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lara31 [8.8K]
3 years ago
5

Suppose that the sun shrank in size but that its mass remained the same. What would happen to the orbit of the earth?

Physics
1 answer:
Minchanka [31]3 years ago
6 0

Answer:

Nothing

Explanation:

The radius of the orbit of the Earth does not depend on the radius of the sun.

In fact, the gravitational attraction between the Earth and the Sun provides the centripetal force that keeps the Earth in orbit:

G\frac{Mm}{r^2} = m\frac{v^2}{r}

where

G is the gravitational constant

M is the mass of the sun

m is the mass of the Earth

r is the radius of the orbit of the Earth

v is the orbital speed of the earth

Re-arranging the equation for r:

r=\frac{GM}{v^2}

Also,

v=\omega r

where \omega is the angular velocity of the Earth's orbit. So we can rewrite the equation as

r=\frac{GM}{\omega^2 r^2}\\r^3 = \frac{GM}{\omega^2}

As we see, the radius of the orbit of the Earth, r, does not depend on the mass of the Sun, so if the sun shrank in size, the orbit remains the same.

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How does the mass of an object affect the gravitational forces experienced by another object?.
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F=(Gmm)/(r^2)

larger the mass stronger the force. An object with a large mass will produce a large force. A small mass will produce a small force.

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2 years ago
An electron moves 4.5 m in the direction of an electric field of strength 325 N/C. What is the change in electrical potential en
Vinvika [58]
Calculate q* E * d 
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6 0
4 years ago
100%
xxMikexx [17]

Answer:

1. The elastic potential energy is 0.0176 Joules

2. The kinetic energy of the pinball the instant it leaves the spring is 0.0176 Joules

3. The speed of the pinball the instant it leaves the spring is approximately 2.42212 m/s

4. The height of the part where the pinball is located on the machine above the ground is approximately 0.213 meters

Explanation:

The spring constant of the pinball machine's plunger, k = 22 N/m

The amount by which the pinball machine's plunger is compressed, x = 0.04 m

The mass of the pinball ball, m = 0.006 kg

1. The elastic potential energy, P.E. = 1/2·k·x²

By substitution, we get;

P.E. = 1/2 × 22 N/m × (0.04 m)² = 0.0176 J

The elastic potential energy, P.E. = 0.0176 J

2. At the instant the pinball leaves the spring, the plunger and therefore the force of the plunger no longer acts on the pinball

Since there are no external forces acting on the pinball to increase the speed of the pinball after it leaves the spring, the velocity reached is its maximum velocity, and therefore, the kinetic energy, K.E. is the maximum kinetic energy which by the conservation of energy, is equal to the initial potential energy

Therefore;

K.E. = P.E. = 0.0176 J

The kinetic energy of the pinball the instant it leaves the spring, K.E.= 0.0176 J

3. The kinetic energy, K.E., is given by the following formula;

K.E. = 1/2·m·v²

Where;

v = The speed or velocity of the object having kinetic energy K.E.

Therefore, from K.E. = 0.0176 J, and by plugging in the values of the variables, we have;

K.E. = 0.0176 J = 1/2 × 0.006 kg × v²

v² = 0.0176 J/(1/2 × 0.006 kg) = 88/15 m²/s²

v = √(88/15 m²/s²) ≈ (2·√330)/15 m/s ≈ 2.42212 m/s

The speed of the pinball the instant it leaves the spring, v ≈ 2.42212 m/s

4. The height of the pinball is given by the following kinematic equation of motion;

v_h² = u² - 2·g·h

Where;

v_h = The velocity of the pinball at the given height = 1.3 m/s

u = v ≈ 2.42212 m/s (The initial velocity of the pinball as it the spring)

g = The acceleration due to gravity ≈ 9.8 m/s²

h = The height of the pinball above the ground

We get;

v_h² = 1.3² = 2.42212² - 2 × 9.8 × h

∴  h = (2.42212² - 1.3²)/(2 × 9.8) ≈ 0.213

The height of the part where the pinball is located on the machine above the ground, h ≈ 0.213 m

5 0
3 years ago
A single-turn circular loop of wire of radius 50 mm lies in a plane perpendicular to a spatially uniform magnetic field. During
Alinara [238K]

Answer:

ε= 7.86 mV,  Current: Anti-clockwise

Explanation:

radius= 50 mm

dt= 0.10 s

Initial magnitude of magnetic field= B1 = 200 mT

Final magnitude of magnetic field = B2 = 300 mT

Ф= B. A= BA cosα

Ф1= B1 * A * cosα

Ф1= (200*10^{-3})* \pi  * (50*10^{-3} )^2*(1)

Ф1= 0.00157 Wb

Ф2= B2 * A * cosα

Ф1= (300*10^{-3})* \pi  * (50*10^{-3} )^2*(1)

Ф2=0.00236 Wb

dФ= Ф2 - Ф1

dФ= 0.00236 - 0.00157

dФ= 0.000786 Wb

ε= \frac{d}{dt} Ф

ε=0.001786/ 0.10

ε=0.00786 v = 7.86 mV

b)

According to lenz's law the induced emf always oppose the cause producing it.

Applied field is out of the paper so the current will flow in anti-clockwise direction to produce north pole pointing toward the paper.

6 0
3 years ago
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