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3 years ago

What is the formula for calculating the efficiency of a heat engine? Answers:

Physics

1 answer:

Setler [38]3 years ago

7 0

Efficiency of heat engine is determined by the ratio of difference in temperature of cold from hot reservoir to the temperature of hot reservoir over temperature of hot reservoir.

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When a trebuchet counterweight is hoisted by soldiers, what form of energy is

vladimir1956 [14]

10. A safe place to use the trebuchet would be away from other buildings and people. A good example of a place would be a large field with no nearby structures.
14. Many factors need to be kept consistent throughout the experiment. One example of a variable that would need to be consistent is the weight and size of the projectile.
15. It is important to do many trials so that you can make sure that the results of each trial are nearly the same. If they are all vastly different, then it means that something has gone wrong.

Sorry I was only able to answer a few questions, but I hope these few answers help! :)

4 0

3 years ago

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Isabella drops a pen off her balcony by accident while celebrating the successful completion of a physics problem.

Studentka2010 [4]

Answer:

2 seconds

Explanation:

v = at + v₀

19.62 m/s = (9.81 m/s²) t + 0 m/s

t = 2 s

6 0

3 years ago

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A converging lens has a focal length of 20 cm. An object 1 cm tall is placed 10 cm from the center of the lens. What is the heig

SCORPION-xisa [38]

Answer: 2 cm

Explanation:

Given , for a converging lens

Focal length : $f=20\ cm$

Height of object : $h=1\ cm$

Object distabce from lens : $u=-10\ cm$

Using lens formula: $\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$ , we get

$\dfrac{1}{20}=\dfrac{1}{v}+\dfrac{1}{10}$ , where v = image distance from the lens.

On solving aboive equation , we get

$\dfrac{1}{v}=\dfrac{1}{20}-\dfrac{1}{10}=\dfrac{1-2}{20}=\dfrac{-1}{20}\Rightarrow\ v=-20\ cm$

Formula of Magnification : $m=\dfrac{v}{u}=\dfrac{h'}{h}$ , where h' is the height of image.

Put value of u, v and h in it , we get

$\dfrac{-20}{-10}=\dfrac{h'}{1}\\\\\Rightarrow\ h'=2\ cm$

Hence, the height of the image is 2 cm.

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2 years ago

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15

Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

$T-mg = m\frac{v^2}{R}$

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

$m \frac{v^2}{R}$ is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

$T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N$

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

$T=m\frac{v^2}{R}$

And by substituting the numerical values, we find

$T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N$

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

$T+mg=m\frac{v^2}{R}$

And substituting the numerical values and re-arranging it, we find

$T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N$

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

$T=0\\mg = m\frac{v^2}{R}$

and re-arranging the equation, we find

$v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s$

7 0

3 years ago

2. A rock is shot straight up into the air with a slingshot that had been stretched 0.30 m. Assume

Luba_88 [7]

Answer:

Explanation:

4 0

3 years ago

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