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garik1379 [7]
3 years ago
8

One cubic foot of water can store 312btu. a home requires 100,000 what is the volume​

Physics
1 answer:
Neko [114]3 years ago
7 0

Volume of water required to store 100,000 Btu of thermal energy is 320.51foot^{3} .

<u>Explanation:</u>

The complete question is : One cubic foot of water can store 312 Btu of thermal energy. On a cold winter day a well-constructed home may require 100,000 Btu of nighttime space heating. What is the volume of water required to store this energy? In this question , it's given that One cubic foot of water can store 312 Btu of thermal energy or 312 Btu takes 1 cubic foot of water ,So

1 Btu takes  \frac{1}{312} cubic foot of water

Therefore, 100,000 Btu takes:

⇒ volume = \frac{1}{312}(100,000) foot^{3}

⇒ volume = \frac{100,000}{312} foot^{3}

⇒ volume = 320.51 foot^{3}

∴ Volume of water required to store 100,000 Btu of thermal energy is 320.51foot^{3} .

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What is the threshold velocity vthreshold(water) (i.e., the minimum velocity) for creating Cherenkov light from a charged partic
VladimirAG [237]

Complete Question

The  complete question is shown on the first uploaded image  

Answer:

A

   v_w  =  2.256 *10^{8} \  m/s

B

  v_e  =  2.21 *10^{8} \  m/s

C

The  correct option is  B  

Explanation:

From the question we are told that

      The refractive  index of water is  n_w  =  1.33

      The  refractive  index of ethanol is  n_e  =  1.36

       

Generally the threshold velocity for creating Cherenkov light   from a charged particle as it travels through water is mathematically evaluated as

       v_w  =  \frac{c}{n_w }

Where  c is the speed of light with value  c =  3.0 *10^{8} \  m/s

       v_w  =  \frac{3.0 *10^{8}}{1.33 }

       v_w  =  2.256 *10^{8} \  m/s

Generally the threshold velocity for creating Cherenkov light   from a charged particle as it travels through water is mathematically evaluated as

            v_e  =  \frac{ c}{n_e }

  =>       v_e  =  \frac{3.0 *10^{8}}{1.36 }

=>          v_e  =  2.21 *10^{8} \  m/s

4 0
3 years ago
You charge an initially uncharged 65.7-mf capacitor through a 39.1-Ï resistor by means of a 9.00-v battery having negligible int
uysha [10]
In a RC-circuit, with the capacitor initially uncharged,  when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
Q(t) = Q_0 (1-e^{-t/\tau})
where t is the time, Q_0 = CV is the maximum charge on the capacitor at voltage V, and \tau = RC is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.

1) Using R=39.1 \Omega and C= 65.7 mF=65.7\cdot 10^{-3}F, the time constant of the circuit is:
\tau = RC=(39.1 \Omega)(65.7 \cdot 10^{-3}F)=2.57 s

2) To find the charge on the capacitor at time t=1.95 \tau, we must find before the maximum charge on the capacitor, which is
Q_0 = CV=(65.7 \cdot 10^{-3}F)(9 V)=0.59 C
And then, the charge at time t=1.95 \tau is equal to
Q(1.95 \tau) = Q_0 (1-e^{-t/\tau})=(0.59 C)(1-e^{-1.95})=0.51 C

3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be Q_0 = 0.59 C. We can see this also from the previous formule, by using t=\infty:
Q(t) = Q_0 (1-e^{-\infty})=Q_0(1-0) = 0.59 C

4 0
3 years ago
Please answer this im not sure
vovikov84 [41]

Answer:

the answer is B

8 0
3 years ago
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Elza [17]

Answer:

maybe its heat sorry if it's wrong

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4 0
3 years ago
Show that the Mass spring system executes simple harmonic motion(SHM)?​
murzikaleks [220]

Explanation:

Show that the motion of a mass attached to the end of a spring is SHM

Consider a mass "m" attached to the end of an elastic spring. The other end of the spring is fixed

at the a firm support as shown in figure "a". The whole system is placed on a smooth horizontal surface.

If we displace the mass 'm' from its mean position 'O' to point "a" by applying an external force, it is displaced by '+x' to its right, there will be elastic restring force on the mass equal to F in the left side which is applied by the spring.

According to "Hook's Law

F = - Kx ---- (1)

Negative sign indicates that the elastic restoring force is opposite to the displacement.

Where K= Spring Constant

If we release mass 'm' at point 'a', it moves forward to ' O'. At point ' O' it will not stop but moves forward towards point "b" due to inertia and covers the same displacement -x. At point 'b' once again elastic restoring force 'F' acts upon it but now in the right side. In this way it continues its motion

from a to b and then b to a.

According to Newton's 2nd law of motion, force 'F' produces acceleration 'a' in the body which is given by

F = ma ---- (2)

Comparing equation (1) & (2)

ma = -kx

Here k/m is constant term, therefore ,

a = - (Constant)x

or

a a -x

This relation indicates that the acceleration of body attached to the end elastic spring is directly proportional to its displacement. Therefore its motion is Simple Harmonic Motion.

5 0
3 years ago
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