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Agata [3.3K]
3 years ago
5

Please help!! Also please explain if possible...

Physics
1 answer:
solmaris [256]3 years ago
3 0

This is third equation of motion. As there is no values given, I assume you just have to keep variable 'a' (representing acceleration) on one side of the equation and all other variables on the other side.

v_{f}^{2}= v_{o}^{2}+2a(x_{f}-x_{o})

v_{f}^{2}- v_{o}^{2}=2a(x_{f}-x_{o})

\frac{v_{f}^{2}- v_{o}^{2}}{2(x_{f}-x_{o})}=a

OR

a=\frac{v_{f}^{2}- v_{o}^{2}}{2(x_{f}-x_{o})}

Here, v_{o} is initial velocity,

v_{f} is final velocity ,

x_{o} is initial position,

x_{f} is final position, and

a is acceleration.

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A boat takes off from the dock at 2.5 m/s and speeds up at 4.2 m/s2 for 6.0 s. How far is the boat traveled?
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The answer is in attachment.

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A lamp has a power of 120 w and is left on for 24 hours. A television has a power of 350 W and is left on for 3 hours. Which sta
soldi70 [24.7K]
<h2>Hello!</h2>

The answer is C. The television uses 6,588,000 J less than the lamp.

<h2>Why?</h2>

First, we must know that 1 Wh( Watt consumed in 1 hour) = 3600 J

Wh is the unit that express the energy amount of consumed in 60 minutes/ 1 hour.

So,

From the enunciate information we can calculate the energy consumed by the lamp and the television.

For the lamp, we have that it left on for 24 hours and it has a power of 120

For the television, we have that it left on for 3 hours and it has a power of 350W

Calculating the energy consumed (Wh) we have:

Wh= Watt*Time= 120 Watt* 24 hours= 2880 Wh

and converting it to Joule, we have:

2880 Wh*\frac{3600 J}{1 Wh}= 10,368,000 J

Doing the same for the television, we have:

Wh= 350 Watt* 3 hours= 1050 Wh

and converting it to Joule, we have:

1050 Wh*\frac{3600 J}{1 Wh}= 3,780,000 J

Finally, we have

10,368,000J(Lamp)-3,780,000J(Television)= 6,588,000J:

Then having calculated both consume, we have that the television uses 6,588,000 J less than the lamp.

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3 years ago
Read 2 more answers
A piston–cylinder device initially contains 2 L of air at 100 kPa and 25°C. Air is now compressed to a final state of 600 kPa an
bagirrra123 [75]

Answer:

a. The energy of the air at the initial and the final states is 0kJ and 0.171kJ respectively

b. 0.171kJ

c. 0.143

Explanation:

a.

Because there are same conditions of the state of air at the surroundings and at the Initial stage, the energy of air at the Initial stage is 0kJ.

Calculating energy at the final state;

We start by calculating the specific volume of air in the environment and at the final state.

U2 = At the final state, it is given by

RT2/P2

U1= At the Initial state, it is given by

RT1/P1

Where R = The gas constant of air is 0.287 kPa.m3/kg

T2 = 150 + 273 = 423K

T1 = 25 + 273 = 298K

P2 = 600KPa

P1 = 100KPa

U2 = 0.287 * 423/600

U2 = 0.202335m³/kg

U1 = 0.287 * 298/100

U1 = 0.85526m³/kg

Then we Calculate the mass of air using ideal gas relation

PV = mRT

m = P1V/RT1 where V = 2*10^-3kg

m = 100 * 2 * 10^-3/(0.287 * 298)

m = 0.00234kg

Then we calculate the entropy difference, ∆s. Which is given by

cp2 * ln(T2/T1) - R * ln(P2/P1)

Where cp2 = cycle constant pressure = 1.005

∆s = 1.005 * ln (423/298) - 0.287 * ln(600/100)

∆s = -0.1622kJ/kg

Energy at the final state =

m(E2 - E1 + Po(U2 - U1) -T0 * ∆s)

E2 and E1 are gotten from energy table as 302.88 and 212.64 respectively

Energy at the final state

= 0.00234(302.88 - 212.64 + 100(0.202335 - 0.85526) - 298 * -0.1622)

Energy at the final state = 0.171kJ

b.

Minimum Work = ∆Energy

Minimum Work = Energy at the final state - Energy at the initial state

Minimum Work = 0.171 - 0

Minimum Work done = 0.171kJ

c. The second-law efficiency of this process is calculated by ratio of meaningful and useful work

= 0.171/1.2

= 0.143

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