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3 years ago

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A rocket fires its engines to launch straight up from rest with an upward acceleration of 5 m/s2 for 10 seconds. After this time

, the engine shuts off and the rocket freely falls straight down back to Earth's surface.
Please help! This is for AP Physics.

2 answers:

IrinaK [193]3 years ago

7 0

Answer:

Hi! These graphs I'm providing below are the answer to this problem, as well as the free response sections. I'll attach them below for you.

Hope this helps! :)

Shtirlitz [24]3 years ago

4 0

EDIT: I didn't realize that I had solved the problem wrong, so I deleted my answer. The other answer to this question is really good so look at that one, not this. I'm so sorry!

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A 1 036-kg satellite orbits the Earth at a constant altitude of 98-km. (a) How much energy must be added to the system to move t

Veronika [31]

Answer:

a) The Energy added should be 484.438 MJ

b) The Kinetic Energy change is -484.438 MJ

c) The Potential Energy change is 968.907 MJ

Explanation:

Let 'm' be the mass of the satellite , 'M'(6× $10^{24}$ be the mass of earth , 'R'(6400 Km) be the radius of the earth , 'h' be the altitude of the satellite and 'G' (6.67× $10^{-11}$ N/m) be the universal constant of gravitation.

We know that the orbital velocity(v) for a satellite -

v= $\sqrt{\frac{Gmm}{R+h} }$ [(R+h) is the distance of the satellite from the center of the earth ]

Total Energy(E) = Kinetic Energy(KE) + Potential Energy(PE)

For initial conditions ,

h = $h_{i}$ = 98 km = 98000 m

∴Initial Energy ( $E_{i}$ ) = $\frac{1}{2}$ m $v^{2}$ + $\frac{-GMm}{(R+h_{i} )}$

Substituting v= $\sqrt{\frac{GMm}{R+h_{i} } }$ in the above equation and simplifying we get,

$E_{i}$ = $\frac{-GMm}{2(R+h_{i}) }$

Similarly for final condition,

h= $h_{f}$ = 198km = 198000 m

∴Final Energy( $E_{f}$ ) = $\frac{-GMm}{2(R+h_{f}) }$

a) The energy that should be added should be the difference in the energy of initial and final states -

∴ ΔE = $E_{f}$ - $E_{i}$

= $\frac{GMm}{2}$ ( $\frac{1}{R+h_{i} }$ - $\frac{1}{R+h_{f} }$ )

Substituting ,

M = 6 × $10^{24}$ kg

m = 1036 kg

G = 6.67 × $10^{-11}$

R = 6400000 m

$h_{i}$ = 98000 m

$h_{f}$ = 198000 m

We get ,

ΔE = 484.438 MJ

b) Change in Kinetic Energy (ΔKE) = $\frac{1}{2}$ m[ $v_{f} ^{2}$ - $v_{i} ^{2}$ ]

= $\frac{GMm}{2}$ [ $\frac{1} {R+h_{f} }$ - $\frac{1} {R+h_{i} }$ ]

= -ΔE

= - 484.438 MJ

c) Change in Potential Energy (ΔPE) = GMm[ $\frac{1}{R+h_{i} }$ - $\frac{1}{R+h_{f} }$ ]

= 2ΔE

= 968.907 MJ

3 0

3 years ago

A magnetic field is directed perpendicular to the plane of a 0.15-m × 0.30-m rectangular coil consisting of 240 loops of wire. T

ValentinkaMS [17]

Answer:

7.344 s

Explanation:

A = 0.15 x 0.3 m^2 = 0.045 m^2

N = 240

e = - 2.5 v

B1 = 0.1 T

B2 = 1.8 T

ΔB = B2 - B1 = 1.8 - 0.1 = 1.7 T

Δt = ?

e = - dФ/dt

e = - N x A x ΔB/Δt

- 2.5 = - 240 x 0.045 x 1.7 / Δt

2.5 = 18.36 / Δt

Δt = 7.344 s

6 0

3 years ago

Drag the right word to the word that means the opposite

Ainat [17]

What do you mean? Is there a picture ?

7 0

3 years ago

How long does it take to travel a distance of 672km at a speed of 95km/h?

Brilliant_brown [7]

Answer:

7.07 hours

Explanation:

divide the distance by the speed

so in this case, divide 672 by 95

6 0

3 years ago

Can someone help me with this please?

Slav-nsk [51]

Answer:

C2, C1, C4, C5 and C6 are in parallel. Therefore, we use the formula Cp = C1 + C2 + ....

Cp = C2 + C1 + C4 + C5 + C6 = ( 7 * 10 ^-3) + (18 * 10^-6) + (0.8F) + (200 * 10^-3 F) + (750 * 10^-6) = 1.008F

Now, Cp will become one capacitor and it will be aligned with C3, therefore it will now become a circuit in series.

We use the formula: 1/Cs = 1/C1 + 1/C2 + .... + ....1/Cn

Thus,

1/Cs = 1/C3 + 1/Cp

1/Cs = 1/(14 * 10^-3 F) + 1/(1.008F)

Cs = 1.4 * 10 ^-2 or if we do not round too much it will give exactly 0.0138 F

So the answer should be a)

8 0

3 years ago