Answer:
A) T1 = 566 k = 293°C
B) T2 = 1132 k = 859°C
Explanation:
A)
The average kinetic energy of the molecules of an ideal gas is givwn by the formula:
K.E = (3/2)KT
where,
K.E = Average Kinetic Energy
K = Boltzman Constant
T = Absolute Temperature
At 10°C:
K.E = K10
T = 10°C + 273 = 283 K
Therefore,
K10 = (3/2)(K)(283)
FOR TWICE VALUE OF K10:
T = T1
Therefore,
2 K10 = (3/2)(K)(T1)
using the value of K10:
2(3/2)(K)(283) = (3/2)(K)(T1)
<u>T1 = 566 k = 293°C</u>
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B)
The average kinetic energy of the molecules of an ideal gas is given by the formula:
K.E = (3/2)KT
but K.E is also given by:
K.E = (1/2)(m)(vrms)²
Therefore,
(3/2)KT = (1/2)(m)(vrms)²
vrms = √(3KT/m)
where,
vrms = Root Mean Square Velocity of Molecule
K = Boltzman Constant
T = Absolute Temperature
m = mass
At
T = 10°C + 273 = 283 K
vrms = √[3K(283)/m]
FOR TWICE VALUE OF vrms:
T = T2
Therefore,
2 vrms = √(3KT2/m)
using the value of vrms:
2√[3K(283)/m] = √(3KT2/m)
2√283 = √T2
Squaring on both sides:
(4)(283) = T2
<u>T2 = 1132 k = 859°C</u>
To help you I need to assume a wavelength and then calculate the momentum.
The momentum equation for photons is:
p = h / λ , this is the division of Plank's constant by the wavelength.
Assuming λ = 656 nm = 656 * 10 ^ - 9 m, which is the wavelength calcuated in a previous problem, you get:
p = (6.63 * 10 ^-34 ) / (656 * 10 ^ -9) kg * m/s
p = 1.01067 * 10^ - 27 kg*m/s which must be rounded to three significant figures.
With that, p = 1.01 * 10 ^ -27 kg*m/s
The answers are rounded to only 2 significan figures, so our number rounded to 2 significan figures is 1.0 * 10 ^ - 27 kg*m/s
So, assuming the wavelength λ = 656 nm, the answer is the first option: 1.0*10^-27 kg*m/s.
The equation for Hall voltage Vh is:
Vh=v*B*w, where v is the velocity of the strip, B is the magnitude of the magnetic field, and w is the width of the strip.
v=25 cm/s = 0.25 m/s
B=5.6 T
w= 1.2 mm = 0.0012 m
We input the numbers into the equation and get:
Vh= 0.25*5.6*0.0012 = 0.00168 V
The maximum Hall voltage is Vh= 0.00168 V.
If a coin is dropped at a relatively low altitude, it's acceleration remains constant. However, if the coin is dropped at a very high altitude, air resistance will have a significant effect. The initial acceleration of the coin will be the greatest. As it falls down, air resistance will counteract the weight of the coin. So, the acceleration will decrease. Although the acceleration decreases, the coin still accelerates, that is why it falls faster. When the air resistance fully counters the weight of the coin, the acceleration will become zero and the coin will fall at a constant speed (terminal velocity). So, the answer should be, The acceleration decreases until it reaches 0. The closest answer is.
a. The acceleration decreases.
<span>Hello!
We have the following data:
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Time (T) = ? (in minutes)
Power (P) = 3 kW → 3000 W
Energy (E) = 9 MJ → 9000000 J or (W/s)
Formula of the consumption of electric energy:
Solving:
How many minutes can it run for? (<span>Let's convert in minutes)
</span>
1 minute --------- 60 seconds
y minute --------- 3000 seconds
<span>Product of extremes equals product of means
</span>
I hope this helps! =)
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