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3 years ago

9. Steradian is the angel which lies in: a) One dimension b) Two dimensions

Physics

2 answers:

Marizza181 [45]3 years ago

8 0

Answer:

Explanation:

m_a_m_a [10]3 years ago

6 0

Answer:

<h2>Question:Steradian is the angel which lies in?</h2>

<h2>Answer:A.)One dimension</h2><h2 />

Explanation:

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<h2>THANK YOU!!!</h2>

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The volume of a gas can be converted to moles by

BaLLatris [955]

B. Because I say so

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3 years ago

A vehicle reaches a speed of 7.5 m/s over 15 seconds. What is its acceleration if it

denpristay [2]

$acceleration = \frac{velocity}{time} = \frac{7.5}{15} = 0.5ms^-^2$

So the answer is option b.

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3 years ago

1. A man throws a ball up with a velocity of 30 m/s. How high does it get?

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Yes the answer is g=5

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3 years ago

A penny has a mass of 2.50 g, a diameter of 19.55 mm, and a thickness of 1.55 mm. Calculate the density of the material from whi

Dmitry [639]

Density = (mass) divided by (volume)

We know the mass (2.5 g). We need to find the volume.

The penny is a very short cylinder.
The volume of a cylinder is (π · radius² · height).
The penny's radius is 1/2 of its diameter = 9.775 mm.
The 'height' of the cylinder is the penny's thickness = 1.55 mm.

Volume = (π) (9.775 mm)² (1.55 mm)

   = (π) (95.55 mm²) (1.55 mm)

   = (π) (148.1 mm³)

   = 465.3 mm³

We know the volume now. So we could state the density of the penny,
but nobody will understand what we have. Here it is:

   mass/volume = 2.5 g / 465.3 mm³ = 0.0054 g/mm³ .

Nobody every talks about density in units of ' gram/(millimeter)³ ' .
It's always ' gram / (centimeter)³ '.
So we have to convert our number for the volume.

   (0.0054 g/mm³) x (10 mm / cm)³

   = (0.0054 x 1,000) g/cm³

   = 5.37 g/cm³ .

This isn't actually very close to what the US mint says for the density
of a penny, but it's in a much better ball park than 0.0054 was.

4 0

3 years ago

Block A has a mass of 0.5kg, and block B has a mass of 2kg. Block is is released at a height of 0.75 meters above B. The coeffic

VikaD [51]

Answer:

0.075 m

Explanation:

The picture of the problem is missing: find it in attachment.

At first, block A is released at a distance of

h = 0.75 m

above block B. According to the law of conservation of energy, its initial potential energy is converted into kinetic energy, so we can write:

$m_Agh=\frac{1}{2}m_Av_A^2$

where

$g=9.8 m/s^2$ is the acceleration due to gravity

$m_A=0.5 kg$ is the mass of the block

$v_A$ is the speed of the block A just before touching block B

Solving for the speed,

$v_A=\sqrt{2gh}=\sqrt{2(9.8)(0.75)}=3.83 m/s$

Then, block A collides with block B. The coefficient of restitution in the collision is given by:

$e=\frac{v'_B-v'_A}{v_A-v_B}$

where:

e = 0.7 is the coefficient of restitution in this case

$v_B'$ is the final velocity of block B

$v_A'$ is the final velocity of block A

$v_A=3.83 m/s$

$v_B=0$ is the initial velocity of block B

Solving,

$v_B'-v_A'=e(v_A-v_B)=0.7(3.83)=2.68 m/s$

Re-arranging it,

$v_A'=v_B'-2.68$ (1)

Also, the total momentum must be conserved, so we can write:

$m_A v_A + m_B v_B = m_A v'_A + m_B v'_B$

where

$m_B=2 kg$

And substituting (1) and all the other values,

$m_A v_A = m_A (v_B'-2.68) + m_B v_B'\\v_B' = \frac{m_A v_A +2.68 m_A}{m_A + m_B}=1.30 m/s$

This is the velocity of block B after the collision. Then, its kinetic energy is converted into elastic potential energy of the spring when it comes to rest, according to

$\frac{1}{2}m_B v_B'^2 = \frac{1}{2}kx^2$

where

k = 600 N/m is the spring constant

x is the compression of the spring

And solving for x,

$x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(2)(1.30)^2}{600}}=0.075 m$

5 0

3 years ago