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3 years ago

Three differences between orbital motion and axial motion of the earth.

Physics

1 answer:

NISA [10]3 years ago

7 0

Answer:

Orbital motion

1.is a gravitational motion.

2.they move in the same direction.

3.move are the same time.

Axial motion

1.is a rotary motion

2.move around it own axis

3. revolution takes one year

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Stars are formed in which of the following?

MAXImum [283]

Stars are formed in <u>nebulas</u>, interstellar clouds of dust and gas.

6 0

3 years ago

Experiment with the battery voltage set to 15 volts, measure the current in a parallel circuit with 1,2,3, and 4 light bulbs. (I

koban [17]

Answer:

1) current = I

2) Resistance = V/I

3) current = 2I

4) resistance = V/2I

5) current = 3I

6) Resistance = V/3I

7) Current = 4I

8) Resistance = V/4I

Explanation:

When one bulb is connected across the battery then let say the current is given as I

Then resistance is given as

$R = \frac{V}{I}$

When two bulbs are in parallel with the battery then

total current becomes twice of initial current

so we have

current = 2I

Resistance of the circuit is now

$R = \frac{V}{2I}$

When three bulbs are in parallel with the battery then

total current becomes three times of initial current

so we have

current = 3I

Resistance of the circuit is now

$R = \frac{V}{3I}$

When four bulbs are in parallel with the battery then

total current becomes four times of initial current

so we have

current = 4I

Resistance of the circuit is now

$R = \frac{V}{4I}$

3 0

3 years ago

A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at 4.03 m/s on the horizontal section of a

seraphim [82]

Answer:

2.38 m/s, 4.31 m/s, lower

Explanation:

Initial energy = final energy

½ m v₀² + ½ I ω₀² = mgh + ½ m v₁² + ½ I ω₁²

Since the ball is rolling without slipping, ω = v / r.

For a hollow sphere, I = ⅔ m r².

½ m v₀² + ½ (⅔ m r²) (v₀ / r)² = mgh + ½ m v₁² + ½ (⅔ m r²) (v₁ / r)²

½ m v₀² + ⅓ m v₀² = mgh + ½ m v₁² + ⅓ m v₁²

⅚ m v₀² = mgh + ⅚ m v₁²

⅚ v₀² = gh + ⅚ v₁²

v₀² = 1.2gh + v₁²

v₁ = √(v₀² − 1.2gh)

Given v₀ = 4.03 m/s, g = 9.80 m/s, h = 0.900 m:

v₁ = √((4.03)² − 1.2 (9.80) (0.900))

v₁ ≈ 2.38 m/s

At the top of the loop, the sum of the forces in the radial direction is:

∑F = ma

W + N = m v² / R

N = m v² / R - mg

N = m (v² / R - g)

Given v = 2.38 m/s, R = 0.450 m, and g = 9.80 m/s²:

N = m ((2.38)² / 0.450 - 9.80)

N = 2.77m

N ≥ 0, so the ball stays on the track.

Initial energy = final energy

Borrowing from part a):

v₂ = √(v₀² − 1.2gh)

This time, h = -0.200 m:

v₂ = √((4.03)² − 1.2 (9.80) (-0.200))

v₂ ≈ 4.31 m/s

Without the rotational energy:

½ m v₀² = mgh + ½ m v₁²

½ v₀² = gh + ½ v₁²

v₀² = 2gh + v₁²

v₁ = √(v₀² - 2gh)

This is less than v₁ we calculated earlier.

6 0

3 years ago

A thin flake of mica (n=1.58) is used to cover one slit of a double-slit interference arrangement. The central point on the wiew

Llana [10]

To develop this problem it is necessary to apply the optical concepts related to the phase difference between two or more materials.

By definition we know that the phase between two light waves that are traveling on different materials (in this case also two) is given by the equation

$\Phi = 2\pi(\frac{L}{\lambda}(n_1-n_2))$

Where

L = Thickness

n = Index of refraction of each material

$\lambda =$ Wavelength

Our values are given as

$\Phi = 7(2\pi)$

$L=t$

$n_1 = 1.58$

$n_2 = 1$

$\lambda = 550nm$

Replacing our values at the previous equation we have

$\Phi = 2\pi(\frac{L}{\lambda}(n_1-n_2))$

$7(2\pi) = 2\pi(\frac{t}{\lambda}(1.58-1))$

$t = \frac{7*550}{1.58-1}$

$t = 6637.931nm \approx 6.64\mu m$

Therefore the thickness of the mica is 6.64μm

3 0

3 years ago

Two identical charges are separated by a distance d. If the distance between them is increased to 3d, what will happen to the fo

belka [17]

A-It will be one-ninth the original force

3 0

3 years ago

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