Ask question

Ask question

All categories

2 years ago

6

Two satellites revolve around the Earth. Satellite A has mass m and has an orbit of radius r. Satellite B has mass 6m and an orb

it of unknown radius rb. The forces of gravitational attraction between each satellite and the Earth is the same. Find rb.

1 answer:

melomori [17]2 years ago

8 0

Answer:

aaaaa

Explanation:

M = Mass of the Earth

m = Mass of satellite

r = Radius of satellite

G = Gravitational constant

$F=G\frac{Mm}{r^2}$

$F=G\frac{M6m}{r_b^2}$

$G\frac{Mm}{r^2}=G\frac{M6m}{r_b^2}\\\Rightarrow \frac{1}{r^2}=\frac{6}{r_b^2}\\\Rightarrow \frac{r_b^2}{r^2}=6\\\Rightarrow \frac{r_b}{r}=\sqrt{6}\\\Rightarrow r_b=2.44948r$

$r_b=2.44948r$

You might be interested in

A 4500-kg spaceship is in a circular orbit 190 km above the surface of Earth. It needs to be moved into a higher circular orbit

Maslowich

Answer:

Explanation:

Total energy of a satellite in an orbit , h height away

= - GMm /2 ( R + h )

When h = 380 km

Total energy of a satellite = $\frac{6.67\times10^{-11}\times5.97\times10^{24}\times 4500}{2\times(6378+380)\times10^3}$

= - 13.25 x 10¹⁰ J

When h = 190 km

Total energy of a satellite =

$\frac{6.67\times10^{-11}\times5.97\times10^{24}\times 4500}{2\times(6378+190)\times10^3}$

= - 13.63 x 10¹⁰ J

Diff

= 38 x 10⁸ J Energy will be required.

8 0

2 years ago

A package is dropped from a helicopter that is descending steadily at a speed v0. After t seconds have elapsed, consider the fol

qaws [65]

Answer:

Part a)

$v = \sqrt{v_o^2 + g^2t^2}$

Part b)

$d = \frac{1}{2}gt^2$

Part c)

$v_f = v_o - gt$

Part d)

$d = \frac{1}{2}gt^2$

Explanation:

Part a)

As we know that speed of package is same as that of helicopter in horizontal direction

So after time "t" the velocity in x direction will remain constant while in Y direction it will go free fall

So we have

$v_y = -gt$

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{v_o^2 + g^2t^2}$

Part b)

Distance from helicopter is same as the distance of free fall

so we will have

$d = \frac{1}{2}gt^2$

Part c)

If helicopter is rising upwards with uniform speed

then final speed of the package after time t is given as

$v_f = v_i + at$

$v_f = v_o - gt$

Part d)

distance from helicopter

$d = \frac{1}{2}gt^2$

8 0

2 years ago

Read 2 more answers

A blue puck has a velocity of 0i – 3j m/s and a mass of 4 kg. A gold puck has a velocity of 12i – 5j m/s and a mass of 6 kg. Wha

Mnenie [13.5K]

By definition, the kinetic energy is given by:
K = (1/2) * m * v ^ 2
where
m = mass
v = speed
We must then find the speed of both objects:
blue puck
v = root ((0) ^ 2 + (- 3) ^ 2) = 3
gold puck
v = root ((12) ^ 2 + (- 5) ^ 2) = 13
Then, the kinetic energy of the system will be:
K = (1/2) * m1 * v1 ^ 2 + (1/2) * m2 * v2 ^ 2
K = (1/2) * (4) * (3 ^ 2) + (1/2) * (6) * (13 ^ 2)
K = <span> 525</span> J
answer
The kinetic energy of the system is<span> <span>525 </span></span>J

6 0

1 year ago

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

Tresset [83]

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is $Q =2.094 C$

Explanation:

From the question we are told that

The diameter of the wire is $d = 0.205cm = 0.00205 \ m$

The radius of the wire is $r = \frac{0.00205}{2} = 0.001025 \ m$

The resistivity of aluminum is $2.75*10^{-8} \ ohm-meters.$

The electric field change is mathematically defied as

$E (t) = 0.0004t^2 - 0.0001 +0.0004$

Generally the charge is mathematically represented as

$Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt$

Where A is the area which is mathematically represented as

$A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2$

So

$\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega$

Therefore

$Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

substituting values

$Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

From the question we are told that t = 5 sec

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

$Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

$Q =2.094 C$

5 0

2 years ago

A microprocessor scans the status of an output I/O device every 20 ms. This is accomplished by means of a timer alerting the pro

Lerok [7]

Answer:

0.0000045 s

Explanation:

f = Frequency = 8 MHz

Clock cycle is given by

$\dfrac{1}{f}=\dfrac{1}{8\times 10^6}=1.25\times 10^{-7}\ s$

Time taken for 12 clock cycles

$12\times 1.25\times 10^{-7}=0.0000015\ s$

Time taken per instruction is 0.0000015 s

In reading and displaying information it requires 3 processes

1 for reading, 1 for searching and 1 for displaying.

$3\times 0.0000015=0.0000045\ s$

Time taken is 0.0000045 s

6 0

1 year ago