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3 years ago

8

Question 4 (1 point)

1 answer:

Valentin [98]3 years ago

4 0

cardiovascular endurance :))

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Which are properties of a liquid? Check all that apply.

Strike441 [17]

Answer:

the property of liquid are

1 they can flow from one place to another if surface is slanted

2 it cannot be compressed

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3 years ago

based on the law of conservation of energy. how can we reasonably improve a machines ability to do work? A.move the machine to a

natima [27]

The correct answer is

C. reduce the friction between its moving parts

In fact, by reducing the friction between the moving parts of the machine, it is possible to reduce the energy wasted due to this friction; therefore, more input energy is converted into useful work, and this will improve the efficiency of the machine.

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4 years ago

Read 2 more answers

flat sheet is in the shape of a rectangle with sides of lengths 0.400 mm and 0.600 mm. The sheet is immersed in a uniform electr

Ksenya-84 [330]

Answer:

$6.29591\times 10^{-6}\ N/C^2$

Explanation:

Flux is given by

$\phi=EAcos\theta$

A = Area

$A=0.4\times 10^{-3}\times 0.6\times 10^{-3}$

E = Electric field = 76.7 N/C

Angle is given by

$\theta=90-20\\\Rightarrow \theta=70^{\circ}$

$\phi=76.7\times 0.4\times 10^{-3}\times 0.6\times 10^{-3}\times cos70\\\Rightarrow \phi=6.29591\times 10^{-6}\ N/C^2$

The flux through the sheet is $6.29591\times 10^{-6}\ N/C^2$

6 0

3 years ago

According to Newton's first law, which characteristic of a moving object would remain constant if there were no other

suter [353]

In object in motion stays in motion; speed

8 0

3 years ago

In a heat engine if 1000 j of heat enters the system the piston does 500 j of work, what is the final internal energy of the sys

nydimaria [60]

Answer : The final energy of the system if the initial energy was 2000 J is, 3500 J

Solution :

(1) The equation used is,

$\Delta U=q+w\\\\U_{final}-U_{initial}=q+w$

where,

$U_{final}$ = final internal energy

$U_{initial}$ = initial internal energy

q = heat energy

w = work done

(2) The known variables are, q, w and $U_{initial}$

initial internal energy = $U_{initial}$ = 2000 J

heat energy = q = 1000 J

work done = w = 500 J

(3) Now plug the numbers into the equation, we get

$U_{final}-(2000J)=(1000J)+(500J)$

(4) By solving the terms, we get

$U_{final}-(2000J)=(1000J)+(500J)$

$U_{final}-(2000J)=1500J$

$U_{final}=2000J+1500J$

$U_{final}=3500J$

(5) Therefore, the final energy of the system if the initial energy was 2000 J is, 3500 J

5 0

3 years ago