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Licemer1 [7]
3 years ago
10

What is the specific heat of an unknown substance if 2000 J of energy are required to raise the temperature of 4 grams of the su

bstance 5 degrees Celsius?
Physics
1 answer:
Dmitriy789 [7]3 years ago
7 0
The energy required to heat a substance is related by the formula:
Q = mCpΔT ; where Q is the energy, m is the mass of the substance, Cp is the specific heat capacity and ΔT is the change in temperature.
2000 = (4)(Cp)(5)
Cp = 100 Joules / g °C
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Gas pressure in a closed system is caused by _____.
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It is caused by she system letting off fumes and making gas pressure.
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A rigid container filled with a gas is placed in ice. What will happen to the pressure of the gas? And what will happen to the v
tankabanditka [31]

The volume of the can remains constant, and as the pressure increases with temperature, the can can't contain the pressure any longer.

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3 years ago
Light with a frequency of 7.30 x 1014 hz lies in the violet region of the visible spectrum. What is the wavelength of this frequ
Leona [35]

From the theory we know that:

c = λ / T

f = 1 / T

Where:

c = 3.10^{8} / m   (the speed of light)

λ is the wavelengh (in meters)

T is the period (in seconds)

f is the frequency (in Hz)

We were told that:

f = 7.30 . 10^{14}

And we want to find out the value of λ.

c = λ / T  

c = λ . 1/T

Swaping 1/T = f

c = λ . f

λ = c / f

λ = 3 . 10^{8} /  7.30 . 10^{14}

λ = 4.12 10^{-7} m

Response: 4.12 10^{-7} m = 412 nm

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6 0
4 years ago
Which of the following best describes a longitudinal wave?
Misha Larkins [42]

Answer:

A part not for sureokay

6 0
3 years ago
Common transparent tape becomes charged when pulled from a dispenser. If one piece is placed above another, the repulsice force
Dmitriy789 [7]

Answer:

Q = 1.095 x 10^-9 C

Let the force experienced by the top piece of tape be F

F = kQ²/r²

r = distance between the two pieces tape = 1.00cm = 1.00 x 10^ -2 m

1/4(pi)*Eo = k = 8.99 x 10^9 Nm²/C²

The electric force of repulsion between the two charges and the weight of the top piece of tape are equal so

F = kQ²/r² = mg

Where m is the mass of the top piece of tape and g is the acceleration due to gravity

On re-arranging the equation above,

Q² = mgr²/k

Q² = ((11.0 x 10^-6) x 9.8 x (1.00x10^-2)²)/(8.99 x 10^9)

Q = 1.095x10^-9 C

Explanation:

The charge Q on both pieces of tape are equal and both act with a force of repulsion on each other.

The force of repulsion between both tapes pushes the top piece of tape upwards. The weight of the top piece of tape acts vertically downward. Since the top tape is in a position of equilibrium, the two forces acting on the top piece of tape must be equal to each other. This assumption is backed up by newton's first law of motion which states that the summation of all forces acting on a body at rest must be equal to zero. That is

Fe (electric force) - Fg (gravitational force) = 0

Fe = Fg

kQ²/r² = mg

On substituting the respective values for all variables except Q and rearranging the equation Q = 1.09 x 10^-9

6 0
4 years ago
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