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bearhunter [10]
2 years ago
12

A rocket that has a mass of 4000 lbm travels at 27,000 ft/sec. What is most nearly its kinetic energy

Physics
1 answer:
solong [7]2 years ago
7 0

Answer:

61440280516.767 J

Explanation:

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Net force = ?
Snezhnost [94]

Answer:

A. the total net force is 14

B. the total net force is 15

Explanation:

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3 years ago
(01.01 LC) Two boxes are 8 cm apart. Which of the following should Janet do to
Trava [24]
Place the boxes 10cm apart, if they are closer that concentrates the mass of the boxes more
4 0
3 years ago
Read 2 more answers
1. Two-point charges, QA = +8 μC and QB = -5 μC, are separated by a distance r = 10 cm. What is the magnitude and direction of t
tiny-mole [99]

Explanation:

Charges,q_1=8\ \mu C=8\times 10^{-6}\ C

q_2=-5\ \mu C=-5\times 10^{-6}\ C

The distance between charges, r = 10 cm = 0.1 m

We need to find the magnitude and direction of the electric force. It is given by :

F=\dfrac{kq_1q_2}{r^2}\\\\F=\dfrac{9\times 10^9\times 8\times 10^{-6}\times 5\times 10^{-6}}{(0.1)^2}\\\\F=36\ N

So, the required force between charges is 36 N and it is towards positive charge i.e. +8 μC.

6 0
3 years ago
Describe in a couple sentences how the index of refraction is related to the speed of light?
lesantik [10]

Answer: n = c / v" "c" is the speed of light in a vacuum, "v" is the speed of light in that substance and "n" is the index of refraction. According to the formula, the index of refraction is the relation between the speed of light in a vacuum and the speed of light in a substance.

Explanation: the relation is the vacuum and the speed of light in a substance.

3 0
3 years ago
A solid conducting sphere with radius R carries a positive total charge Q. The sphere is surrounded by an insulating shell with
Illusion [34]

Answer:

Explanation:

Volume of the insulating shell is,

V_{shell}=\frac{4}{3}\pi(R^3_2-R^3_1)

Charge density of the shell is,

\rho=\frac{Q_{shell}}{\frac{4}{3}\pi(R^3_2-R^3_1)}

Here, R_2 =2R, R_1 =R \,and\, Q_{shell} =-Q

\rho=\frac{Q_{shell}}{\frac{4}{3}\pi((2R)^3-R^3)}=\frac{-3Q}{28\piR^3}

B)

The electric field is E=\frac{1}{4\pi\epsilon_0}\frac{Qr}{R^3}

For 0 <r<R the electric field is zero, because the electric field inside the conductor is zero.

C)

For R <r <2R According to gauss law

E(4\pi r^2)=\frac{Q}{\epsilon_0}+\frac{4\pi\rho}{3\epsilon_0}(r^3-R^3)

substitute \rho=\frac{-3Q}{28\piR^3}

E=\frac{2}{7\pi\epsilon_0}\frac{Q}{r^2}-\frac{Qr}{28\piR^3}

D)

The net charge enclosed for each r in this range is positive and the electric field is outward

E)

For r>2R

Charge enclosed is zero, so electric field is zero

8 0
3 years ago
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