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2 years ago

If a heat engine takes in 4565 kJ and gives up 2955 kJ during one cycle, what is the engine’s efficiency?

Physics

1 answer:

vodomira [7]2 years ago

5 0

The engine efficiency is 64.73 %

Explanation:

Given data

To find the engine’s efficiency we have the formula,

Energy input- 4565 KJ

Energy output - 2955KJ

Efficiency= energy output/ energy input ×100%

=2955/4565

=0.6473 ×100

η =64.73 %

The engine efficiency is 64.73 %

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DerKrebs [107]

Melting ice would damage this polar bears habitat meaning the polar bear may decrease

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2 years ago

A toy cannon uses a spring to project a 5.24-g soft rubber ball. The spring is originally compressed by 5.01 cm and has a force

salantis [7]

Answer:

Speed will be equal to 1.40 m/sec

Explanation:

Mass of the rubber ball m = 5.24 kg = 0.00524 kg

Spring is compressed by 5.01 cm

So x = 5.01 cm = 0.0501 m

Spring constant k = 8.08 N/m

Frictional force f = 0.031 N

Distance moved by ball d = 15.8 cm = 0.158 m

Energy gained by spring

$KE=\frac{1}{2}kx^2=\frac{1}{2}\times 8.08\times 0.0501^2=0.0101J$

Energy lost due to friction

$W=Fd=0.031\times 0.158=0.0048J$

So remained energy to move the ball = 0.0101 - 0.0048 = 0.0052 J

This energy will be kinetic energy

$\frac{1}{2}mv^2=0.0052$

$\frac{1}{2}\times 0.00524\times v^2=0.0052$

v = 1.40 m/sec

7 0

2 years ago

EASY BRAINLIEST PLEASE HELP!!

Rudiy27

Answer:

I think the awnser is B (but don't qoute me on that) if its right then yay but if its wrong im sorry

Explanation:

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2 years ago

Read 2 more answers

I was driving along at 20 m/s, trying to change a CD and not watching where I was going. When I looked up, I found myself 45 m f

cricket20 [7]

Answer:

$a=4.44\frac{m}{s^2}$

Explanation:

First we have to find the time required for train to travel 60 meters and impact the car, this is an uniform linear motion:

$t=\frac{d}{v}\\\\t=\frac{60m}{30\frac{m}{s}}=2s$

The reaction time of the driver before starting to accelerate was 0.50 seconds. So, remaining time for driver is 1.5 seconds.

Now, we have to calculate the distance traveled for the driver in this 0.5 seconds before he start to accelerate. Again, is an uniform linear motion:

$d=vt\\d=20\frac{m}{s}(0.5s)=10m$

The driver cover 10 meters in this 0.5 seconds. So, the remaining distance to be cover in 1.5 seconds by the driver are 35 meters. We calculate the minimum acceleration required by the car in order to cross the tracks before the train arrive, Since this is an uniformly accelerated motion, we use the following equation:

$d=v_0t+\frac{1}{2}at^2\\a=\frac{2(d-v_0t)}{t^2}\\a=\frac{2(35m-20\frac{m}{s}*1.5s}{(1.5s)^2}\\a=4.44\frac{m}{s^2}$

7 0

3 years ago

What speed must an electron have if its momentum is to be the same as that of an x-ray photon with a wavelength of 0.30?

lorasvet [3.4K]

The momentum of the x-ray photon is p = h/lambda . Lambda is the wavelength (0.30nm=3x10^(-9)m) and h is Planck's constant,(h=6.62607004 × 10-34 m2 kg / s).The momentum is: 2.2 x 10^(-25).

The momentum can be calculated also as: p=mv, where m is the mass of the electron and v is the speed.

So v=p/m,p is known,and also the mass of the electron (m=9.10938356 × 10-31 kilograms).

v=2.2 x 10^(-25)/9.10938356 × 10-31 kilograms=0.24 x 10^6 m/s

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2 years ago