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inn [45]
3 years ago
12

Disadvantages of friction

Physics
1 answer:
MrRa [10]3 years ago
7 0
Heat, kinetic energy loss.
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A geosynchronous satellite moves in a circular orbit around the Earth and completes one circle in the same time T during which t
andreyandreev [35.5K]

Answer:

Explanation:

The time period of geosynchronous satellite must be equal to T .

The radius of its orbit will be (  R+ h )

orbital velocity  V₀ =  \sqrt{\frac{GM}{( R+h)} }

Time period T = 2π( R + h ) / V₀

= 2π( R + h ) x \sqrt{\frac{( R+h)}{GM } }

\frac{T^\frac{2}{3}(GM)^\frac{1}{3}  }{(2\pi )^\frac{2}{3} } = R +h

h = \frac{T^\frac{2}{3}(GM)^\frac{1}{3}  }{(2\pi )^\frac{2}{3} } - R.

7 0
4 years ago
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Differences between weightlessness in space and weightlessness in earth​
zhuklara [117]

Answer:

it depends on a person's own weight

6 0
3 years ago
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A cable is wrapped several times around a uniform solid cylinder with diameter 0.12 m and mass
Naya [18.7K]

Answer:

Which of the following was not a reaction to John Brown's execution because of his failed raid on Harper's Ferry ???????

Explanation:

8 0
3 years ago
"A short-wave radio antenna is supported by two guy wires, 150 ft and 170 ft long. Each wire is attached to the top of the anten
Nonamiya [84]

Answer:

The anchor points are 78.37 ft and 111.99 ft

Explanation:

If you look at the attached (Fig 1) you will find that the union of antenna and its guy wires forms two right triangles. To solve problems that involve this kind of triangles, you can apply trigonometric functions (sine, cosine, etc) and Pythagoras Theorem. Trigonometric functions states the relation between angles, sides and hypotenuse of a right triangle. If you look Fig2, considering α angle, "b" is the opposite side, "a" the adjacent side and "c" the hypotenuse. Then

a) Sine (α) = b/c it means opposite side/hypotenuse

b) Cosine (α)= a/c, it means adjacent side/hypotenuse

c) Tangent (α) = b/a opposite side/adjacent side.

Pythagoras theorem states that if you called "a" and "b", the sides of the right triangle, and "c" the hypotenuse, then:

                                      a² + b² = c²    

As the problem states the lengths 150 ft and 170 ft represents the value of the hypotenuse of each triangle and 65° is one of the angles of the triangle with 150 ft hypotenuse. So you can solve this using sin (65°) to find the height of the antenna (h) and then the two distances (x and y,).

Sine (65°) = h/ 150 ft ⇒ Sine (65°) x 150ft = h ⇒ h = 127.9 ft.

To find x : Cosine (65°) = x/ 150 ft ⇒ Cosine (65°) x 150 ft = x

⇒ x = 78.37 ft.

And finally, to find y we can apply Pythagoras theorem

(170 ft)² = (127.9 ft)² + y² ⇒ y² = (170 ft)² - (127.9 ft)² ⇒ y = 111.99 ft

Summarizing, the anchor points are 78.37 ft and 111.99 ft

4 0
3 years ago
A particle leaves the origin with a speed of 3.6 106 m/s at 34 degrees to the positive x axis. It moves in a uniform electric fi
Andrej [43]

Answer:

E = -4556.18 N/m

Explanation:

Given data

u = 3.6×10^6 m/sec

angle = 34°

distance x = 1.5 cm = 1.5×10^-2 m  (This data has been assumed not given in

Question)

from the projectile motion the horizontal distance traveled by electron is

x = u×cosA×t

⇒t = x/(u×cos A)

We also know that force in an electric field is given as

F = qE

q= charge , E= strength of electric field

By newton 2nd law of motion

ma = qE

⇒a = qE/m

Also, y = u×sinA×t - 0.5×a×t^2

⇒y = u×sinA×t - 0.5×(qE/m)×t^2

if y = 0 then

⇒t = 2mu×sinA/(qE) = x/(u×cosA)

Also, E = 2mu^2×sinA×cosA/(x×q)

Now plugging the values we get

E = 2×9.1×10^{-31}×3.6^2×10^{12}×(sin34°)×(cos34°)/(1.5×10^{-2}×(-1.6)×10^{-19})

E = -4556.18 N/m

4 0
3 years ago
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