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Salsk061 [2.6K]
2 years ago
5

You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 1.

70×107 m and its rotation period to be 22.3 hours. You have previously determined that the planet orbits 2.70×1011 m from its star with a period of 402 earth days. Once on the surface you find that the free-fall acceleration is 12.5 m/s2 .
A) What is the mass of the planet?
B) What is the mass of the star?
Physics
1 answer:
Arlecino [84]2 years ago
4 0

Answer:

sorry I don't know but you

Explanation:

can ask quora it's an app

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A positive point charge Q is fixed on a very large horizontal frictionless tabletop. A second positive point charge q is release
Scilla [17]

Answer:

(A) As it moves farther and farther from Q, its speed will keep increasing.

Explanation:

When a positive charge Q is fixed on a horizontal frictionless tabletop and a second charge q is released near to it then according to the Coulombs law the force acting on it decreases with the square of the distance between them.

Mathematically:

F=\frac{1}{4\pi.\epsilon_0} \times \frac{Q.q}{r^2}

where:

r = distance between the charges

\epsilon_0= permittivity of free space

By the Newtons' second law of motion if the we know that the acceleration is directly proportional to the force applied. So as  the distance between the charges increases the its acceleration also decreases therefore now the charge feels less acceleration but still continues to accelerate with a fading magnitude.

7 0
3 years ago
How does an object's acceleration change if the net
topjm [15]
C.the acceleration is doubled
5 0
3 years ago
Consider a series RLC circuit where R = 855 Ω and C = 6.25 μF. However, the inductance L of the inductor is unknown. To find its
sashaice [31]

Answer:

L= 0.059 mH

Explanation:

Given that

R = 855 Ω and C = 6.25 μF

V= 84 V

Frequency

ω = 51900 1/s

We know that

\omega=\sqrt{\dfrac{1}{LC}}

L=Inductance

C=Capacitance

ω =angular Frequency

ω² L C =1

(51900)² x L x 6.25 x 10⁻⁶ = 1

L= 5.99 x 10⁻⁵ H

L= 0.059 mH

6 0
3 years ago
An iron nail is driven into a block of ice by a single blow of a hammer. The hammerhead has a mass of 0.5 kg and an initial spee
larisa [96]

Answer:

The ice melts mass is:

m_g=7.6x10^{-3} g

Explanation:

Kinetic Energy  

K_E = 1/2*m*v^2

K_E = 1/2*0.5kg*(3.2m/s)^2

K_E=2.56 kg*m^2/s^2

Heat gained by ice= mass(g) x 80 cal

( 1 cal = 4.184 *10^7er or g cm^2/ sec^2)

Assuming no loss in heat,  in the motion so both continue with temperature 0~C

To find so the mass (gm) of ice melted

m_g= 1/2 *(0.5*1000)*(3.2m/s)^2* 100*100 / (80*4.184*10^7)

m_g=7.6x10^{-3} g

5 0
3 years ago
0.250 moles of gas are in a piston.The gas does 109 J of work while 240 J of heat are added. What is the change in internal ener
irina [24]

Answer:

+131Joules

Explanation:

Energy can be expressed using below expresion.

ΔE = (q + w).........eqn(1)

q will be + be if heat is gained hence, q= 240 J

work "w" will be - ve if work is done by the system, hence w= -109 J

Then substitute into eqn(1)

Change in Internal energy=

= (240 -109 )

= +131J

6 0
3 years ago
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