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2 years ago

1. Calculate the total heat content of 10 kg of ice at -23°C.?

Physics

1 answer:

vesna_86 [32]2 years ago

8 0

To solve the problem, we must know the heat capacity of ice and water.

For Cp = 2090 J/kg C

H = mCpT

H = (10 kg) ( 2090 J/ Kg C) ( -23 C)

H = - 480700 J

For water Cp = 4180 j/kg C

H = (100 kg) ( 4180 J/kg C) ( 60 C)

H = 2508000 J

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In an RC-circuit, a resistance of R=1.0 "Giga Ohms" is connected to an air-filled circular-parallel-plate capacitor of diameter

solniwko [45]

Answer:

$\tau = 1\ ms$

Explanation:

First we need to find the capacitance of the capacitor.

The capacitance is given by:

$C = \epsilon_0 * area / distance$

Where $\epsilon_0$ is the air permittivity, which is approximately 8.85 * 10^(-12)

The radius is 12/2 = 6 mm = 0.006 m, so the area of the capacitor is:

$Area = \pi * radius^{2}\\Area = \pi * 0.006^2\\Area = 113.1 * 10^{-6}\ m^2$

So the capacitance is:

$C = \frac{8.85 * 10^{-12} * 113.1 * 10^{-6}}{0.001}$

$C = 10^{-12}\ F = 1\ pF$

The time constant of a rc-circuit is given by:

$\tau = RC$

So we have that:

$\tau = 10^{9} * 10^{-12} = 10^{-3}\ s = 1\ ms$

7 0

2 years ago

Which part of an Adam is actively exchanged or shared in a chemical bond?

Lemur [1.5K]

Atom* the particles are (Electrons)

5 0

2 years ago

A 40.0 kg wheel, essentially a thin hoop with radius 0.810 m, is rotating at 438 rev/min. It must be brought to a stop in 21.0 s

garik1379 [7]

Answer:

(a) Workdone = -27601.9J

(b) Average required power = 1314.4W

Explanation:

Mass of hoop,m =40kg

Radius of hoop, r=0.810m

Initial angular velocity Winitial=438rev/min

Wfinal=0

t= 21.0s

Rotation inertia of the hoop around its central axis I= mr²

I= 40 ×0.810²

I=26.24kg.m²

The change in kinetic energy =K. E final - K. E initail

Change in K. E =1/2I(Wfinal² -Winitial²)

Change in K. E = 1/2 ×26.24[0-(438×2π/60)²]

Change in K. E= -27601.9J

(a) Change in Kinetic energy = Workdone

W= 27601.9J( since work is done on hook)

(b) average required power = W/t

=27601.9/21 =1314.4W

4 0

2 years ago

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Pls help its about baseball and energy (one question) '^'

avanturin [10]

The answer would be A. As the bat is swung, it gains kinetic energy. But once it hits the ball, it loses, or transfers, it’s kinetic energy to the ball.

5 0

3 years ago

A copper rod at 25°C is 2.5 m long. How long would it take a sound to move through the rod from one end to another? how would I

andrew-mc [135]

Length of the copper rod = 2.5 meters
Speed at which sound travels through copper = 3560 meter per second
Let us assume the time taken
by sound to cover the given distance = x seconds
We already know that
Speed = Distance/ Time
Then
Time = Distance/ Speed
x = 2.5/3560 seconds
= 0.0007 seconds.
This can ve done by hitting one end of the rod and then receiving the sound at the other end and using the stop clock to measure the time taken.

5 0

3 years ago

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