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3 years ago

9

In which phase is the full illuminated face of the Moon visible on Earth? A) New moon Eliminate B) Full moon C) Waning gibbous D

) Waxing crescent

1 answer:

Nitella [24]3 years ago

4 0

Full moon

New cant be seen

Gibbous is 3/4

Crescent is 1/4

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A mass m attached to a spring of constant k is oscillating on a frictionless surface. A second mass of mass m is dropped on top

elixir [45]

Answer:

E. The period of oscillation increases.

Explanation:

The period of oscillation is:

T = 2π√(m/k)

Frequency is the inverse of period (f = 1/T), so as period increases, frequency decreases.

Increasing the mass will increase the period and decrease the frequency.

8 0

2 years ago

An object is placed at O ona number line. It moves 3 units to the right, then 4 units to the left, and then 6 units to

kvv77 [185]

Answer:

You have a displacement of 5 units to the right.

Explanation:

First you go three to the right which lands on the 3 mark. Then you move it 4 to the left which substracts 4, landing the object at -1. Finally you move 6 to the right, and you finish at marker 5. Since displacement is not total distance but just final distance from the start point directly to end point, it is only a displacement of 5.

6 0

2 years ago

Read 2 more answers

A certain frictionless simple pendulum having a length L and mass M swings with period T. If both L and M are doubled, what is t

vampirchik [111]

The new period is D) √2 T

$\texttt{ }$

<h3>Further explanation</h3>

Let's recall Elastic Potential Energy and Period of Simple Pendulum formula as follows:

$\boxed{E_p = \frac{1}{2}k x^2}$

where:

<em>Ep = elastic potential energy ( J )</em>

<em>k = spring constant ( N/m )</em>

<em>x = spring extension ( compression ) ( m )</em>

$\texttt{ }$

$\boxed{T = 2\pi \sqrt{ \frac{L}{g} }}$

where:

<em>T = period of simple pendulum ( s )</em>

<em>L = length of pendulum ( m )</em>

<em>g = gravitational acceleration ( m/s² )</em>

Let us now tackle the problem!

$\texttt{ }$

<u>Given:</u>

initial length of pendulum = L₁ = L

initial mass = M₁ = M

final length of pendulum = L₂ = 2L

final mass = M₂ = 2M

initial period = T₁ = T

<u>Asked:</u>

final period = T₂ = ?

<u>Solution:</u>

$T_1 : T_2 = 2\pi \sqrt{ \frac{L_1}{g} }} : 2\pi \sqrt{ \frac{L_2}{g} }}$

$T_1 : T_2 = \sqrt{L_1} : \sqrt{L_2}$

$T : T_2 = \sqrt{L} : \sqrt{2L}$

$T : T_2 = 1 : \sqrt{2}$

$\boxed {T_2 = \sqrt{2}\ T}$

$\texttt{ }$

<h3>Learn more</h3>

Kinetic Energy : brainly.com/question/692781

Acceleration : brainly.com/question/2283922

The Speed of Car : brainly.com/question/568302
Young Modulus : brainly.com/question/9202964
Simple Harmonic Motion : brainly.com/question/12069840

$\texttt{ }$

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Elasticity

3 0

3 years ago

Read 2 more answers

A grindstone increases in angular speed from 6.00 rad/s to 12.20 rad/s in 16.00 s. Through what angle does it turn during that t

Akimi4 [234]

Answer:

Explanation:

Given that the grand stone has initial angular velocity of

w(ini)= 6rad/

And it has a final angular velocity of

w(fin)=12.20rad/sec

Time taken is t=16s

Using equation of angular motion

To get angular acceleration (α)

w(fin)=w(ini)+αt

12.20=6+16α

16α=12.20-6

16α=6.2

α=6.2/16

α=0.3875rad/sec²

The angular acceleration is 0.39rad/s²

Angle that he turn using

w(fin)²=w(ini)²+2αθ

12.2²=6²+2×0.3875θ

12.2²-6²=0.775θ

0.775θ=112.84

Then, θ=112.84/0.775

θ=145.6radian

The angular displacement is 145.6rad

6 0

2 years ago

Most stars in the Milky Way's halo are _________. blue or white in color very young very old found inside molecular clouds

Alina [70]

Answer:

Most stars in the Milky Way's halo are <u>very old</u>

<u>very oldThere is no recycling of gas in the halo, so halo stars are quite old.</u>

6 0

2 years ago