = (3 kg) * v

Since the total initial momentum is zero:

p_b = p_r

That is:

v = (550 m/s ) (0.0250 kg/ 3 kg ) = 4.583 m/s

The kinetic energy gained by the rifle is:

K = (1/2) m v^2 = (1/2) *(3 kg) *(4.583 m/s)^2 = 31.505 J

We use the same formula as in a), but with m=28kg instead of 3 kg

v = (550 m/s ) (0.0250 kg/ 28 kg ) = 0.491 m/s

Again, the same formula as b, but with m=28 and v=0.491 m/s

K = 3.375 J

p_player = (110 kg)(8 m/s) = 880 kg m/s

p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s

I believe that the kinetic energy is more related to the problem than the momentum. The relation between these two quantities is:

K = p^2/(2m)

usiing this relation, we get:

K_player = 3520 J

K_ball = 128.125 J

Therefore the kinetic energy of the player is around 27 time larger than the kinetic energy of the ball, that being said, the pain of being tackled by that player is around 27 times greater that being hit by the ball!

4 0

3 years ago

A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro

svetoff [14.1K]

Answer:

The equation of motion is $x(t)=-$ $\frac{1}{3} cos4\sqrt{6t}$

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is $32ft/s^2$

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet $x=\frac{4}{12} =\frac{1}{3}feet$

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

$W=kx$ ⇒ $k=\frac{W}{x}$

The spring constant , $k=\frac{24}{1/3}$

= 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

$m\frac{d^2x}{dt} +kx=0$

$\frac{3}{4} \frac{d^2x}{dt} +72x=0$

$\frac{d^2x}{dt} +96x=0$

Auxiliary equation is, $m^2+96=0$

$m=\sqrt{-96}$

= $\frac{+}{} i4\sqrt{6}$

Thus , the solution is $x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}$

$x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6t}$

The mass is released from the rest x'(0) = 0

$=-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6(0)}$ =0

$c_2$ $4\sqrt{6} =0$

$c_2=0$

Therefore , $x(t)=c_1$ $cos 4\sqrt{6t}$

Since , the mass is released from the rest from 4 inches

$x(0)= -4$ inches

$c_1 cos 4\sqrt{6(0)} =-\frac{4}{12}$ feet

$c_1=-\frac{1}{3}$ feet

Therefore , the equation of motion is $-\frac{1}{3} cos4\sqrt{6t}$

7 0

3 years ago

A proton in a high-energy accelerator is given a kinetic energy of 50.0 GeV. Determine (a) the momentum and (b) the speed of the

koban [17]

(a) The momentum of the proton is determined as 5.17 x 10⁻¹⁸ kgm/s.

(b) The speed of the proton is determined as 3.1 x 10⁹ m/s.

<h3>Momentum of the proton</h3>

The momentum of the proton is calculated as follows;

K.E = ¹/₂mv²

where;

m is mass of proton = 1.67 x 10⁻²⁷ kg
v is speed of the proton = ?

<h3>Speed of the proton</h3>

v² = 2K.E/m

v² = (2 x 50 x 10⁹ x 1.602 x 10⁻¹⁹ J)/(1.67 x 10⁻²⁷)

v² = 9.6 x 10¹⁸

v = 3.1 x 10⁹ m/s

<h3>Momentum of the proton</h3>

P = mv = (1.67 x10⁻²⁷ x 3.1 x 10⁹) = 5.17 x 10⁻¹⁸ kgm/s

Learn more about momentum here: brainly.com/question/7538238

#SPJ4

4 0

2 years ago

Brainliest!! Please help with number 4 and please give me the equation too!! Brainliest!!!

Thepotemich [5.8K]

Answer:

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Explanation:

8 0

3 years ago

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lyudmila [28]

Answer:

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Explanation:

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3 years ago