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Assoli18 [71]
3 years ago
8

A satellite is orbiting the earth. If a payload of material is added until it doubles the satellite's mass, the earth's pull of

gravity on this satellite will double but the satellite's orbit will not be affected. A satellite is orbiting the earth. If a payload of material is added until it doubles the satellite's mass, the earth's pull of gravity on this satellite will double but the satellite's orbit will not be affected. True False
Physics
2 answers:
adelina 88 [10]3 years ago
7 0

Answer:

it is True

Explanation:

I hope this helps

Fofino [41]3 years ago
5 0

Answer:

Explanation:

True, provided that the extra mass is also already in the same orbit itself.

If the extra mass is NOT in the same orbit, it's called a collision and the orbits of both will be altered due to the changes in momentum and kinetic energy.

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Which of the following is not an element of installment credit?
olasank [31]

Answer:

I believe it's A. An installment credit has an interest rate, repayment terms and fees, and fixed payment.

3 0
3 years ago
A wall has a negative charge distribution producing a uniformhorizontal electric field. A small plastic ball of mass .01kg carry
Sauron [17]

Answer:

a)  E = -4 10² N / C , b) x = 0.093 m, c)     a = 10.31 m / s², θ=-71.9⁰

Explanation:

For that exercise we use Newton's second Law, in the attached we can see a free body diagram of the ball

X axis

             F_{e} - T_{x} = m a

Axis y

            T_{y} - W = 0

Initially the system is in equilibrium, so zero acceleration

            Fe = T_{x}  

            T_{y} = W

Let us search with trigonometry the components of the tendency

            cos θ = T_{y} / T

            sin θ = T_{x}  / T

           T_{y} = cos θ

           T_{x}  = T sin θ

We replace

            q E = T sin θ

            mg = T cosθ

             

a) the electric force is

                F_{e} = q E

                E = F_{e} / q

                E = -0.032 / 80 10⁻⁶

                E = -4 10² N / C

b) the distance to this point can be found by dividing the two equations

                q E / mg = tan θ

                θ = tan⁻¹ qE / mg

Let's calculate

              θ = tan⁻¹ (80 10⁻⁶ 4 10² / 0.01 9.8)

              θ = tan⁻¹ 0.3265

               θ = 18 ⁰

               sin 18 = x/0.30

               x =0.30 sin 18

               x = 0.093 m

c) The rope is cut, two forces remain acting on the ball, on the x-axis the electric force and on the axis and the force gravitations

X axis

           F_{e} = m aₓ

            aₓ = q E / m

           aₓ = 80 10⁻⁶ 4 10² / 0.01

           aₓ = 3.2 m / s²

Axis y

           W = m a_{y}

           a_{y} = g

           a_{y} = 9.8 m/s²

The total acceleration is can be found using Pythagoras' theorem

             a = √ aₓ² + a_{y}²

             a = √ 3.2² + 9.8²

             a = 10.31 m / s²

The Angle meet him with trigonometry

               tan θ = a_{y} / aₓ

               θ = tan⁻¹ a_{y} / aₓ

               θ = tan⁻¹ (-9.8) / 3.2

               θ = -71.9⁰

Movement is two-dimensional type with acceleration in both axes

8 0
3 years ago
If you push a crate across a factory floor at constant speed in a constant direction, what is the magnitude of the force of fric
poizon [28]

Answer:

The magnitude of the force of friction equals the magnitude of my push

Explanation:

Since the crate moves at a constant speed, there is no net acceleration and thus, my push is balanced by the frictional force on the crate. So, the magnitude of the force of friction equals the magnitude of my push.

Let F = push and f = frictional force and f' = net force

F - f = f' since the crate moves at constant speed, acceleration is zero and thus f' = ma = m (0) = 0

So, F - f = 0

Thus, F = f

So, the magnitude of the force of friction equals the magnitude of my push.

3 0
3 years ago
A grandfather clock has a pendulum that consists of a thin brass disk of radius r = 13.62 cm and mass 1.199 kg that is attached
Feliz [49]

Answer:

Explanation:

Expression for time period of a pendulum is as follows

T = 2\pi\sqrt{\frac{l}{g} }

l is length of pendulum from centre of bob and g is acceleration due to gravity

Given

Time period T = 1.583

g = 9.846

Substituting the values

1.583 = 2\pi\sqrt{\frac{l}{9.846} }

l = \frac{(1.583)^2\times9.846}{4\times(\frac{22}{7})^2 }

l = .6244 m

= 62.44 cm

Length of rod  = length of pendulum - radius of bob

= 62.44 - 13.62

= 48.82 cm

= .488 m

8 0
3 years ago
Which wave has a greater frequency?
DedPeter [7]
The wave diagramed in blue.
8 0
3 years ago
Read 2 more answers
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