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4 years ago

If Pete ( mass=90.0kg) weights himself and finds that he weighs 30.0 pounds, how far away from the surface of the earth is he

Physics

1 answer:

shutvik [7]4 years ago

5 0

Answer: 9938.8 km

Explanation:

1 pound-force = 4.48 N

30.0 pounds-force = 134.4 N

The force of gravitation between Earth and object on the surface of is given by:

$F = \frac{GMm}{R^2} = mg$

Where M is the mass of the Earth, m is the mass of the object, R (6371 km) is the radius of the Earth.

At height, h above the surface of the Earth, the weight of the object:

$(mg)'= \frac{GMm}{(R+h)^2}$

we need to find "h"

taking the ratio of two:

$\frac{mg}{(mg)'}=\frac{(R+h)^2}{R^2}\\ \Rightarrow \frac{90kg \times 9.8 m/s^2}{134.4 N}=\frac{(R+h)^2}{R^2}\\ \Rightarrow 6.56 R^2= (R+h)^2 \Rightarrow h= (2.56-1)R\\ \Rightarrow h = 1.56 R = 1.56 \times 6371 km = 9938. 8 km$

Hence, Pete would weigh 30 pounds at 9938.8 km above the surface of the Earth.

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A hot-air balloon is ascending at the rate of 10 m/s and is 74 m above the ground when a package is dropped over the side. (a) H

Reika [66]

Answer:

The answer to your question is:

a) t1 = 2.99 s ≈ 3 s

b) vf = 39.43 m/s

Explanation:

Data

vo = 10 m/s

h = 74 m

g = 9.81 m/s

t = ? time to reach the ground

vf = ? final speed

a) h = vot + (1/2)gt²

74 = 10t + (1/2)9.81t²

4.9t² + 10t -74 = 0 solve by using quadratic formula

t = (-b ± √ (b² -4ac) / 2a

t = (-10 ± √ (10² -4(4.9(-74) / 2(4.9)

t = (-10 ± √ 1550.4 ) / 9.81

t1 = (-10 + √ 1550.4 ) / 9.81 t2 = (-10 - √ 1550.4 ) / 9.81

t1 = (-10 ± 39.38 ) / 9.81 t2 = (-10 - 39.38) / 9.81

t1 = 2.99 s ≈ 3 s t2 = is negative then is wrong there are

no negative times.

b) Formula vf = vo + gt

vf = 10 + (9.81)(3)

vf = 10 + 29.43

vf = 39.43 m/s

4 0

3 years ago

2. A wave on a rope has a wavelength of 2.0 m and a frequency of 2.0 Hz. What is the speed of the

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Answer:

4 m/s or 4 meters per second.

Explanation:

In order to calculate the speed of wave, you multiply the wavelength in meters and the frequency of the Wave in Hertz. 2 times 2 equals 4. The wave speed is always in m/s considering that the wavelength is also in meters.

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4 years ago

If for every action force an equal and opposite reaction force exlsts, how can anything ever be accelerated?

klasskru [66]

Answer:

Explanation:

The net force is the force between action and reaction and when this forces are not the same an acceleration is spurred.

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3 years ago

Salmon often jump waterfalls to reach their

natta225 [31]

Answer:

5.0 m/s

Explanation:

The horizontal motion of the salmon is uniform, so the horizontal component of the salmon's velocity is constant and it is

$v_x = u cos \theta$

where u is the initial speed and $\theta=37.7^{\circ}$ . The horizontal distance travelled by the salmon is

$d=v_x t = (ucos \theta)t$

where d = 1.95 m and t is the time needed to reach the final point.

Re-arranging for t,

$t=\frac{d}{v_x}=\frac{d}{u cos \theta}$ (1)

Along the vertical direction, the equation of motion is

$y=h+u_y t -\frac{1}{2}gt^2$

where:

y = 0.311 m is the final height reached by the salmon

h = 0 is the initial height

$u_y = u sin \theta$ is the vertical component of the initial velocity of the salmon

$g=9.81 m/s^2$ is the acceleration of gravity

t is the time

Substituting t as found in eq.(1), we get the equation

$y=(u sin \theta) \frac{d}{u cos \theta}- \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}$

and we can solve this formula for u, the initial speed of the salmon:

$y=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}\\\\u=\sqrt{\frac{gd^2}{2(dtan \theta -y)cos^2 \theta}}=\sqrt{\frac{(9.81)(1.95)^2}{2((1.95)(tan 37.7^{\circ}) -0.311)cos^2 37.7^{\circ}}}=5.0 m/s$