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sladkih [1.3K]
3 years ago
12

A rotating paddle wheel is inserted in a closed pot of water. The stirring action of the paddle wheel heats the water. During th

e process, 30 kJ of heat is transferred to the water and 5 kJ is lost to the surrounding air. The paddle wheel work amounts to 500 J. Determine the final energy of the system if its initial energy is 10 kJ.
Physics
1 answer:
fgiga [73]3 years ago
3 0

Answer:

the final energy of the system is 35.5 kJ.

Explanation:

Given;

initial energy of the system, E₁ = 10 kJ

heat transferred to the system, q₁  30 kJ

Heat lost to the surrounding, q₂ = 5kJ

heat gained by the system, Q = q₁ - q₂ = 30 kJ - 5kJ = 25 kJ

work done on the system, W = 500 J = 0.5 kJ

Apply first law of thermodynamic,

ΔU = Q + W

where;

ΔU  is change in internal energy

Q is the heat gained by the system

W is work done on the system

ΔU = 25kJ + 0.5 kJ

ΔU = 25.5 kJ

The final energy of the system is calculated as;

E₂ = E₁ + ΔU

E₂ = 10 kJ + 25.5 kJ

E₂ =  35.5 kJ.

Therefore, the final energy of the system is 35.5 kJ.

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What electrical force dies a Uranium nucleus exert on one of its inner electrons, located at a distance of 175 picometers (= 1.7
AlekseyPX

Answer:

correct option is d) 7.0 x 10^-7 N

Explanation:

given data

distance = 175 picometers = 1.75 × 10^{-10}  m

to find out

electrical force

solution

we know atomic no of uranium is 92

and charge on electron is = 1.6 × 10^{-19} C

and electrical force is express as

electrical force = \frac{1}{4 \pi \epsilon _o} \frac{q1q2}{r^2}    .............1

put here value we get

electrical force = 9*10^9 \frac{92*(1.6*10^{-19})^2}{(1.75*10^{-10})^2}

electrical force =  6.921 × 10^{-7} N

so correct option is d) 7.0 x 10^-7 N

5 0
3 years ago
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A 9V battery is directly connected to each of 3 LED bulbs. Select the statement that accurately describes this circuit.
dem82 [27]

Answer:

where are the statements?

Explanation:

8 0
3 years ago
Which process produces rising air when mountains push the air upward?
kicyunya [14]

Answer;

A. orographic lifting

Explanation;

Orographic lifting is a process that takes place when an air mass is forced from a low elevation to a higher elevation as it moves over rising terrain.

This can be explained by;  When air is blocked by mountains, it cannot go through these mountains, As it ascends or moves up the mountain, the air then cools as it rises and when it cools to its saturation point, the water vapor condenses and cloud forms.

These clouds formed  are known as orographic clouds, that develop in response to the lifting forced by the topography the earth.

5 0
3 years ago
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A 2.00 kg block hangs from a spring balance calibrated in Newtons that is attached to the ceiling of an elevator.(a) What does t
tresset_1 [31]

Answer:

Part a)

Reading = 2.00 kg

Part b)

Reading = 2.00 kg

Part c)

Reading = 4.04 kg

Part d)

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

Explanation:

Part a)

When elevator is ascending with constant speed then we will have

F_{net} = 0

T - mg = 0

T = mg

So it will read same as that of the mass

Reading = 2.00 kg

Part b)

When elevator is decending with constant speed then we will have

F_{net} = 0

T - mg = 0

T = mg

So it will read same as that of the mass

Reading = 2.00 kg

Part c)

When elevator is ascending with constant speed 39 m/s and acceleration 10 m/s/s then we will have

F_{net} = ma

T - mg = ma

T = mg + ma

Reading is given as

Reading = \frac{mg + ma}{g}

Reading = 2.00\frac{9.81 + 10}{9.81}

Reading = 4.04 kg

Part d)

Here the speed of the elevator is constant initially

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

3 0
3 years ago
In the Bohr model of hydrogen, the electron moves in a circular orbit around the nucleus. (a) Determine the orbital frequency of
Airida [17]

Answer:

(a) 6.567 * 10^15 rev/s or hertz

(b) 8.21 * 10^14 rev/s or hertz

Explanation:

Fn= 4π^2k^2e^4m * z^2/(h^3*n^3)

Where Fn is frequency at all levels of n.

Z = 1 (nucleus)

e = 1.6 * 10^-19c

m = 9.1 * 10^-31 kg

h = 6.62 * 10-34

K = 9 * 10^9 Nm2/c2

(a) for groundstate n = 1

Fn = 4 * π^2 * (9*10^9)^2*(1.6*10^-19)^4* (9.1 * 10^-31) * 1 / (6.62 * 10^-31)^3 = 6.567 * 10^15 rev/s

(b) first excited state

n = 1

We multiple the groundstate answer by 1/n^3

6.567 * 10^15 rev/s/ 2^3

F2 = 8.2 * 10^ 14 rev/s

3 0
3 years ago
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