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3 years ago

Four satellites are in orbit around the Earth. The heights and the masses of

Physics

1 answer:

Kaylis [27]3 years ago

4 0

Answer:

satellite B

Explanation:

A .F= G (mM)/r²

B .F= G (2mM)/r² = 2G (Mm)r²

C .F= G (3mM)/(2r)² = ¾G (mM)/r²

D .F= G (4mM)/(2r)² = G (mM)/r²

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Suppose the microphone on an AM radio is not working properly. Which best explains how this will affect the process of amplitude

Anna007 [38]

You can have it your own definition of

3 0

3 years ago

Read 2 more answers

The interior space of large box is kept at 30 C. The walls of the box are 3 m high and have a ‘sandwich’ construction consisting

White raven [17]

Answer:

$\frac{\dot Q}{A} =20.129\ W.m^{-2}$

$T_1=27.58\ ^{\circ}C$ & $T_2=2.41875\ ^{\circ}C$

Explanation:

Given:

interior temperature of box, $T_i=30^{\circ}C$

height of the walls of box, $h=3\ m$

thickness of each layer of bi-layered plywood, $x_p=1.25\ cm=0.0125\ m$

thermal conductivity of plywood, $k_p=0.104\ W.m^{-1}.K^{-1}$

thickness of sandwiched Styrofoam, $x_s=5\ cm=0.05\ m$

thermal conductivity of Styrofoam, $k_s=0.04\ W.m^{-1}.K^{-1}$

exterior temperature, $T_o=0^{\circ}C$

From the Fourier's law of conduction:

$\dot Q=\frac{dT}{(\frac{x}{kA}) }$

$\dot Q=\frac{dT}{R_{th} }$ ....................................(1)

Now calculating the equivalent thermal resistance for conductivity using electrical analogy:

$R_{th}=R_p+R_s+R_p$

$R_{th}=\frac{x_p}{k_p.A}+\frac{x_s}{k_s.A}+\frac{x_p}{k_p.A}$

$R_{th}=\frac{1}{A} (\frac{x_p}{k_p}+\frac{x_s}{k_s}+\frac{x_p}{k_p})$

$R_{th}=\frac{1}{A} (\frac{0.0125}{0.104}+\frac{0.05}{0.04}+\frac{0.0125}{0.104})$

$R_{th}=\frac{1.4904}{A}$ .....................(2)

Putting the value from (2) into (1):

$\dot Q=\frac{30-0}{\frac{1.4904}{A} }$

$\dot Q=\frac{30\ A}{1.4904}$

$\frac{\dot Q}{A} =20.129\ W.m^{-2}$ is the heat per unit area of the wall.

The heat flux remains constant because the area is constant.

For plywood-Styrofoam interface from inside:

$\frac{\dot Q}{A} =k_p.\frac{T_i-T_1}{x_p}$

$20.129=0.104\times \frac{30-T_1}{0.0125}$

$T_1=27.58\ ^{\circ}C$

&For Styrofoam-plywood interface from inside:

$\frac{\dot Q}{A} =k_s.\frac{T_1-T_2}{x_s}$

$20.129=0.04\times \frac{27.58-T_2}{0.05}$

$T_2=2.41875\ ^{\circ}C$

4 0

3 years ago

Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 168 cm , but its circumference is decreasi

DedPeter [7]

Answer:

103.1 V

Explanation:

We are given that

Initial circumference=C=168 cm

$\frac{dC}{dt}=-15cm/s$

Magnetic field,B=0.9 T

We have to find the magnitude of the emf induced in the loop after exactly time 8 s has passed since the circumference of the loop started to decrease.

Magnetic flux= $\phi=BA=B(\pi r^2)$

Circumference,C= $2\pi r$

$r=\frac{C}{2\pi}$

$r=\frac{168}{2\pi}$ cm

$\frac{dr}{dt}=\frac{1}{2\pi}\frac{dC}{dt}=\frac{1}{2\pi}(-15)=-\frac{15}{2\pi} cm/s$

$\int dr=-\int \frac{15}{2\pi}dt$

$r=-\frac{15}{2\pi}t+C$

When t=0

$r=\frac{168}{2\pi}$

$\frac{168}{2\pi}=C$

$r=-\frac{15}{2\pi}t+\frac{168}{2\pi}$

E= $-\frac{d\phi}{dt}=-\frac{d(B\pi r^2)}{dt}=-2\pi rB\frac{dr}{dt}$

$E=-2\pi(-\frac{5}{2\pi}t+\frac{168}{2\pi})B\times -\frac{15}{2\pi}$

t=8 s

B=0.9

$E=2\pi\times \frac{15}{2\pi}\times 0.9(-\frac{15}{2\pi}(8)+\frac{168}{2\pi})$

$E=103.1 V$

6 0

3 years ago

A 90 kg man stands in a very strong wind moving at 17 m/s at torso height. As you know, he will need to lean in to the wind, and

victus00 [196]

Answer:

a) $t=195.948N.m$

b) $\phi=13.6 \textdegree$

Explanation:

From the question we are told that:

Density $\rho=1.225kg/m^2$

Velocity of wind $v=14m/s$

Dimension of rectangle:50 cm wide and 90 cm

Drag coefficient $\mu=2.05$

Generally the equation for Force is mathematically given by

$F=\frac{1}{2}\muA\rhov^2$

$F=\frac{1}{2}2.05(50*90*\frac{1}{10000})*1.225*17^2$

$F=163.29$

Therefore Torque

$t=F*r*sin\theta$

$t=163.29*1.2*sin90$

$t=195.948N.m$

Generally the equation for torque due to weight is mathematically given by

$t=d*Mg*sin90$

Where

$d=sin \phi$

Therefore

$t=sin \phi*Mg*sin90$

$195.948=833sin \phi$

$\phi=sin^{-1}\frac{195.948}{833}$

$\phi=13.6 \textdegree$

5 0

3 years ago

List 5 reason why water is not thermometric liquid

kolezko [41]

Answer:

(1) it is transparent so it makes the reading difficult. (2) it is volatile. (3) it is a poor conductor of heat. (4) it has a higher specific heat capacity, so it absorbs more heat from the body with which it is kept in contact. (5) Water cannot be used in thermometer because of its higher freezing point and lower boiling point than other liquids . If water is used in a thermometer , it will start phase change at 0oC and 100oC and will not measure temperature , out of this range .

PLEASE READ CAREFULLY AND PLEASE MARK AS BRAINLIEST :)

5 0

3 years ago