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almond37 [142]
3 years ago
11

A satellite with a mass of 5.6 E 5 kg is orbiting the Earth in a circular path. Determine the satellite's velocity if it is orbi

ting at a distance of 6.8 E 5 m above the Earth's surface. Earth's mass = 5.98 E 24 kg; Earth's radius = 6.357 E 6 m.
6,800 m/s
7,200 m/s
7,500 m/s
7,900 m/s
Physics
1 answer:
solniwko [45]3 years ago
7 0

Answer:

7500 m/s

Explanation:

Centripetal acceleration = gravity

v² / r = GM / r²

v = √(GM / r)

Given:

G = 6.67×10⁻¹¹ m³/kg/s²

M = 5.98×10²⁴ kg

r = 6.8×10⁵ + 6.357×10⁶ = 7.037×10⁶ m

v = √(6.67×10⁻¹¹ (5.98×10²⁴) / (7.037×10⁶))

v = 7500

The orbital velocity is 7500 m/s.

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Shalnov [3]

B: 11 N.s is the answer

7 0
3 years ago
(PLEASE HELP THANK YOU) A car is going 8 meters per second on an access road into a highway and then accelerates at 1.8 meters p
jonny [76]

Answer:

20.96 m/s

Explanation:

Apply the kinematic equation:

Vf=Vi+at

Vi=8m/s

a=1.8m/s^2

t=7.2s

Putting this all in should give you your answer of 12.96m/s

6 0
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

5 0
4 years ago
Somebody please help me with this question..!
Rudik [331]
8.1) Here, Pressure increases with depth in the ocean at higher rate then that of increase in altitude in atmosphere. So, the rate of change of pressure is different in these journeys.So, your most correct answer would be option D.

8.2) Difference in the shape of lines is due to different density of air & water. So, that physical property and your answer would be option A

So, in Short Answers of your questions are:
8.1) - Option D
8.2) - Option A

Hope this helps!


3 0
3 years ago
In both the camera and the __________, light enters a narrow opening and is projected onto a photosensitive surface. Group of an
Citrus2011 [14]

Answer: The HUMAN EYE

Explanation:

The human eye is made up of different parts which ranges from controlling the amount of light that enters the eye to the focusing of the image that is formed. The camera is a device which is both mechanically and electronically operated which shares a number of similarities with the eye.

In the human eye, the IRIS helps to regulate the amount of rays passing through the pupil to the lens by either contracting or dilating in light or dark environment respectively. While in the camera, the DIAPHRAGM controls the amount of light entering the camera.

The PUPIL serves as the passage for light into the eye while in the camera, the APERTURE does the same.

The photosensitive surface in the eye is the YELLOW SPOT while in the camera, the photosensitive surface is the PHOTOGRAPHIC FILM.

5 0
3 years ago
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