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olga2289 [7]
3 years ago
12

A planet of mass 7.00 1025 kg is in a circular orbit of radius 6.00 1011 m around a star. The star exerts a force on the planet

of constant magnitude 6.51 1022 N. The speed of the planet is 2.36 104 m/s.
(a) In half a "year" the planet goes half way around the star. What is the distance that the planet travels along the semicircle?
distance = m
(b) During this half "year", how much work is done on the planet by the gravitational force acting on the planet?
work = J
(c) What is the change in kinetic energy of the planet?
?K = J
(d) What is the magnitude of the change of momentum of the planet?
Physics
1 answer:
BARSIC [14]3 years ago
7 0

Answer:

A) 1.88 * 10^17 m

B) 1.22 * 10^34 J

C) 1.95 * 10^34 J

Explanation:

Parameters given:

Mass of planet = 7.00 * 10^25 kg

Radius of orbit = 6.00 * 10^11 m

Force exerted on planet = 6.51 * 10^22 N

Velocity of planet = 2.36 * 10^4 m/s

A) The distance traveled by the planet is half of the circumference of the orbit (which is circular).

The circumference of the orbit is

C = 2 * pi * R

R = radius of orbit

C = 2 * 3.142 * 6.0 * 10¹¹

C = 3.77 * 10¹² m

Hence, distance traveled will be:

D = 0.5 * 3.77 * 10¹²

D = 1.88 * 10 ¹² m/s

B) Work done is given as:

W = F * D

W = 652 * 10²² * 1.88 * 10¹¹

W = 1.22 * 10³⁴ J

C) Change in Kinetic energy is given as:

K. E. = 0.5 * m * v²

K. E. = 0.5 * 7 * 10^25 * (2.36 * 10^4)²

K. E. = 1.95 * 10³⁴ J

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Answer:

thanks for the points liar

Explanation:

4 0
3 years ago
A long, fine wire is wound into a coil with inductance 5 mH. The coil is connected across the terminals of a battery, and the cu
Effectus [21]

Answer:

Compared with the current in the first coil, the current in the second coil is unchanged.

Explanation:

All coils, inductors, chokes and transformers create a magnetic field around themselves consist of an Inductance in series with a Resistance forming an LR Series Circuit.

The steady state of current in the LR circuit is:

I= V/R (1 - e^-Rt/L)

Where I= current

R= Resistance

V= Voltage

Where R/L is the time constant.

For a conducting wire, it has a very small resistance. The time constant will be in microseconds. The current will be in a steady state after few second. The current is independent on the inductance and dependent on the resistance. The length of wire and the resistance here are the same. Therefore, the current remains unchanged.

5 0
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German physicist Werner Heisenberg related the uncertainty of an object's position (Δx) to the uncertainty in its velocity (Δ???
fredd [130]

Answer:

\Delta x = 5.47 \times 10^{-9} m

Explanation:

As we know by the principle of uncertainty that the product of uncertainty in position and uncertainty in momentum is given as

\Delta x \times \Delta P = \frac{h}{4\pi}

so here we know that

\Delta v = 0.01 \times 10^6 m/s

m = 9.11 \times 10^{-31} kg

so we have

\Delta x \times (9.11 \times 10^{-31})(0.01 \times 10^6) = \frac{6.26 \times 10^{-34}}{4\pi}

\Delta x = 5.47 \times 10^{-9} m

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What is the direction of the field halfway between two horizontal parallel wires if the top wire has a current of 4 A to the lef
Alex777 [14]

Answer:

A) Out of the page.

Explanation:

Right-hand rule points the direction of the magnetic field at any point.

<u>Top wire</u>: Current is to the left. Point your thumb to the left and curl your other fingers around the wire. The tips of the four fingers points the direction of the field at that point. In this case, out of the page.

<u>Bottom wire</u>: Current is to the right. Point your thumb to the right and curl your other fingers around the wire. The tips of the four finger points out of the page again.

So, the total field produced by both wires is directed out of the page.

Another method to figure out the direction is the mathematical method.

Use the B-field formula:

d\vec{B} = \frac{\mu_0}{4\pi}\frac{Id\vec{l}\times \^r}{r^2}

The cross product between the direction of the current and the target position gives the direction of the B-field. If the left is -x direction and downwards is the -y direction, then

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