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2 years ago

8

A metal sample of mass M requires a power input P to just remain molten. When the heater is turned off, the metal solidifies in

a time T. The heat of fusion of this metal is

1 answer:

vovangra [49]2 years ago

4 0

Answer:

L = Pt/M

Explanation:

Power, P= Q/t = mL/t

we know that, (Q=m×l)

Now ⇒l= Pt/M

Thus l= Pt/M

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The unusually bright centers found in some galaxies are called ______.

Nimfa-mama [501]

<span>Active Galactic Nuclei.</span>

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2 years ago

How does the mass affect the speed as it rolls down a slope?

VikaD [51]

As we use the Kinetic energy and the equation is 1/2mv^2, changing its mass will change its speed and its energy. So more mass, more speed more energy. also the gravitational potential energy; mass x gravity x height; more mass and more height more speed as it go down to the slope! Hope it helps!

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2 years ago

34. A train, starting from rest, accelerates along the platform at a uniform rate of 0.6 m/s2. A passenger standing on the platf

ira [324]

Answer:

4.08 s

Explanation:

Let the passenger took "t" time to catch the train

so in this case the total distance moved by the train + 5 m = total distance moved by the passenger

so we will have

distance moved by train is given as

$d_1 = \frac{1}{2}(0.6) t^2$

also the distance moved by passenger

$d_2 = \frac{1}{2}(1.2) t^2$

so we will have

$d_1 + 5 = d_2$

$0.3 t^2 + 5 = 0.6 t^2$

$0.3 t^2 = 5$

t = 4.08 s

3 0

2 years ago

how to find the change in spring potential energy of the block-spring system from when the block is released to when the block h

Serhud [2]

Answer: Subtract the kinetic energy of the block at x=0.02mx=0.02m from the kinetic energy of the block at x=0.00mx=0.00m.

Explanation:

7 0

1 year ago

The half-life of plutonium 239 is 24,200 years. Assume that the decay rate is proportional to the amount. Determine the amount o

kotykmax [81]

Answer:

time taken is equal to 14,156 years

Explanation:

we know,

$Y=Ae^{-kt}$

at t = 0

Y(0) = A

given that half life of plutonium 239 = 24,200

$\dfrac{A}{2}=Ae^{-kt}\\0.5=e^{-kt}\\k\times 24200 = ln(2)\\k = \dfrac{ ln(2)}{24200}$

$Y=Ae^{-kt}$

$\frac{3}{2} = e^{-kt}\\ln(1.5)=-\dfrac{ ln(2)}{24200}\times t\\t=-\dfrac{ln(1.5)\times 24200}{ ln(2)}\\t=14,156 \ years$

hence time taken is equal to 14,156 years

5 0

2 years ago