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3 years ago

What would happen to the moon if the earth stopped exerting the force of gravity onto it?

1 answer:

nataly862011 [7]3 years ago

5 0

It would drift away from the Earth and enter its own orbit around the sun. It would become the 9th planet, in an orbit very close to Earth's orbit. (Our moon is larger than Pluto, and almost as large as Mercury.)

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A roller coaster car speed is 4 m/s, but 3 seconds later its speed is 22 m/s. What is its average acceleration?

Juli2301 [7.4K]

The velocity increased from 4 m/s to 22 m/s in 3 seconds. 18 m/s in 3 seconds so the average acceleration is change in velocity divided by time. 18 m/s divided by 3 seconds = 6 m/s^2

6 0

3 years ago

An electron in an unknown energy level of a hydrogen atom transitions to the n=2 level and emits a photon with wavelength 410 nm

lyudmila [28]

Answer:

The initial energy level = 6

Explanation:

Photon wavelength is proportional to energy. The wavelength of emitted photons is related to the energy levels of the atom as given by the Rydberg formula:

ₕ₁₂

(1/λ) = Rₕ [(1/n₂²) − (1/n₁²)]

where n₂ = final energy level = 2

n₁ = initial energy level = ?

Rₕ = Rydberg's constant = 1.097 × 10⁷ m⁻¹

λ = wavelength = 410 nm = 410 × 10⁻⁹ m

1/(410 × 10⁻⁹) = (1.097 × 10⁷) [(1/2²) − (1/n₁²)]

0.223 = [(1/4) − (1/n₁²)]

(1/n₁²) = 0.02778

n₁² = 1/0.02778 = 36

n₁ = 6.

7 0

3 years ago

The acceleration reaches its minimal value of zero at the top of the trajectory. *

Serhud [2]

it is true. the trajectory reaches the value of zero at the top

4 0

3 years ago

A 2.5 g marshmallow is placed in one end of a 40 cm pipe, as shown in the figure above. A person blows into the left end of the

frosja888 [35]

Answer:

Your answer would be

A person 40 cm- blows into the left end of the pipe to eject the marshmallow from the right end. ... A strain of sound waves is propagated along an organ pipe and gets reflected from an. play · like-icon ... The velocity of sound in air is 340ms^(-1). ... The two pipes are submerged in sea water, arranged as shown in figure. Pipe.Explanation:

I belive this is the answer sorry if im wrong!

3 0

3 years ago

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A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the ma

raketka [301]

Let, the maximum height covered by projectile be $\sf{H_m}$

$\purple{ \longrightarrow \bf{h_m = \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} }}$

Projectile is thrown with a velocity = v
Angle of projection = θ

Velocity of projectile at a height half of the maximum height covered be $\sf{v_0}$

$\qquad$ ______________________________

Then –

$\qquad$ $\pink{ \longrightarrow \bf{ \dfrac{h_m}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }}$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }$

$\qquad$ $\longrightarrow \bf{v_0 = \sqrt{ \dfrac{ {v}^{2} }{2} } = \dfrac{v}{ \sqrt{2} } }$

Now, the vertical component of velocity of projectile at the height half of $\sf{h_m}$ will be –

$\qquad$ $\longrightarrow \bf{v_{(y)}=v_0 \: sin \theta }$

$\qquad$ $\longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }$

Therefore, the vertical component of velocity of projectile at this height will be–

☀️ $\qquad$ $\pink {\bf{ \dfrac{v \: sin \: \theta}{ \sqrt{2} }} }$

6 0

2 years ago

Read 2 more answers