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2 years ago

All heat engines involve _____. the refrigeration cycle evaporation and condensation combustion

Physics

2 answers:

AURORKA [14]2 years ago

7 0

The answer is Combustion took the test and made a 93

pishuonlain [190]2 years ago

4 0

Combustion is the answer i just took the test

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A quantity must be divided by multiples of ten when converting from a larger unit to a smaller unit.

sineoko [7]

Answer:

what is the answers? i cant help you without the answers

Explanation:

5 0

3 years ago

A 75-m-long train begins uniform acceleration from rest. the the train has a speed of 23 m/s when it passes a railway worker tan

Lapatulllka [165]

Answer:

acceleration a = 1.04 m/s2

Explanation:

Assume the train has a speed of 23m/s when the last car passes the railway workers. Once this happens the last car would have traveled a total distance of the 180m distance between the railway worker standing 180 m from where the front of the train started plus the 75m distance from the first car to the last car:

s = 75 + 180 = 255 m

We can use the following equation of motion to find out the distance traveled by the car:

$v^2 - v_0^2 = 2as$ where v = 23 m/s is the velocity of the car when it passes the worker, $v_0$ = 0m/s is the initial velocity of the car when it starts, a m/s2 is the acceleration, which we are looking for.

$23^2 - 0^2 = 2*a*255$

$510a = 529$

$a = 529 / 510 = 1.04 m/s^2$

8 0

3 years ago

A bullet with mass 1.0kg and velocity 180 m/s is brought to rest in 0.02 s by a sandbag.assuming constant acceleration in the sa

kotykmax [81]

Hello
The bullet is moving by uniformly accelerated motion.
The initial velocity is $v_i=180~m/s$ , the final velocity is $v_i=0~m/s$ , and the total time of the motion is $\Delta t=0.02~s$ , so the acceleration is given by
$a= \frac{v_f-v_i}{\Delta t} = -9000~m/s^2$
where the negative sign means that is a deceleration.
Therefore we can calculate the total distance covered by the bullet in its motion using
$S=v_i t + \frac{1}{2}at^2 = 180~m/s \cdot 0.02~s + \frac{1}{2}(-9000~m/s^2)(0.02~s)^2=1.8~m$
So, the bullet penetrates the sandbag 1.8 meters.

5 0

3 years ago

If you walk 1.2 km north and then 1.6 km east, what are the magnitude and direction of your resultant displacement?

SVETLANKA909090 [29]

Assume north and east as two sides of a right angled triangle. magnitude of the distance is then given by the length of the hypotenuse which is $\sqrt{a^2 + b^2}$

where a = 1.2 km north

and b = 1.6 km east

magnitude = 2 km

Direction is given by the angle between them, that is atan(a/b) = 36.86 deg north of east = 53.1 deg east of north.

8 0

3 years ago

A 70.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 33.5 m/s. A second person, with a mass of 55.0 kg, c

saw5 [17]

Answer:

The final velocity of the thrower is $\bf{3.88~m/s}$ and the final velocity of the catcher is $\bf{0.029~m/s}$ .

Explanation:

Given:

The mass of the thrower, $m_{t} = 70~Kg$ .

The mass of the catcher, $m_{c} = 55~Kg$ .

The mass of the ball, $m_{b} = 0.0480~Kg$ .

Initial velocity of the thrower, $v_{it} = 3.90~m/s$

Final velocity of the ball, $v_{fb} = 33.5~m/s$

Initial velocity of the catcher, $v_{ic} = 0~m/s$

Consider that the final velocity of the thrower is $v_{ft}$ . From the conservation of momentum,

$&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s$

Consider that the final velocity of the catcher is $v_{fc}$ . From the conservation of momentum,

$&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s$

Thus, the final velocity of thrower is $3.88~m/s$ and that for the catcher is $0.029~m/s$ .

8 0

3 years ago