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Deffense [45]
3 years ago
10

Earth's orbit around the sun is not perfectly circular. The nearest point of its orbit is 1.471E8 kilometers. The mass of the su

n is 1.989E30 kilograms. The mass of the earth is 5.97E24 kilograms. The universal gravitational constant is 6.6738E-11 N(m/kg)2. (a) What is the force exerted by the sun on the earth they are closest together? (b) What is the corresponding force of the earth on the sun?
Physics
1 answer:
BARSIC [14]3 years ago
7 0

Answer:

a) -3.66233E22 N

b) 3.66233E22 N

Explanation:

Closest distance between Earth and Sun = 1.471E8 kilometers = r

Mass of Earth = 5.97E24 kilograms = m

Mass of the sun = 1.989E30 kilograms = M

a) Universal gravitational constant = 6.6738E-11 N(m/kg)2

F_{Mm}=-G\frac{Mm}{r^2}\\\Rightarrow F=-6.6738E-11 N \frac{5.97E24\times 1.989E30}{(1.471E11)^2}\\\Rightarrow F=-3.66233E22\ N

Force exerted by the sun on the earth they are closest together is -3.66233E22 N

b) The magnitude of the force that the earth will exert on the sun will be same but the direction will be different represented by the change of the sign

F_{mM}=-F_{Mm}\\\Rightarrow F_{mM}=-(-3.66233E22\ N)\\\Rightarrow F_{mM}=3.66233E22\ N

Force exerted by the earth on the sun when they are closest together is 3.66233E22 N.

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The field strength needed to produce a 24.0 V peak emf is 0.73T.

To find the answer, we need to know about the expression of emf.

What's the expression of peak emf produced in a rotating rectangular loops?

  • The peak emf produced in a rotating loops= N×B×A×w
  • N= no. of turns of the loop, B= magnetic field, A= area of loop and w= angular frequency
  • So, B = emf/(N×A×w)
<h3>What's the magnetic field applied to the loop, when rectangular coil with 300 turns of dimensions 5.00 cm by 5.22 cm rotates at 400 rpm produce a 24.0 V peak emf?</h3>
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  • Emf= 24V, w= 2π×400 rpm= 2π×(400rps/60) = 42 rad/s
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B= 24/(300×0.00261×42) = 0.73T

Thus, we can conclude that the magnetic field is 0.73T.

Learn more about the electromagnetic force here:

brainly.com/question/13745767

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2 years ago
How do kinetic and potential energy transfer to one throughout a roller coaster ride?
mojhsa [17]

Answer:

As the cars ascend the next hill, some kinetic energy is transformed back into potential energy. Then, when the cars descend this hill, potential energy is again changed to kinetic energy. This conversion between potential and kinetic energy continues throughout the ride.

Explanation:

hope it helps U

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2 years ago
The movement of an object at a constant speed around a circular radius is known as
forsale [732]

Answer:

Uniform Circular Motion

Explanation:

Uniform circular motion can be described as the motion of an object in a circle at a constant speed. As an object moves in a circle, it is constantly changing its direction. At all instances, the object is moving tangent to the circle.

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An electron accelerated from rest through a voltage of 780 v enters a region of constant magnetic field. part a part complete if
maxonik [38]
The electron is accelerated through a potential difference of \Delta V=780 V, so the kinetic energy gained by the electron is equal to its variation of electrical potential energy:
\frac{1}{2}mv^2 =  e \Delta V
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m is the electron mass
v is the final speed of the electron
e is the electron charge
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Re-arranging this equation, we can find the speed of the electron before entering the magnetic field:
v= \sqrt{ \frac{2 e \Delta V}{m} } = \sqrt{ \frac{2(1.6 \cdot 10^{-19}C)(780 V)}{9.1 \cdot 10^{-31} kg} }=1.66 \cdot 10^7 m/s


Now the electron enters the magnetic field. The Lorentz force provides the centripetal force that keeps the electron in circular orbit:
evB=m \frac{v^2}{r}
where B is the intensity of the magnetic field and r is the orbital radius. Since the radius is r=25 cm=0.25 m, we can re-arrange this equation to find B:
B= \frac{mv}{er}= \frac{(9.1 \cdot 10^{-31}kg)(1.66 \cdot 10^7 m/s)}{(1.6 \cdot 10^{-19}C)(0.25 m)} =3.8 \cdot 10^{-4} T
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3 years ago
A block of mass 500g is pulled from rest on ahorizontal frictionless bench by a steady force F and travels 8m in 2s find the acc
Korvikt [17]

<u>We are Given:</u>

Mass of the block (m) = 500 grams or 0.5 Kg

Initial velocity of the block (u) = 0 m/s

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Time taken to cover 8 m (t)= 2 seconds

Acceleration of the block (a) = a m/s²

<u>Solving for the acceleration:</u>

From the seconds equation of motion:

s = ut + 1/2* (at²)

<em>replacing the variables</em>

8 = (0)(2) + 1/2(a)(2)²

8 = 2a

a = 4 m/s²

Therefore, the acceleration of the block is 4 m/s²

3 0
3 years ago
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