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A heat engine performs (245 + A) J of work in each cycle while also delivering (142 + B) J of heat to the cold reservoir. Find t

Ganezh [65]

Answer:

The value is $\eta = 54.4 \%$

Explanation:

From the question we are told that

The work input is $W = ( 245 + A ) \ J$

The heat delivered is $Q = (142 + B) \ J$

The value of A is A = 14

The value of B is B = 72

Generally the efficiency of the heat engine is mathematically represented as

$\eta = \frac{W}{Q_t}$

Here $Q_t$ is the total out energy produce by the heat engine and this is mathematically represented as

$Q_t= Q + W$

=> $Q_t= 245 + A + 142 + B$

=> $Q_t= 390 + A+B$

So

$\eta = \frac{245 + A }{390 + A+ B}$

=> $\eta = 0.544$

=> $\eta = 0.544 *100$

=> $\eta = 54.4 \%$

5 0

2 years ago

Does the charge that flows into the capacitor during the charging go all the way through the capacitor and back to the battery,

snow_lady [41]

The capacitor is used to store electric charge.That is what makes capacitors special. 
The charge that flows into the capacitor is stored on the plate of the capacitor that the source voltage is connected to. When current flows into a capacitor, the charges get “stuck” on the plates because they can’t get past the insulating dielectric. One plate is positively charged and the other negatively The stationary charges on these plates create an electric field. When charges group together on a capacitor like this, the cap is storing electric energy just as a battery might store chemical energy.

8 0

3 years ago

The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.32 with the floor. If t

coldgirl [10]

Answer:

The shortest braking distance is 35.8 m

Explanation:

To solve this problem we must use Newton's second law applied to the boxes, on the vertical axis we have the norm up and the weight vertically down

On the horizontal axis we fear the force of friction (fr) that opposes the movement and acceleration of the train, write the equation for each axis

Y axis

N- W = 0

N = W = mg

X axis

-Fr = m a

-μ N = m a

-μ mg = ma

a = μ g

a = - 0.32 9.8

a = - 3.14 m/s²

We calculate the distance using the kinematics equations

Vf² = Vo² + 2 a x

x = (Vf² - Vo²) / 2 a

When the train stops the speed is zero (Vf = 0)

Vo = 54 km/h (1000m/1km) (1 h/3600s)= 15 m/s

x = ( 0 - 15²) / 2 (-3.14)

x= 35.8 m

The shortest braking distance is 35.8 m

7 0

2 years ago

At 15°C air is transmitted at 340 m/s. Express this speed in Kilometers per hour.

EleoNora [17]

Answer:

1224km/hr

Explanation:

To convert from m/s to km/hr

1000m = 1km

Divide both sides by 1000

1m = 1/1000 km................. (1)

60×60 seconds = 1 hr

3600s = 1hr

Divide both sides by 3600

1s = 1/3600 .............(2)

Divide (2) by (1)

1m/s = 1/1000 ÷ 1/3600 km/hr

1m/s = 1/1000 × 3600/1 km/hr

1m/s = 3600/1000 km/hr

1m/s = 3.6 km/hr .............(3)

To convert 340m/s to km/hr

Multiply (3) by 340

1× 340m/s = 3.6 × 340 km/hr

340m/s = 1224km/hr

I hope this was helpful, please mark as brainliest

7 0

3 years ago

Read 2 more answers

A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio

DanielleElmas [232]

Answer:

$\alpha =-2.2669642\times^{-10}rad/s^2$

Explanation:

Angular acceleration is defined by $\alpha =\frac{\Delta \omega}{\Delta t}=\frac{\omega_f-\omega_i}{\Delta t}$

Angular velocity is related to the period by $\omega=\frac{2\pi}{T}$

Putting all together:

$\alpha =\frac{\frac{2\pi}{T_f}-\frac{2\pi}{T_i}}{\Delta t}=\frac{2\pi}{\Delta t}(\frac{1}{T_f}-\frac{1}{T_i})$

Taking our initial (i) point now and our final (f) point one year later, we would have:

$\Delta t=1\ year=(365)(24)(60)(60)s=31536000
s$

$T_i=0.0786s$

$T_f=0.0786s+7.03\times10^{-6}s$

So for our values we have:

$\alpha =\frac{2\pi}{\Delta t}(\frac{1}{T_f}-\frac{1}{T_i})=\frac{2\pi}{31536000s}(\frac{1}{0.0786s+7.03\times10^{-6}s}-\frac{1}{0.0786s})=-2.2669642\times^{-10}rad/s^2$

Where the minus sign indicates it is decelerating.

8 0

3 years ago