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3 years ago

13

If the effects of heat and friction are ignored, the amount of work output is always _______ the amount of work input, even when

using a simple machine. A. the same as B. less than C. It changes by situation. D. greater than

1 answer:

vodomira [7]3 years ago

3 0

Answer:

A

Explanation:

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You are asked to help design a new type of loop-the-loop ride. Instead of rolling down a long hill to generate the speed to go a

Trava [24]

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.

5 0

4 years ago

A charge Q is distributed uniformly around the perimeter of a ring of radius R. Determine the electric potential difference betw

il63 [147K]

Answer:

the electric potential difference between the point at the center of the ring and a point on its axis ΔV is $( 0.8356 )$ $\frac{kQ}{R}$

Explanation:

Given the data in the question;

electric potential at the center of the ring V₀ = kQ / R

electric potential on the axis point Vr = kQ / √( R² + x² )

at a distance 6R from the center,

point at x = 6R

so distance circumference r = √( R² + (6R)² )

so

electric potential on the axis point Vr = kQ / √( R² + (6R)² )

Vr = kQ / R√37

Now

ΔV = V₀ - Vr

we substitute

ΔV = ( kQ / R) - ( kQ / R√37 )

ΔV = kQ/R( 1 - 1/√37 )

ΔV = kQ/R( 1 - 0.164398987 )

ΔV = kQ/R( 0.8356 )

ΔV = $( 0.8356 )$ $\frac{kQ}{R}$ { where k = $\frac{1}{4\pi e_0}$ }

Therefore, the electric potential difference between the point at the center of the ring and a point on its axis ΔV is $( 0.8356 )$ $\frac{kQ}{R}$

5 0

3 years ago

Which best describes the energy change that takes place during deposition?

alina1380 [7]

The correct answer for above statement is:

A. Heat energy is released by the substance.

Explanation:

The reverse of deposition is sublimation and therefore generally deposition is termed desublimation. One example of deposition is that the method by that, in sub-freezing air, water vapor changes on to ice while not initial changing into a liquid.It is a process of releasing energy.

4 0

4 years ago

Read 2 more answers

Q C Blaise Pascal duplicated Torricelli's barometer using a red Bordeaux wine, of density 984kg/ m³ , as the working liquid (Fig

levacccp [35]

The pressure within Earth's atmosphere is referred to as atmospheric pressure or barometric pressure. 10.701 was the height h of the wine column for normal atmospheric pressure.

<h3>What is Atmospheric pressure?</h3>

The pressure within Earth's atmosphere is referred to as atmospheric pressure or barometric pressure. The definition of the standard atmosphere is 101,325 Pa, or the same as 1013.25 millibars, 760 mm Hg, 29.9212 inches Hg, or 14.696 psi.

The force per unit area that an atmospheric column exerts is known as atmospheric pressure, often known as barometric pressure (that is, the entire body of air above the specified area). A weather indicator is atmospheric pressure. There will typically be clouds, wind, and precipitation when a low-pressure system enters a region. Fair, quiet weather is frequently a result of high pressure systems.

We have the density for the red Bordeaux wine given $\rho=965 \frac{\mathrm{kg}}{\mathrm{m}^3}$ , the atmospheric pressure on the Torricelli's barometer is given by:

$P_{a t m}=\rho g h$

Solving for the height of wine in the column we have this:

$h=\frac{P_{\text {anm }}}{\rho g}$

And replacing we have:

$h=\frac{101300 \mathrm{~Pa}}{965 \frac{\mathrm{kg}}{\mathrm{m}^3 9.81 \frac{\mathrm{m}}{\mathrm{m}^2}}}=10.701 \mathrm{~m}$$$

So the height of the red Bordeaux wine would be $\mathrm{h}=10.701 \mathrm{~m}$ . A very high value on this case if we compare with the usual values for this variable.

To learn more about Atmospheric pressure refer to:

brainly.com/question/19587559

#SPJ4

6 0

2 years ago

A rotating wheel requires 2.90-s to rotate through 37.0 revolutions. Its angular speed at the end of the 2.90-s interval is 97.2

goldenfox [79]

Answer:

Angular acceleration, $\alpha =20.32\ rad/s^2$

Explanation:

It is given that,

Displacement of the rotating wheel, $\theta=37\ rev=232.47\ radian$

Time taken, t = 2.9 s

Initial speed of the wheel, $\omega_i=0$

Final speed of the wheel, $\omega_f=97.2\ rad/s$

Let $\alpha$ is the angular acceleration of the wheel. Using the third equation of kinematics to find it as :

$\alpha=\dfrac{\omega_f^2-\omega_i^2}{2\theta}$

$\alpha=\dfrac{(97.2)^2}{2\times 232.47}$

$\alpha =20.32\ rad/s^2$

So, the angular acceleration of the wheel is $20.32\ rad/s^2$ . Hence, this is the required solution.

6 0

3 years ago