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Katyanochek1 [597]

3 years ago

14

Does heat always give of heat

2 answers:

Marysya12 [62]3 years ago

6 0

If you meant doest heat always give off heat than yes

iren2701 [21]3 years ago

3 0

Yes it does (not to be mean its kinda stupid for you to ask)

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Tính chỉ số ampe kế thay ampe kế bằng vô kế có điện trở rất lớn thì vô kế chỉ bao nhiêu

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Picture a ball traveling at a constant speed around the inside of a circular Structure. is the ball acceleration? Explain why?

baherus [9]

Yes. On a circular path, the direction of motion is constantly changing. Change of direction is acceleration, even at constant speed.

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A force of 20N acts upon a body whose weight is 9.8 N. What is the mass of the body and how much is its acceleration?

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Compare between chromosphere, corona and photosphere

gladu [14]

Answer:

Explained

Explanation:

Photosphere: The lowest layer of the sun is called photo sphere . It is about 300 miles thick from the surface. It is the source of solar flares. It is marked by bright bubbling granules of plasma.

chromosphere emits a reddish glow as the super heated hydrogen burns off but the red rim can only be seen during total solar eclipse.

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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