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3 years ago

Help me please please

Physics

1 answer:

shutvik [7]3 years ago

4 0

I’m sorry I wish I knew the answer but I don’t I need this for points I’m so sorry

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Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o

Juliette [100K]

(a) 0.473

The potential energy of a satellite orbiting around Earth is given by

$U=-\frac{GMm}{R+h}$

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

R is the Earth's radius

h is the altitude of the satellite above the Earth's surface

So the potential energy of satellite A is

$U_A=-\frac{GMm}{R+h_A}$

while potential energy of satellite B is

$U_B=-\frac{GMm}{R+h_B}$

Therefore the ratio of the potential energy of satellite B to that of satellite A is

$\frac{U_B}{U_A}=\frac{R+h_A}{R+h_B}$

and using

hA = 5920 km

hB = 19600 km

R = 6370 km

we find

$\frac{U_B}{U_A}=\frac{6370+5920}{6370+19600}=0.473$

(b) 0.473

The kinetic energy of a satellite orbiting around Earth instead is given by

$K=\frac{GMm}{2(R+h)}$

So the kinetic energy of satellite A is

$K_A=\frac{GMm}{2(R+h_A)}$

while kinetic energy of satellite B is

$K_B=\frac{GMm}{2(R+h_B)}$

Therefore the ratio of the kinetic energy of satellite B to that of satellite A is

$\frac{K_B}{K_A}=\frac{R+h_A}{R+h_B}$

which is identical to before, so it gives

$\frac{K_B}{K_A}=\frac{6370+5920}{6370+19600}=0.473$

(c) Satellite B

The total energy of a satellite in orbit is given by

$E=U+K = -\frac{GMm}{R+h}+\frac{GMm}{2(R+h)}=-\frac{GMm}{2(R+h)}$

We see that the total energy is:

1) negative (because the satellite is on a bound orbit)

2) inversely proportional to the distance of the satellite from the Earth's center, R+h

So the magnitude of the fraction in the equation is larger for the satellite which is closer to the Earth's surface (satellite A), but since the energy is negative, this means that the total energy of this satellite is smaller than that of satellite B. So, satellite B has a greater total energy.

(d) $1.03\cdot 10^7 J$

We have to calculate the total energy of each satellite.

Given:

$G=6.67\cdot 10^{-11}$

$M=5.98\cdot 10^{24} kg$

m = 12.0 kg

$R+h_A = 6370 km+5920 km=12290 km = 12.3 \cdot 10^6 m$

$R+h_B = 6370 km+19600 km=25970 km = 26.0 \cdot 10^6 m$

We find:

$E_A = - \frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})(12.0)}{2(12.3\cdot 10^6)}=-1.95\cdot 10^{7} J$

$E_B = - \frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})(12.0)}{2(26.0\cdot 10^6)}=-9.2\cdot 10^{6} J$

So the difference in total energy is

$E_B-E_A = -9.2\cdot 10^6 - (-1.95\cdot 10^7) =1.03\cdot 10^7 J$

6 0

3 years ago

Suppose a ball has a potential energy of 5 J when you drop it. What would be its kinetic energy just as it hit the ground? (Igno

mihalych1998 [28]

If you simply drop it ... so that it's not moving until you let it go ... then
the kinetic energy it has when it hits the ground is exactly the potential
energy it had while you were holding it.

5 J potential energy in your hand ===> 5 J kinetic energy when it lands

8 0

3 years ago

g A small car travels up the hill with a speed of v = 0.2 s (m/s) where s is the distance measured from point A in meters. Deter

jasenka [17]

Answer:

The magnitude of the acceleration of the car when $s = 50\,m$ is $2\,\frac{m}{s^{2}}$ .

Explanation:

The acceleration can be obtained by using the following differential equation:

$a = v \cdot \frac{dv}{ds}$

Where $a$ , $v$ and $s$ are the acceleration, speed and distance masured in meters per square second, meters per second and meters, respectively.

Given that $v = 0.2\cdot s$ , its first derivative is:

$\frac{dv}{ds} = 0.2$

The following expression is obtained by replacing each term:

$a = 0.2\cdot 0.2\cdot s$

$a = 0.04\cdot s$

The magnitude of the acceleration of the car when $s = 50\,m$ is:

$a = 0.04\cdot (50\,m)$

$a = 2\,\frac{m}{s^{2}}$

5 0

3 years ago

Two horizontal pipes have the same diameter, but pipe B is twice as long as pipe A. Water undergoes viscous flow in both pipes,

zheka24 [161]

Answer:

c) 2Q

Explanation:

From the given information:

The pressure inside a pipe can be expressed by using the formula:

$\Delta P = \dfrac{128 \mu L Q}{\pi D^4}$

Since the diameter in both pipes is the same, we can say:

$D = D_A = D_B$

where;

length of the first pipe A $L_A = L$ and the length of the second pipe B $L_B = 2L$

Since the difference in pressure is equivalent in both pipes:

Then:

$\dfrac{128 \mu L_1Q_1}{\pi D_1^4} = \dfrac{128 \mu L_2Q_2}{\pi D_2^4}$

$\dfrac{ L_1Q_1}{D_1^4} = \dfrac{ L_2Q_2}{D_2^4}$

$\dfrac{ LQ_1}{D^4} = \dfrac{ 2LQ}{D^4}$

$\mathbf{Q_1 = 2Q}$

6 0

3 years ago

What causes a bolt of lightning

Sonbull [250]

Answer:

The friction between two or more clouds cause them to heat up then it turns to static electricty and then hits the ground i hope you are happy anfd i hope this is correct :)

4 0

3 years ago

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