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Sergeeva-Olga [200]
3 years ago
15

Three different planet-star systems, which are far apart from one another, are shown above. The masses of the planets are much l

ess than the masses of the stars.
In System A , Planet A of mass Mp orbits Star A of mass Ms in a circular orbit of radius R .

In System B , Planet B of mass 4Mp orbits Star B of mass Ms in a circular orbit of radius R .

In System C , Planet C of mass Mp orbits Star C of mass 4Ms in a circular orbit of radius R .
(a) The gravitational force exerted on Planet A by Star A has a magnitude of F0 . Determine the magnitudes of the gravitational forces exerted in System B and System C .

___ Magnitude of gravitational force exerted on Planet B by Star B

___ Magnitude of gravitational force exerted on Planet C by Star C
(b) How do the tangential speeds of planets B and C compare to that of Planet A ? In a clear, coherent paragraph-length response that may also contain equations and/or drawings, provide claims about

why the tangential speed of Planet B is either greater than, less than, or the same as that of Planet A , and
why the tangential speed of Planet C is either greater than, less than, or the same as that of Planet A .
Physics
1 answer:
alex41 [277]3 years ago
3 0

a) 4F0

b) Speed of planet B is the same as speed of planet A

Speed of planet C is twice the speed of planet A

Explanation:

a)

The magnitude of the gravitational force between two objects is given by the formula

F=G\frac{m_1 m_2}{r^2}

where

G is the gravitational constant

m1, m2 are the masses of the 2 objects

r is the separation between the objects

For the system planet A - Star A, we have:

m_1=M_p\\m_2 = M_s\\r=R

So the force is

F_A=G\frac{M_p M_s}{R^2}=F_0

For the system planet B - Star B, we have:

m_1 = 4 M_p\\m_2 = M_s\\r=R

So the force is

F=G\frac{4M_p M_s}{R^2}=4F_0

So, the magnitude of the gravitational force exerted on planet B by star B is 4F0.

For the system planet C - Star C, we have:

m_1 = M_p\\m_2 = 4M_s\\r=R

So the force is

F=G\frac{M_p (4M_s)}{R^2}=4F_0

So, the magnitude of the gravitational force exerted on planet C by star C is 4F0.

b)

The gravitational force on the planet orbiting around the star is equal to the centripetal force, therefore we can write:

G\frac{mM}{r^2}=m\frac{v^2}{r}

where

m is the mass of the planet

M is the mass of the star

v is the tangential speed

We can re-arrange the equation solving for v, and we find an expression for the speed:

v=\sqrt{\frac{GM}{r}}

For System A,

M=M_s\\r=R

So the tangential speed is

v_A=\sqrt{\frac{GM_s}{R}}

For system B,

M=M_s\\r=R

So the tangential speed is

v_B=\sqrt{\frac{GM_s}{R}}=v_A

So, the speed of planet B is the same as planet A.

For system C,

M=4M_s\\r=R

So the tangential speed is

v_C=\sqrt{\frac{G(4M_s)}{R}}=2(\sqrt{\frac{GM_s}{R}})=2v_A

So, the speed of planet C is twice the speed of planet A.

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The image mentioned is in the attachment

Answer: a) P = 2450 Pa;

b) P = 2940 Pa;

c) F = 4.9 N

Explanation:

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The image shows a block applying pressure on the large side of the piston. The force applied is due to gravitation, so:

P = \frac{F}{A}

P = \frac{m.g}{A}

P = \frac{300.9.8}{1.2}

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The pressure generated by the block is P = 2450 Pa.

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P_{staticfluid} = 600.9.8.0.5

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The pressure in the fluid at 50 cm deep is P_{staticfluid} = 2940 Pa.

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\frac{F}{A_s} = \frac{F_b}{A_b}

F = \frac{F_b}{A_b}.A_s

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The force necessary to be equilibrium is F = 4.9 N.

4 0
3 years ago
98 Points and brainlyest for 5 Science questions please I need it doe before 2:30 ET!!!
Marina CMI [18]
Picture #1:
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GPE = (2 kg) x (9.8 m/s²) x (40 m) = 784 joules

KE = (1/2) (mass) (speed²)
KE = (1/2) (2 kg) (5 m/s)²
KE = (1 kg) (25 m²/s²)  =  25 joules

Picture #2:
KE = (1/2) (mass) (speed²)
KE = (1/2) (2 kg) (10 m/s)²
KE = (1 kg) (100 m²/s²)  =  100 joules

Picture #3:
GPE = (mass) x (gravity) x (height)
GPE = (20 kg) x (9.8 m/s²) x (2 m) = 392 joules

KE = (1/2) (mass) (speed²)
KE = (1/2) (20 kg) (5 m/s)²
KE = (10 kg) (25 m²/s²)  =  250 joules

Picture #4:
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Height = (98 joules) / (1 kg x 9.8 m/s²)
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Picture #5:
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39,200 Joules = (mass) x (9.8 m/s²) x (20 m)
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An oscillating block - spring system has a mechanical energy of 1.10 j, and amplitude of 11.0 cm, and a maximum speed of 1.7 m/s
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The total mechanical energy of the block-spring system is given by the sum of the potential energy and the kinetic energy of the block:
E=U+K= \frac{1}{2}kx^2 + \frac{1}{2}mv^2
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At the point of maximum displacement of the spring, the velocity of the block is zero: v=0, so the kinetic energy is zero and the mechanical energy is just potential energy of the spring:
E= \frac{1}{2}kA^2 (1)
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An ambulance is rapidly approaching you at a stop light. What happens to the frequency and pitch of the sound as the ambulance d
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Answer:

A )

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as we hear a higher frequency , it makes the <em>pitch higher</em> too

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3 years ago
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Answer:

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(B² x e x d²) / V  =  M

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