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kow [346]
3 years ago
11

A system experiences a change in internal energy of 36 kJ in a process that involves a transfer of 14 kJ of heat into the surrou

ndings. Simultaneously, which of the following is true?
a. Q= 14 u=36
b. U=Q-W
c. 14-36=22
d. 22 kJ of work is done on the system.
Physics
1 answer:
Andre45 [30]3 years ago
7 0

Answer:

b. U = Q - W

Explanation:

Given;

change in internal energy, ΔU = 36 kJ

heat transferred to the surroundings Q  = 14 kJ

Apply first law of thermodynamic; the change in internal energy is equal to heat added to the system minus work done by the system.

ΔU = Q - W

Since heat was lost to surroundings, Q = - Q

ΔU = (-Q) - W

36 kJ = -14 kJ - W

36 kJ + 14 kJ = - W

50 kJ = - W

W = - 50 kJ (the negative sign shows that work has been done on the system)

Thus, 50 kJ of work is done on the system.

The only correct answer in the given options is "b" U = Q-W

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A child has a mass of 35 kg. The child is running across a fiend and has a speed of 3 m/s. What is the kinetic energy of the chi
Sladkaya [172]

Answer:

Explanation:

Given the following data;

Mass = 35 kg

Velocity = 3 m/s

To find the kinetic energy of the child;

K.E = ½mv²

4 0
3 years ago
Can you answer this math homework? Please!
steposvetlana [31]
Using the count data and observational data you acquired, calculate the number of CFUs in the original sample
5 0
3 years ago
A cyclist traveling at constant speed of 12m/s when he passes a stationary bus.The bus starts moving just as the cyclist passes
Bogdan [553]

Answer:

A.) 8 seconds

B.) 16 seconds

C.) 48 m

Explanation:

Given that a cyclist traveling at constant speed of 12 m/s

and the bus accelerates uniformly at 1.5ms²

A.) The bus has the following parameters

Acceleration a = 1.5 m/s^2

Initial velocity U = 0. Since the bus is starting from rest.

Final velocity V = 12 m/s

Use equation one of linear motion.

V = U + at

Substitute V, U and a into the formula

12 = 0 + 1.5t

1.5t = 12

t = 12/1.5

t = 8 seconds

Therefore, the bus reach the same speed as the cyclist at 8 seconds.

B.) For the cyclist moving at constant speed, acceleration a = 0. Using second equation of motion

h = Ut + 1/2at^2

Since a = 0, the equation is reduced to:

h = Ut.

Also, for the bus,

h = Ut + 1/2at^2

Equate the two equations since the h is the same

Ut = Ut + 1/2at^2

Substitute all the parameters into the formula

12t = 0 + 1/2 × 1.5t^2

12t = 0.75t^2

0.75t = 12

t = 12/0.75

t = 16 seconds

Therefore, the bus takes 16 seconds to catch the cyclist

C.) Use third equation of linear motion.

V^2 = U^2 + 2as

Where s = distance

Substitute V, U and a into the formula

12^2 = 0 + 2 × 1.5 S

144 = 3S

S = 144/3

S = 48 m

8 0
3 years ago
at certain times the demand for electric energy is low and electric energy is used to pump water to a reservoir 45 m above the g
Readme [11.4K]

The mass of water that must be raised is 5.25\cdot 10^7 kg

Explanation:

Since the process is 70% efficiency, the power in output to the turbine can be written as

P_{out} = 0.70 P_{in}

where P_{in} is the power in input.

The power in input can be written as

P_{in} = \frac{W}{t}

where

W is the work done in lifting the water

t = 3 h = 10,800 s is the time elapsed

The work done in lifting the water is given by

W=mgh

where

m is the mass of water

g=9.8 m/s^2 is the acceleration of gravity

h = 45 m is the height at which the water is lifted

Combining the three equations together, we get:

P_{out} = 0.70 \frac{mgh}{t}

Where

P_{out} = 150 MW = 150\cdot 10^6 W

And solving for m, we find:

m=\frac{Pt}{0.70gh}=\frac{(1.50\cdot 10^6)(10800)}{(0.70)(9.8)(45)}=5.25\cdot 10^7 kg

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

3 0
3 years ago
An investigator collects a sample of a radioactive isotope with an activity of 490,000 Bq.48 hours later, the activity is 110,00
Daniel [21]

Answer:

The correct answer is "22.27 hours".

Explanation:

Given that:

Radioactive isotope activity,

= 490,000 Bq

Activity,

= 110,000 Bq

Time,

= 48 hours

As we know,

⇒ A = A_0 e^{- \lambda t}

or,

⇒ \frac{A}{A_0}=e^{-\lambda t}

By taking "ln", we get

⇒ ln \frac{A}{A_0}=- \lambda t

By substituting the values, we get

⇒ -ln \frac{110000}{490000} = -48 \lambda

⇒    -1.4939=-48 \lambda

                 \lambda = 0.031122

As,

⇒ \lambda = \frac{ln_2}{\frac{T}{2} }

then,

⇒ \frac{ln_2}{T_ \frac{1}{2} } =0.031122

⇒ T_\frac{1}{2}=\frac{ln_2}{0.031122}

         =22.27 \ hours  

3 0
3 years ago
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