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AURORKA [14]
3 years ago
15

The motivation for Isaac Newton to discover his laws of motion was to explain the properties of planetary orbits that were obser

ved by Tycho Brahe and analyzed by Johannes Kepler. A good starting point for understanding this (as well as the speed of the space shuttle and the height of geostationary satellites) is the simplest orbit - a circular one. This problem concerns the properties of circular orbits for a satellite orbiting a planet of mass M. Use G for the universal gravitational constant. Find the orbital period T.
Physics
1 answer:
Mandarinka [93]3 years ago
3 0

Answer:

Explanation:

We know that , for an object to remain in  circular motion , a force towards centre is required which is called centripetal force. In the circular motion of

satellites around planet , this force is provided by the gravitational attraction between satellite and planet.

If M be the mass of planet and m be the mass of satellite,  G be gravitational constant and R be the distance between planet and satellite or R be the radius of orbit

Gravitational force = G Mm / R²

If v be the velocity with which satellite is orbiting

centripetal force

= m v² /R

Centripetal force = gravitational attraction

m v² /R  =  G Mm / R²

v = \sqrt{\frac{GM}{R} }

Time period = time the satellite takes to make one rotation

= distance / orbital velocity

= 2πR/ v

= \frac{2\pi R\sqrt{R} }{\sqrt{GM} }

T = \frac{2\pi R^\frac{3}{2} }{\sqrt{GM} }

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A 20 cm-long wire carrying a current of 6 A is immersed in a uniform magnetic field of 3 T. If the magnetic field is oriented at
SOVA2 [1]

Answer:

the  magnitude of the force that the wire will experience = 1.8 N

Explanation:

The force on a current carrying wire placed in a magnetic field is :

F = Idl × B

where:

I = current flowing through the wire

dl = length of the wire

B = magnetic field

We can equally say that :

|F| = IdlBsin \theta

where : sin θ is the angle at which the orientation from the magnetic field  to the wire occurs = 30°

Then;

|F| = B\ I \ L \ sin \theta

Given that:

L = 20 cm = 0.2 m

I = 6 A

B = 3 T

θ = 30°

Then:

F = 3 × 6 × 0.2 sin 30°

F = 1.8 N

Therefore, the  magnitude of the force that the wire will experience = 1.8 N

6 0
3 years ago
What is the minimum force required to increase the energy of a car by 84 J over a distance of 38 m? Assume the force is constant
konstantin123 [22]

Answer:

2.210N

Explanation:

Workdone = Force x distance

Distance = 38m , Workdone = 84J

Hence 84J = Force x 38m

Force = 84J / 38m

Force = 2.210N =2.2N

4 0
3 years ago
a 20-kg object travelling at 20 m/s collides head on with an 18-kg object travelling at 17 m/s.If they were locked together afte
djverab [1.8K]

Answer:

2.47 m/s

Explanation:

Momentum = Mass X Velocity

If they were locked together, it means its a perfectly inelastic collision. Therefore,

Total momentum before = Total momentum after

Total momentum before = (20 X 20) - (18 X 17)

= 94

Total momentum after = 94

Y = Object speed after collision

94 = (20+18)Y

Y = 2.47368421 m/s

3 0
2 years ago
Determine a formula for the magnitude of the force F exerted on the large block (Mc) so that the mass Ma does not move relative
SVEN [57.7K]

Answer:

The magnitude of the force F is given by

F =  (M_{a} + M_{b} + M_{c} ) *(M_{b}*g/(\sqrt{M_{a} ^{2}-M_{b} ^{2}}))

Explanation:

Given there are three blocks of masses M_{a}, M_{b} and M_{c} (ref image in attachment)

When all three masses move together at an acceleration a, the force F is given by

F =  (M_{a} + M_{b} + M_{c} ) *a    ................(equation 1)

Also it is given that M_{a} does not move with respect to M_{c}, which gives tension T  is exerted on pulley  by M_{a} only, Hence tension T is

T = M_{a} *a    ..........(equation 2)

There is also also tension exerted by M_{b}. There are two components here: horizontal due to acceleration a and vertical component due to gravity g. Thus tension is given by

T = M_{b} \sqrt{a^{2} +g^{2} }   ................(equation 3)

From equation 2 and 3, we get

M_{a} *a  = M_{b} \sqrt{a^{2} +g^{2} }  

Squaring both sides we get

M_{a} ^{2} *a^{2} = M_{b} ^{2} * (a^{2}+g^{2})

M_{a} ^{2} *a^{2} = (M_{b} ^{2} * a^{2})+ (M_{b} ^{2} *g^{2})

(M_{a} ^{2}  -  M_{b} ^{2}) * a^{2} = M_{b} ^{2} *g^{2}

a^{2} = M_{b} ^{2} *g^{2}/(M_{a} ^{2}  -  M_{b} ^{2})

Taking square root on both sides, we get acceleration a

a = M_{b}*g/(\sqrt{M_{a} ^{2}-M_{b} ^{2}})

Hence substituting the value of a in equation 1, we get

F =  (M_{a} + M_{b} + M_{c} ) *(M_{b}*g/(\sqrt{M_{a} ^{2}-M_{b} ^{2}}))

3 0
3 years ago
All objects emit what radiation
Eduardwww [97]
If an object is not at Absolute Zero, then it is
absorbing and radiating thermal (heat) energy.
5 0
2 years ago
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