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Amanda [17]
3 years ago
5

HELP ASAP WHAT ARE PROPERTIES OF WATER

Physics
1 answer:
iren [92.7K]3 years ago
3 0
Pls mark as brainiest!!

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A solid Cute hug sides 0.50m
irina [24]

Answer:

gg

Explanation:

hh

3 0
3 years ago
Practical steam engines utilize 450ºC steam, which is later exhausted at 270ºC.
Naily [24]

(a) 0.249 (24.9 %)

The maximum efficiency of a heat engine is given by

\eta = 1-\frac{T_C}{T_H}

where

Tc is the low-temperature reservoir

Th is the high-temperature reservoir

For the engine in this problem,

T_C = 270^{\circ}C+273=543 K

T_H = 450^{\circ}C+273=723 K

Therefore the maximum efficiency is

\eta = 1-\frac{T_C}{T_H}=1-\frac{543}{723}=0.249

(b-c) 0.221 (22.1 %)

The second steam engine operates using the exhaust of the first. So we have:

T_H = 270^{\circ}C+273=543 K is the high-temperature reservoir

T_C = 150^{\circ}C+273=423 K is the low-temperature reservoir

If we apply again the formula of the efficiency

\eta = 1-\frac{T_C}{T_H}

The maximum efficiency of the second engine is

\eta = 1-\frac{T_C}{T_H}=1-\frac{423}{543}=0.221

8 0
3 years ago
If opposing forces acting on an object are equal, the net force is
makkiz [27]

Answer:

0 N

Explanation:

suppose, you push a box with 5 N, and another person pushes the box on the opposite side of the box with 5 N, the net force (resultant ) is 0 N, the box will not move if it wasn't moving

hope this helps

8 0
2 years ago
When a satellite is in orbit around the earth the force of gravity on the satellite-
mojhsa [17]

Answer:

Is always towards the center of the Earth

Explanation:

As a satellite moves around the Earth in a circular orbit, the direction of the force of gravity is always towards the center of the Earth. At an altitude of 100 km, you would be so high that you would see black sky and stars if you looked upwards.

5 0
3 years ago
Suppose a particle moves along a straight line with velocity v(t)=t2e−2tv(t)=t2e−2t meters per second after t seconds. How many
dimulka [17.4K]

Explanation:

It is given that,

Velocity of the particle moving in straight line is :

v(t)=t^2e^{-2t}\ m/s

We need to find the distance (x)  traveled by the particle during the first t seconds. It is given by :

x=\int\limits {v.dt}

x=\int\limits {t^2e^{-2t}dt}

Using by parts integration, we get the value of x as :

x=\dfrac{-(2t^2+2t+1)e^{-2t}}{4}\ meters

Hence, this is the required solution.

6 0
3 years ago
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