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harina [27]
2 years ago
12

According to the law of universal gravitation, any two objects are attracted to each other. The strength of the gravitational fo

rce depends on the masses of the objects and their distance from each other.
Many stars have planets around them. If there were no gravity attracting a planet to its star, the planet's motion would carry it away from the star. However, when this motion is balanced by the gravitational attraction to the star, the planet orbits the star.

Two solar systems each have a planet the same distance from the star. The planets have the same mass, but Planet A orbits a more massive star than Planet B.


Which of the following statements is true about the planets?
A.
Planet B is more attracted to its star than Planet A.
B.
Planet B will keep orbiting its star longer than Planet A.
C.
Planet A orbits its star faster than Planet B.
D.
Planet A has a longer year than Planet B.

Physics
1 answer:
hodyreva [135]2 years ago
7 0

Answer:

C. Plant A orbits its star faster than Plant B

Explanation:

Did it on study island

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PLEASE HELP!!!! The displacement vectors A and B, when added together, give the resultant vector R, so that R = A + B. Use the d
GuDViN [60]

Answer:

Ax = 0

Ay = 6 m

Bx = 8 cos phi = cos 34 = 6.63 m

By = 8 sin phi = 8 sin (-34) = -4.47 m

Rx = Ax + Bx = 0 + 6.63 = 6.63 m

Ry = Ay + By = 6 - 4.47 = 1.53 m

R = (6.63^2 + 1.53^2)^1/2 = 6.80 m

tan theta = Ry / Rx = 1.53 / 6.8 = ,225

theta = 12.7 deg

7 0
3 years ago
If an electron moves in a circle of radius 21 cm perpendicular to a B field of 0.4 T, what are the speed of the electron and the
kodGreya [7K]

Answer:

a)

v = 4.048 *10^6 m/s

b)  

Angular frequency =  1.92 * 10^7

Explanation:

As we know

v =  \frac{qBr}{m}

q is the charge on the electron = 3.2 * 10^{-19} C

B is the magnetic field in Tesla = 0.4 T

r is the radius of the circle = 0.21 m

mass of the electrons = 6.64 * 10^{-27} Kg

a)

Substituting the given values in above equation, we get -

v = \frac{3.2 * 10^{-19}*0.4*0.21}{6.64 * 10^{-27}} \\v = 4.048 *10^6m/s

b)  

Angular frequency =

\frac{4.048 * 10^6 }{0.21} \\1.92 * 10^7

8 0
2 years ago
A boulder with a mass of 50 kg, sits on a cliff that is 50 m high. The boulder is pushed off of the cliff. How much kinetic ener
rosijanka [135]
50 +50 =100 Since it’s sitting on a 50m cliff that’s high with a mass of 50 kg it would be adding because once it goes down it’s adding speed
8 0
3 years ago
A linear network has a current input 7.5 cos(10t + 30°) A and a voltage output 120 cos(10t + 75°) V. Determine the associated im
Leona [35]

Answer:

16∠45° Ω

Explanation:

Applying,

Z = V/I................... Equation 1

Where Z = Impedance, V = Voltage output, I = current input.

Given: V = 120cos(10t+75°), = 120∠75°,  I = 7.5cos(10t+30) = 7.5∠30°

Substitute these values into equation 1

Z = 120cos(10t+75°)/7.5cos(10t+30)

Z = 120∠75°/ 7.5∠30°

Z = 16∠(75°-30)

Z = 16∠45° Ω

Hence the impedance of the linear network is 16∠45° Ω

8 0
3 years ago
A jet airliner moving initially at 548 mph
sasho [114]

Let's choose the "east" direction as positive x-direction. The new velocity of the jet is the vector sum of two velocities: the initial velocity of the jet, which is

v_1 =548 mph along the x-direction

v_2 = 343 mph in a direction 67^{\circ} north of east.

To find the resultant, we must resolve both vectors on the x- and y- axis:

v_{1x}= 548 mph

v_{1y}=0

v_{2x} = (343 mph)( cos 67^{\circ})=134.0 mph

v_{2y} = (343 mph)( sin 67^{\circ})=315.7 mph

So, the components of the resultant velocity in the two directions are

v_{x}=548 mph+134 mph=682 mph

v_{y}=0 mph+315.7 mph=315.7 mph

So the new speed of the aircraft is:

v=\sqrt{v_x^2+v_y^2}=\sqrt{(682 mph)^2+(315.7 mph)^2}=751.5 mph

3 0
3 years ago
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