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alina1380 [7]
3 years ago
6

A satellite orbits Earth. The only force on the satellite is the gravitational force exerted by Earth. How does the satellite’s

acceleration compared to the gravitational field at the location of the satellite?
Physics
2 answers:
bulgar [2K]3 years ago
4 0

Answer:

here we can say that acceleration of the satellite is same as the gravitational field due to Earth at that location

Explanation:

As we know that gravitational field is defined as the force experienced by the satellite per unit of mass

so we will have

E = \frac{F}{m}

now in order to find the acceleration of the satellite we know by Newton's II law

F = ma

so we will have

a = \frac{F}{m}

so here we can say that acceleration of the satellite is same as the gravitational field due to Earth at that location

PSYCHO15rus [73]3 years ago
4 0

Answer:

The gravitational field and the acceleration point in the same direction.

The magnitudes of the acceleration and the gravitational field strength are equal.

These are the two answers

Explanation:

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"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 coil of area 0.320 m2 is rotating at 100 rev/s with the axis of rotation perpendicular to a 0.430 T magnetic field. If the coi
weqwewe [10]

Answer:

The maximum  emf generated in the coil is 60527.49 V

Explanation:

Given;

area of coil, A = 0.320 m²

angular frequency, f = 100 rev/s

magnetic field, B = 0.43 T

number of turns, N = 700 turns

The maximum emf generated in the coil is calculated as,

E = NBAω

where;

ω is the angular speed = 2πf

E = NBA(2πf)

Substitute in the given values and solve for E

E = 700 x 0.43 x 0.32 x 2π x 100

E = 60527.49 V

Therefore, the maximum  emf generated in the coil is 60527.49 V

8 0
3 years ago
A room is heated as a result of solar radiation coming in through the windows. Is this a heat or work interaction for the room?
taurus [48]

Answer:

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3 0
3 years ago
Newton’s third law of motion says that for every action there is a(n) and opposite reaction.
Alika [10]

Answer:

Formally stated, Newton's third law is: For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object.

Explanation:

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8 0
3 years ago
Can mechanical waves travel in a vacuum?<br> O yes<br> O no
olga2289 [7]

No mechanical waves cannot travel in a vacuum

8 0
2 years ago
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