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kobusy [5.1K]
1 year ago
15

Kimonoski takes a 9-minute shower every day. The shower uses about 1.8 gal per minute of water. He also uses 23 gallons of hot w

ater per day for clothes and dish washers. The hot water heats the water from 60 to 110 F. What is the total energy required per week for hot water?
Physics
1 answer:
ioda1 year ago
3 0

Answer:

Q_{week} = 458884.6\, BTU

Explanation:

The weekly water consumption of Kimonoski is:

m_{bath,week} = (62.4\,\frac{lbm}{ft^{3}})\cdot (1.8\,\frac{gal}{min} )\cdot (\frac{0.134\,ft^{3}}{1\,gal} )\cdot (\frac{1\,min}{60\,s} )\cdot (9\,min)\cdot (\frac{60\,s}{1\,min} )\cdot (7\,\frac{days}{week} )\cdot (1\,week)

m_{bath.week} = 948.205\,lbm

m_{others, week} = (62.4\,\frac{lbm}{ft^{3}})\cdot (23\,gal)\cdot (\frac{0.134\,ft^{3}}{1\,gal} )\cdot (7\,\frac{days}{week} )\cdot (1\,week)

m_{others, week} = 1346.218\,lbm

m_{week} = m_{bath,week} + m_{others, week}

m_{week} = 2294.423\,lbm

The total energy required per week for hot water is:

Q_{week} = m_{week}\cdot c_{p,water}\cdot \Delta T

Q_{week} =(2294.423\,lbm)\cdot (1\,\frac{BTU}{lbm\cdot ^{\textdegree}F} )\cdot (50^{\textdegree}F)

Q_{week} = 458884.6\, BTU

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Alchen [17]

Answer:

It continue to move forward at a constant velocity which will be slower than before the front thruster was fired.

Explanation:

Before the front thruster was fired, the spacecraft was already moving at a particular velocity.

After the malfunction, the front thruster is fired and then the force exerted by that front thruster slows the spacecraft down, as we are told.

By using the rear thruster to exert a force equal to that from the front thrusters, a force equal in magnitude to that of the front thrusters is added, cancelling out the effect of the front thrusters. Because the spacecraft is already moving at a slower speed at this point compared to the beginning, it continues to move at that speed.

8 0
2 years ago
If period of the pendulum in preceding sample problem were 24s how tall would the tower be ?
frutty [35]

Answer:

So length of pendulum is 143.129 m

Explanation:

We have given period of simple pendulum is 2 sec

We have to find the length of simple pendulum

Let the length of pendulum is l

Acceleration due to gravityg=9.8m/sec^2 is

Time period is given by T=2\pi \sqrt{\frac{l}{g}}

So 24=2\times 3.14\times  \sqrt{\frac{l}{9.8}}

\sqrt{\frac{l}{9.8}}=3.821

Squaring both side

{\frac{l}{9.8}}=14.60

l =143.129 m

So length of pendulum is 143.129 m

8 0
1 year ago
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A 1 mg ball carrying a charge of 2 x 10-8 C hangs from a
Fed [463]

Answer:

σ = 0.255*10^-3 C/m²

Explanation:

The Electric field Intensity act due to plate = σ/ε₀, where σ is surface charge density of plate.

At equilibrium ,

Upward force = downward force

Tcosθ = mg ----(1)

Assuming that the Forward force = backward force, then

Tsinθ = σq/ε₀

[ ∵ F = qE , ∴ F = qσ/ε₀ ] -----(2)

Dividing equation (2) by (1)

Tsinθ/Tcosθ = qσ/ε₀mg

⇒Tanθ = qσ/ε₀mg

σ = ε₀mg tanθ/q

Now substituting the values of

σ = (8.85*10^-12 * 1 * tan 30) / 2*10^-8

σ = (8.85*10^-12 * 0.5774) / 2*10^-8

σ = 5.11*10^-12 / 2*10^-8

σ = 0.255*10^-3 C/m²

7 0
1 year ago
Sir Marvin decided to improve the destructive power of his cannon by increasing the size of his cannonballs. Sir Seymour kept hi
maria [59]
We really can't tell from the given information. 
We don't know HOW MUCH Marv enlarged his cannonballs,
or HOW MUCH faster Seymour's balls became.

If we assume that they both, let's say, DOUBLED something,
then Seymour accomplished more, and the destructive capability
of his balls has increased more. 

I say that because the destructive capability of a cannonball is
pretty much just its kinetic energy when it arrives and hits the target.
Now, we all know the equation for kinetic energy.

                K.E.  =  (1/2) (mass) (speed-SQUARED) .

We can see right away that if Marv started shooting balls with
double the mass but at the same speed, then they have double
the kinetic energy of the old ones.

But if Seymour started shooting the same balls with double the SPEED,
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That's 4 times as much destructive capability as before.  

So we can say that when it comes to cannons and their balls and
smashing things to bits and terrorizing your opponents, if making
a bigger mess is better, then more mass is better, but more speed
is better-squared.
5 0
2 years ago
g A projectile is launched with speed v0 from point A. Determine the launch angle ! which results in the maximum range R up the
Svetlanka [38]

Answer:

The range is maximum when the angle of projection is 45 degree.

Explanation:

The formula for the horizontal range of the projectile is given by

R = \frac{u^{2}Sin2\theta }{g}

The range should be maximum if the value of Sin2θ is maximum.

The maximum value of Sin2θ is 1.

It means 2θ = 90

θ = 45

Thus, the range is maximum when the angle of projection is 45 degree.

If the angle of projection is 0 degree

R = 0

It means the horizontal distance covered by the projectile is zero, it can move in vertical direction.

If the angle of projection is 30 degree.

R = \frac{u^{2}Sin60 }{9.8}

R = 0.088u^2

If the angle of projection is 45 degree.

R = \frac{u^{2}Sin90 }{g}

R = u^2 / g

5 0
2 years ago
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