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shusha [124]
3 years ago
14

Estimate the smallest possible period of a satellite in a circular orbit around earth. (mass and radius of earth is 5.98 x 1024

kg and 6.37 x 106 m. the universal gravitational constant g = 6.67 x 10-11 n.m2/kg2 .)
Physics
1 answer:
myrzilka [38]3 years ago
3 0
Given: Mass of earth Me = 5.98 x 10²⁴ Kg

           Radius of earth r = 6.37 x 10⁶ m

            G = 6.67 x 10⁻¹¹ N.m²/Kg²

Required: Smallest possible period T = ?

Formula: F = ma;  F = GMeMsat/r²     Centripetal acceleration ac = V²/r

               but V = 2πr/T

equate T from all equation.

F = ma

GMeMsat/r² = Msat4π²/rT²    

GMe = 4π²r³/T²

T² = 4π²r³/GMe  

T² = 39.48(6.37 x 10⁶ m)³/6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)

T² = 1.02 x 10²² m³/3.99 x 10¹⁴ m³/s²

T² = 25,563,909.77 s²

T = 5,056.08 seconds    or around   1.4 Hour

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3 years ago
A 1.0 m × 1.5 m double-pane window consists of two 20-mm-thick layers of glass (k = 0.78 W/m·K) that are separated by a 13-mm ai
jeka94

Answer:

\dot Q=105.042\ W is the heat loss by conduction.

Explanation:

Given:

  • area of window pane,A=1.5\ m^2
  • thickness of the glass pane, x=0.02\ mm
  • thickness of the air gap, x'=0.013\ m
  • thermal conductivity of the glass, k=0.78\ W.m^{-1}.K^{-1}
  • thermal conductivity of the air, k_a=0.025\ W.m^{-1}.K^{-1}
  • inside temperature, T_i=20^{\circ}C
  • outside temperature, T_o=-20^{\circ}C

<u>From the Fourier's law of conduction we have the rate of heat transfer as:</u>

\dot Q=k.A.\frac{dT}{dx}

\dot Q=dT\div \frac{dx}{k.A}

where:

k = thermal conductivity of the material

A = area subjected to the conduction

dT = temperature difference across the two surfaces

dx = thickness of the surface

\frac{dx}{k.A} = regarded as thermal resistance on electrical analogy

According to question here the  heat transfer occurs due to conduction of the air.

Here we have two surfaces with air sandwiched between them. So we find an equivalent resistance:

R_e=2\times \frac{x}{k.A}+ \frac{x'}{k_a.A}

R_e=2\times \frac{0.02}{0.78\times 1.5} +\frac{0.013}{0.025\times 1.5}

R_e=0.3808\ K.W^{-1}

Therefore:

\dot Q=40\div 0.3808

\dot Q=105.042\ W

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4 years ago
Use 4 to 5 complete sentences, explain the concepts of beats and how beats are produces.
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Answer:

Beats is a phenomenon which occurs when two sound waves with different frequencies overlap each other.

Explanation:

When these two sound waves overlap, a third frequency is formed. The places where the two waves reinforces each other, constructive interference takes place and a louder sound is produced whereas, at points where destructive interference takes place a softer sound is produced.

These alternative loud and soft sounds are called beats.

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4 years ago
During a tennis match, a player serves the ball at 27.4 m/s, with the center of the ball leaving the racquet horizontally 2.34 m
Maurinko [17]

a. The ball's horizontal and vertical positions at time t are given by

x=\left(27.4\,\dfrac{\rm m}{\rm s}\right)t

y=2.34\,\mathrm m-\dfrac g2t^2

The ball reaches the net when x=12.0\,\rm m:

12.0\,\mathrm m=\left(27.4\,\dfrac{\rm m}{\rm s}\right)t\implies t=0.438\,\rm s

At this time, the ball is at an altitude of

2.34\,\mathrm m-\dfrac g2\left(0.438\,\mathrm s\right)^2=1.40\,\mathrm m

which is 1.40 m - 0.900 m = 0.500 m above the net.

b. The change in angle gives the ball the new position functions

x=\left(27.4\,\dfrac{\rm m}{\rm s}\right)\cos(-5.00^\circ)t

y=2.34\,\mathrm m+\left(27.4\,\dfrac{\rm m}{\rm s}\right)\sin(-5.00^\circ)t-\dfrac g2t^2

The ball reaches the net at time t such that

\left(27.4\,\dfrac{\rm m}{\rm s}\right)\cos(-5.00^\circ)t=12.0\,\mathrm m\implies t=0.440\,\mathrm s

at which point the ball's vertical position would be

2.34\,\mathrm m+\left(27.4\,\dfrac{\rm m}{\rm s}\right)\sin(-5.00^\circ)\left(0.440\,\mathrm s\right)-\dfrac g2\left(0.440\,\mathrm s\right)^2=0.343\,\mathrm m

so that the ball does not clear the net with 0.343 m - 0.900 m = -0.557 m.

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