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andreev551 [17]
3 years ago
14

Eva is in a closed, dark room. She uses her arm muscles to turn on a lamp. When she moves her hand closer to the lamp, the light

heats her hand. The light bounces off the pages of a book and into her eyes.
If this room is a closed system, what happens to the total amount of energy in the room?

A.
It remains constant.

B.
It decreases.

C.
It increases, then decreases.

D.
It increases.
Physics
2 answers:
guapka [62]3 years ago
5 0
Its a tricky one but my guess would be D if she turned on the lamp causing energy if would have increased the amount in the specificed room x
12345 [234]3 years ago
4 0

Well, we know that the total energy in a closed system remains constant.

The problem with the story of Eva is that she is not in a closed system. 
If the dark room were really a closed system, then she could press the
button or turn the switch all day, and the lamp could not light.  It needs
electrical energy coming in from somewhere in order to turn on.

Let's say that Eva used her arm muscles to strike a match and light the
candle on the table.  Then we would have have food energy, muscle
energy, chemical energy in the match, chemical energy in the candle,
heat and light energy coming out of the candle, heat energy soaking into
her hand, light energy bouncing off of the book and into her eyes ... all
going on during the story, and the sum total of all of them would remain
constant.
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lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L}

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

f_{(1  \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}}  }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L}  = 1.3225 \times f_1

f_{(1  \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1

f_{(1  \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz  = 740.6 \  Hz

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, L_{new} = 2/3·L

The value of the fundamental frequency will then be given as follows;

f_{(1  \, new)} =  \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \times \dfrac{2 \times L}{3} }  =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz =  840 \ Hz

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

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3 years ago
How is item A different from Item B?
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Explanation:

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8 0
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If you're going 80 mph how long does it take to go 80 miles
Elza [17]
1 mile. Is this a joke lol
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3 years ago
According to Bernoulli's fluid formula a An increase in the speed will lower the internal pressure b An increase in the speed wi
lorasvet [3.4K]

Answer:

a An increase in the speed will lower the internal pressure

Explanation:

Bernoulli's fluid formula

P_1+\frac{1}{2}\rho v_1^2+\rho gh_1=P_1+\frac{1}{2}\rho v_2^2+\rho gh_2

where

P = Pressure

ρ = Density of fluid

g = Acceleration due to gravity

h = Height

v = Velocity of fluid

If there is no change in height then we get

P_1+\frac{1}{2}\rho v_1^2=P_1+\frac{1}{2}\rho v_2^2\\\Rightarrow P+\frac{1}{2}\rho v^2=constant

According to the Bernoulli's principle when the speed of the fluid is larger in a region of streamline flow the pressure is smaller in that region. From the above equation it can be seen that increase in speed should simultaneously reduce pressure in order for their sum to be constant.

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A car moving in a straight line starts at X=0 at t=0. It passesthe point x=25.0 m with a speed of 11.0 m/s at t=3.0 s. It passes
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Answer:

Average velocity v = 21.18 m/s

Average acceleration a = 2 m/s^2

Explanation:

Average speed equals the total distance travelled divided by the total time taken.

Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)

Average acceleration equals the change in velocity divided by change in time.

Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

Where;

v1 and v2 are velocities at time t1 and t2 respectively.

And x1 and x2 are positions at time t1 and t2 respectively.

Given;

t1 = 3.0s

t2 = 20.0s

v1 = 11 m/s

v2 = 45 m/s

x1 = 25 m

x2 = 385 m

Substituting the values;

Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)

v = (385-25)/(20-3)

v = 21.18 m/s

Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

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a = 2 m/s^2

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