What three factors affect the viscosity of magma

Explain how gravity and friction work to slow down a kicked soccer ball.

goblinko [34]

Use your feet to stop it since it is soccer you can't use your hands!!!! P.S. you can't use gravity.

5 0

3 years ago

A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm

bogdanovich [222]

Answer:

A) 2.75 m/s B) 0.1911 m C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = $\sqrt\frac{k}{M}$

= $\sqrt\frac{21}{0.1}$

= 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic Motion is given as:

$x = A sin (\omega t + \phi)\\$

Differentiating this to find speed of the block immediately before the collision:

$v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\$

As bullet strikes at equilibrium position so,

φ = 0

t= 2nπ

⇒ cos (ω₀t + φ) = 1

⇒ $v= A\omega_{o}$

$v=(.19)(14.49)\\v= 2.75 ms^{-1}$

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

$x= Bsin(\omega t + \phi)$

To find B, consider law of conservation of energy

$K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2} \\P.E = \frac{1}{2} kB^{2}$

$\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m$

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

$T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}$

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

$T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec$

5 0

3 years ago

The 630-nm light from a helium-neon laser irradiates a grating. The light then falls on a screen where the first bright spot is

dimaraw [331]

Answer:

464.8 nm

Explanation:

The second wavelength of light can be calculated using the next equation:

$\lambda = \frac{x*d}{L}$

<u>Where:</u>

<em>λ : is the wavelength of light</em>

<em>x: is the distance from the central maximum</em>

<em>d: is the distance between the spots </em>

<em>L: is the lenght from the screen to the bright spot</em>

For the first wavelength of light we have:

$\lambda_{1} = \frac{x_{1}*d}{L}$

$630 \cdot 10^{-9} m = \frac{0.61 m*d}{L}$

$\frac{d}{L} = \frac{630 \cdot 10^{-9} m}{0.61 m} = 1.033 \cdot 10^{-6}$ (1)

For the second wavelength of light we have:

$\lambda_{2} = \frac{x_{2}*d}{L}$

$\lambda_{2} = 0.45 m*\frac{d}{L}$ (2)

By entering equation (1) into equation (2) we have:

$\lambda_{2} = 0.45 m* 1.033 \cdot 10^{-6} = 4.648 \cdot 10^{-7} m = 464.8 nm$

Therefore, the second wavelength is 464.8 nm

I hope it helps you!

3 0

3 years ago

A block of wood is tied to the bottom of a large container of water so that the block is completely submerged. The density of th

jeka57 [31]

Answer:

58.9 N

Explanation:

The wood is buoyed up by the mass of the water is displaces

wood volume = .030 m^3 = 30 000 cm^3

mass of water displaced = 30 000 g = 30 kg

wood mass = 800 kg/m^3 * .030 m^3 = 24 kg

so tension in string = mg = ( 30 kg - 24 kg) * 9.81 m/s^2 = 58.9 N

6 0

2 years ago

plz halp! For the PV graph shown above, explain whether heat is added to the gas or removed from the gas during the process indi

Delicious77 [7]

Answer:

There's no PV graph. kindly provide it so we help

5 0

3 years ago