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3 years ago

Is motion on earth different from space

1 answer:

8 0

It's not that the laws of motion are any different on Earth than in space. But Earth's gravitational field has such an overwhelming force it masks their precise effects. And gravity is integral to all sorts of phenomena that we take for granted.

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True or False:

mart [117]

This is a true statement

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3 years ago

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What is the difference between resistance and resistivity

Crank

Answer:

google it up

Explanation:

it's there

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3 years ago

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The magnitude of the magnetic flux through the surface of a circular plate is 6.80 10-5 T · m2 when it is placed in a region of

PIT_PIT [208]

Answer:

B = 4.1*10^-3 T = 4.1mT

Explanation:

In order to calculate the strength of the magnetic field, you use the following formula for the magnetic flux trough a surface:

$\Phi_B=S\cdot B=SBcos\alpha$ (1)

ФB: magnetic flux trough the circular surface = 6.80*10^-5 T.m^2

S: surface area of the circular plate = π.r^2

r: radius of the circular plate = 8.50cm = 0.085m

B: magnitude of the magnetic field = ?

α: angle between the direction of the magnetic field and the normal to the surface area of the circular plate = 43.0°

You solve the equation (1) for B, and replace the values of the other parameters:

$B=\frac{\Phi_B}{Scos\alpha}=\frac{6.80*10^{-5}T.m^2}{(\pi (0.085m)^2)cos(43.0\°)}\\\\B=4.1*10^{-3}T=4.1mT$

The strength of the magntetic field is 4.1mT

4 0

3 years ago

A trooper is moving due south along the freeway at a speed of 30.1 m/s when a red car passes the trooper. The red car moves with

romanna [79]

Answer:

The correct answer to the following question will be "41.87 m".

Explanation:

The given values are:

The speed of trooper = $30.1 \ m/s$

The velocity of red car = $45.4 \ m/s$

Now,

A red car goes as far as possible until the speed or velocity of the troops is the same as that of of the red car at

$t=\frac{45-30}{2.7}$ (∵ $time=\frac{distance}{time}$ )

$t=5.56 \ sec$

then,

The distance covered by trooper,

$t1=30\times 5.56+\frac{1}{2}\times 2.7\times (5.56)^2$

$=208.33 \ m$

The distance covered by red car,

= $45\times 5.56$

= $250.2 \ m$

Maximum distance = $250.2-208.33$

= $41.87 \ m$

6 0

3 years ago

The average distance of Enceladus from Saturn is 238,000 km; the average distance of Titan from Saturn is 1,222,000 km. How much

aleksklad [387]

Answer:

Titan takes 11.634 times longer to orbit Saturn as compared to Enceladus.

Explanation:

We have been given that the average distance of Enceladus from Saturn is 238,000 km; the average distance of Titan from Saturn is 1,222,000 km.

We will use Kepler's Law to solve our given problem.

$\frac{(T_1)^2}{(T_2)^2}=\frac{(r_1)^3}{(r_2)^3}$

Upon substituting our given values, we will get:

$\frac{(T_1)^2}{(T_2)^2}=\frac{(238,000)^3}{(1,222,000)^3}$

$\frac{(T_1)^2}{(T_2)^2}=\frac{13481272000000000}{1824793048000000000}$

$\frac{(T_1)^2}{(T_2)^2}=\frac{13481272}{1824793048}$

$\frac{(T_1)^2}{(T_2)^2}=0.0073878361246365$

Taking square root of both sides, we will get:

$\frac{T_1}{T_2}=\sqrt{0.0073878361246365}$

$\frac{T_1}{T_2}=0.0859525225030452495$

$\frac{T_2}{T_1}=\frac{1}{0.0859525225030452495}$

$\frac{T_2}{T_1}=11.634329870476699\approx 11.634$

$T_2=11.634\cdot T_1$

This implies that time period of Titan about Sturn is 11.634 times more compared to time period of Enceladus about Saturn.

So, basically Titan takes 11.634 times longer to orbit Saturn as compared to Enceladus.

8 0

3 years ago