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Sergio039 [100]

3 years ago

11

The work energy principle states that the change in kinetic energy of an object is equal to the net work done on the object. If

you drop an object in class from some height what will the final velocity of the object be? Assume there is no air resistance, and use the insert math equation (sqrt x above) to put in your answer.

1 answer:

sattari [20]3 years ago

6 0

Answer:

$v=\sqrt{2gh}\ m/s$

Explanation:

From work energy theorem

Work done by all forces = Change in kinetic energy

Lets take

m= mass of object

h=height from the ground surface

initial velocity of object = 0 m/s

The final velocity of object is v

Work done by gravitational force = m g . h

The final kinetic energy = 1/2 m v²

So

Work done by all forces = Change in kinetic energy

m g h = 1/2 m v² - 0

v² = 2 g h

$v=\sqrt{2gh}\ m/s$

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denis23 [38]

It’s will be B because the circuit had a open or close so if that doesn’t work than it’s because it’s open

5 0

3 years ago

Read 2 more answers

The FM radio band in most places goes from frequencies of about 89 MHz to 106 MHz. How long is the wavelength of the radiation a

ohaa [14]

Answer:

2.83 m

Explanation:

The relationship between frequency and wavelength for an electromagnetic wave is given by

$\lambda=\frac{c}{f}$

where

$\lambda$ is the wavelength

$c=3.0\cdot 10^8 m/s$ is the speed of light

$f$ is the frequency

For the FM radio waves in this problem, we have:

$f_1=89 MHz=89\cdot 10^6 Hz$ is the minimum frequency, so the maximum wavelength is

$\lambda_2=\frac{c}{f_1}=\frac{3\cdot 10^8}{89\cdot 10^6}=3.37 m$

The maximum frequency is instead

$f_2=106 MHz=106\cdot 10^6 Hz$

Therefore, the minimum wavelength is

$\lambda_1=\frac{c}{f_2}=\frac{3\cdot 10^8}{106\cdot 10^6}=2.83 m$

So, the wavelength at the beginning of the range is 2.83 m.

8 0

2 years ago

A high-speed flywheel in a motor is spinning at 450 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and

alexira [117]

Answer:

A) $\omega_f=17.503\ rad.s^{-1}$

B) $t=55.6822\ s$

C) $\theta=1312\ rad$

Explanation:

Given:

mass of flywheel, $m=40\ kg$

diameter of flywheel, $d=0.72\ m$

rotational speed of flywheel, $N_i=450\ rpm \Rightarrow \omega_i=\frac{450\times 2\pi}{60} =15\pi\ rad.s^{-1}$

duration for which the power is off, $t_0=35\ s$

no. of revolutions made during the power is off, $\theta=180\times 2\pi=360\pi\ rad$

<u>Using equation of motion:</u>

$\theta=\omega_i.t+\frac{1}{2} \alpha.t^2$

$360\pi=15\pi\times 35+\frac{1}{2} \times \alpha\times35^2$

$\alpha=-0.8463\ rad.s^{-2}$

Negative sign denotes deceleration.

A)

Now using the equation:

$\omega_f=\omega_i+\alpha.t$

$\omega_f=15\pi-0.8463\times 35$

$\omega_f=17.503\ rad.s^{-1}$ is the angular velocity of the flywheel when the power comes back.

B)

Here:

$\omega_f=0\ rad.s^{-1}$

Now using the equation:

$\omega_f=\omega_i+\alpha.t$

$0=15\pi-0.8463\times t$

$t=55.6822\ s$ is the time after which the flywheel stops.

C)

Using the equation of motion:

$\theta=\omega_i.t+\frac{1}{2} \alpha.t^2$

$\theta=15\pi\times 55.68225-0.5\times 0.8463\times 55.68225^2$

$\theta=1312\ rad$ revolutions are made before stopping.

3 0

3 years ago

Which of The following are classiFcatons describing how rock and oTher maTerial movedownslope?a. Fall, b. slide, c. slump, d. cr

nlexa [21]

Answer:

correct answer is Fall slide, slump, creep, flow

Explanation:

solution

we know that Movement of particle under the influence of gravity

so rock and other material move down as gravity.

first rock particle fall down because falls occur very rapidly with high slope after that they slide on the slope and after sliding they slump and it occurs when the rupture surface is curved after slump process they creep.

after creeping, it can flow particle as it occurs slowly with the low slope with water.

so correct answer is Fall slide, slump, creep, flow

7 0

3 years ago

The average distance of the planet mercury from the sun is 0.39 times the average distance of the earth from the sun. How long i

Stels [109]

Answer:

$T_1=0.24y$

Explanation:

Using Kepler's third law, we can relate the orbital periods of the planets and their average distances from the Sun, as follows:

$(\frac{T_1}{T_2})^2=(\frac{D_1}{D_2})^3$

Where $T_1$ and $T_2$ are the orbital periods of Mercury and Earth respectively. We have $D_1=0.39D_2$ and $T_2=1y$ . Replacing this and solving for

$T_1^2=T_2^2(\frac{D_1}{D_2})^3\\T_1^2=(1y)^2(\frac{0.39D_2}{D_2})^3\\T_1^2=1y^2(0.39)^3\\T_1^2=0.059319y^2\\T_1=0.24y$

5 0

2 years ago