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2 years ago

A space shuttle orbits Earth 1 time in 90 minutes. How many times does it orbit Earth in 6 hours?

1 answer:

8 0

I'm pretty sure it would be 4 because the first two times to goes around would be 3 hours plus another two times to make 3 more hours making the amount of times it orbited Earth 4

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Identify and sketch all the external forces acting on the chair. because the chair can be represented as a point particle of mas

SOVA2 [1]

There are two external force acts on the chair.

1. The force due to earth gravity, acts in the downward direction.

2. Reaction force of the gravity, which acts in the Upward direction (Normal Force).

On every object, there is a force acts due to gravity of earth, which pulls the object towards the centre of earth, known as gravity force, always acts in the downward direction. Mathematically it's given as

F=mg

here, m is the mass of the object, and g is the acceleration of gravity.

To balance this gravity force, a counter force acts in the opposite direction, whose magnitude is equal to the force of gravity

8 0

2 years ago

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Suppose a straight 1.00-mm-diameter copper wire could just "float" horizontally in air because of the force due to the Earth’s m

insens350 [35]

To solve this problem it is necessary to apply the concepts related to the Force since Newton's second law, as well as the concept of Electromagnetic Force. The relationship of the two equations will allow us to find the magnetic field through the geometric relations of density and volume.

$F_{mag}= BIL$

Where,

B = Magnetic Field

I = Current

L = Length

<em>Note: $F_{mag}$ is a direct adaptation of the vector relation $F=q \times V \times B$ </em>

From Newton's second law we know that the relation of Strength and weight is determined as

$F_g = mg$

Where,

m = Mass

g = Gravitational Acceleration

For there to be balance the two forces must be equal therefore

$F_{mag} = F_g$

$BIL = mg$

Our values are given as,

Diameter $(d) = 1.0mm = 1*10^{-3}m$

Radius $(r) = \frac{d}{2} = \frac{1*10^{-3}}{2} = 0.5*10^{-3}m$

Magnetic Field $(B) = 5.0*10^{-5} T$

From the relationship of density another way of expressing mass would be

$\rho = \frac{m}{V} \rightarrow m = \rho V$

At the same time the volume ratio for a cylinder (the shape of the wire) would be

$V = \pi r^2 L \rightarrow L =Length, r= Radius$

Replacing this two expression at our first equation we have that:

$BIL = mg$

$BIL = ( \rho V)g$

$BIL = ( \rho \pi r^2 L)g$

Re-arrange to find I

$I = \frac{( \rho \pi r^2 L)g}{BL}$

$I = \frac{( \rho \pi r^2 )g}{B}$

We have for definition that the Density of copper is $8.9*10^3 Kg/m^3$ , gravity acceleration is $9.8m/s^2$ and the values of magnetic field (B) and the radius were previously given, then: