Ask question

Ask question

All categories

Anuta_ua [19.1K]

3 years ago

12

Two satellites are in circular orbits around Earth. Satellite A has speed vA. Satellite B has an orbital radius nine times that

of satellite A. What is the speed of Satellite B?
A. vA.9
B. vA/3
C. 3vA
D. 9vA

1 answer:

zaharov [31]3 years ago

5 0

Answer:

option B

Explanation:

given,

Satellite B has an orbital radius nine times that of satellite A.

R' = 9 R

now, orbital velocity of the satellite A

$v_a =\sqrt{\dfrac{GM}{R}}$ ........(1)

now, orbital velocity of satellite B

$v_b=\sqrt{\dfrac{GM}{R'}}$

$v_b=\sqrt{\dfrac{GM}{9R}}$

$v_b=\dfrac{1}{3}\sqrt{\dfrac{GM}{R}}$

from equation 1

$v_b=\dfrac{v_a}{3}$

hence, the correct answer is option B

You might be interested in

A spacecraft at rest has moment of inertia of 100 kg-m^2 about an axis of interest. If a 1 newton thruster with a 1 meter moment

snow_lady [41]

Answer:

18 radians

Explanation:

The computation is shown below:

As we know that

Torque = Force × Moment arm

= 1N × 1M

= 1N-M

Torque = $I\alpha$

$\alpha = \frac{torque}{I}\\\\= \frac{1}{100}\\\\= 0.01 rad/s^2$

Now

$\theta = w_ot + \frac{1}{2} \alpha t^2\\\\w_o = 0\\\\\theta = 0 \times 60 + \frac{1}{2} \times 0.01 \times 60^2\\\\= 18\ radians$

Here t = 1 minutes = 60 seconds

3 0

3 years ago

A 0.106-A current is charging a capacitor that has square plates 6.00 cm on each side. The plate separation is 4.00 mm. (a) Find

FrozenT [24]

Answer:

The time rate of change of flux is $1.34 \times 10^{10}$ $\frac{V}{s}$

Explanation:

Given :

Current $I = 0.106$ A

Area of plate $A = 36 \times 10^{-4}$ $m^{2}$

Plate separation $d = 4 \times 10^{-3}$ m

(A)

First find the capacitance of capacitor,

$C = \frac{\epsilon _{o} A }{d}$

Where $\epsilon _{o} = 8.85 \times 10^{-12}$

$C = \frac{8.85 \times 10^{-12 } \times 36 \times 10^{-4} }{4 \times 10^{-3} }$

$C = 7.9 \times 10^{-12}$ F

But $C = \frac{Q}{V}$

Where $Q = It$

$C = \frac{It}{V}$

$V = \frac{It}{C}$

Now differentiate above equation wrt. time,

$\frac{dV}{dt} = \frac{I}{C}$

$= \frac{0.106}{7.9 \times 10^{-12} }$

$= 1.34 \times 10^{10}$ $\frac{V}{s}$

Therefore, the time rate of change of flux is $1.34 \times 10^{10}$ $\frac{V}{s}$

8 0

3 years ago

To counter the effects of centrifugal force and reduce vehicle traction it is important to to counter the effects of centrifugal

tatiyna

Answer: Add an incline or grade to the road track.

Explanation:
Refer to the figure shown below.

When a vehicle travels on a level road in a circular path of radius r, a centrifugal force, F, tends to make the vehicle skid away from the center of the circular path.
The magnitude of the force is
F = mv²/r
where
m = mass of the vehicle
v = linear (tangential) velocity to the circular path.

The force that resists the skidding of the vehicle is provided by tractional frictional force at the tires, of magnitude
μN = μW = μmg
where
μ = dynamic coefficient of friction.

At high speeds, the frictional force will not overcome the centrifugal force, and the vehicle will skid.

When an incline of θ degrees is added to the road track, the frictional force is augmented by the component of the weight of the vehicle along the incline.
Therefore the force that opposes the centrifugal force becomes
μN + Wsinθ = W(sinθ + μ cosθ).

5 0

3 years ago

Consider a space shuttle which has a mass of about 1.0 x 105 kg and circles the Earth at an altitude of about 200.0 km. Calculat

svetlana [45]

Answer:

1.6675×10^-16N

Explanation:

The force of gravity that the space shuttle experiences is expressed as;

g = GM/r²

G is the gravitational constant

M is the mass = 1.0 x 10^5 kg

r is the altitude = 200km = 200,000m

Substitute into the formula

g = 6.67×10^-11 × 1.0×10^5/(2×10^5)²

g = 6.67×10^-6/4×10^10

g = 1.6675×10^{-6-10}

g = 1.6675×10^-16N

Hence the force of gravity experienced by the shuttle is 1.6675×10^-16N

6 0

2 years ago

A new restaurant is interested in determining the best time-temperature combination for roasting a five-pound cut of lamb. The t

Leto [7]

Answer:

C

Explanation:

(c) The two cuts that are being roasted for each time-temperature combination are an example of replication.

In the question it is given that From 10 identical cuts of lamb, 2 are randomly selected to roast using each of the time-temperature combinations in the same oven. Here it is an act of copying the exact sahpe size of the lamb in all cuts, which is nothing but replication. Moreover, this replication can help in proper comparision.

7 0

3 years ago