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3 years ago

1 An airplane travels east at a velocity of 250 m/s and at the same time is blown north

Physics

1 answer:

mart [117]3 years ago

4 0

Answer:

Hi.Thank you for the points.

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You can obtain a rough estimate of the size of a molecule by the following simple experiment. Let a droplet of oil spread out on

Marat540 [252]

Answer:

The diameter of the molecule of oil is $1.405 10 ^{-9} \ m$

Explanation:

We define density as

$\rho = \frac{mass}{volume}$

So, the volume for our oil will be

$volume = \frac{mass}{\rho}$

$volume = \frac{7.62 \ 10^{-7} \ kg}{ 912 \ \frac{kg}{m^3} }$

$volume = 8.355 \ 10 ^{-10} \ m^3$

the volume for a cylinder with radius r and height h is

$volume = \pi r^2 h$

So, we can obtain the height of the droplet of oil as:

$h = \frac{volume}{\pi r^2}$

the radius is

$r=43.5 \ cm = 0.435 \ m$

$h = 1.405 10 ^{-9} \ m^3$

And this is the diameter of the oil molecule.

6 0

3 years ago

A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacemen

PilotLPTM [1.2K]

Answer:

The average angular acceleration is 0.05 radians per square second.

Explanation:

Let suppose that Ferris wheel accelerates at constant rate, the angular acceleration as a function of change in angular position and the squared final and initial angular velocities can be clear from the following expression:

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha\cdot (\theta-\theta_{o})$

Where:

$\omega_{o}$ , $\omega$ - Initial and final angular velocities, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$\theta_{o}$ , $\theta$ - Initial and final angular position, measured in radians.

Then,

$\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}$

Given that $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\theta-\theta_{o} = 4.9\,rad$ , the angular acceleration is:

$\alpha = \frac{\left(0.70\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(4.9\,rad\right)}$

$\alpha = 0.05\,\frac{rad}{s^{2}}$

Now, the time needed to accelerate the Ferris wheel uniformly is described by this kinematic equation:

$\omega = \omega_{o} + \alpha \cdot t$

Where $t$ is the time measured in seconds.

The time is cleared and obtain after replacing every value:

$t = \frac{\omega-\omega_{o}}{\alpha}$

If $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\alpha = 0.05\,\frac{rad}{s^{2}}$ , the required time is:

$t = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{0.05\,\frac{rad}{s^{2}} }$

$t = 14\,s$

Average angular acceleration is obtained by dividing the difference between final and initial angular velocities by the time found in the previous step. That is:

$\bar \alpha = \frac{\omega-\omega_{o}}{t}$

If $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $t = 14\,s$ , the average angular acceleration is:

$\bar \alpha = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{14\,s}$

$\bar \alpha = 0.05\,\frac{rad}{s^{2}}$

The average angular acceleration is 0.05 radians per square second.

4 0

3 years ago

Which has the most kinetic energy?

lara31 [8.8K]

Answer:

I think A golf ball shot out of a small cannon

Explanation:

3 0

2 years ago

Two clouds collide and form another, more massive cloud. One cloud is stationary, while the other is traveling at 1 m/s. After t

sweet [91]

Answer: 3

Explanation:

Given

One cloud is traveling at rate of $u_2=1\ m/s$

combined velocity of the two is $v=0.25\ m/s$

Suppose the masses of the clouds be $m_1,m_2$

Conserving momentum

$\Rightarrow m_1u_1+m_2u_2=\left(m_1+m_2\right)v\\\text{Divide whole equation by }m_2\\\Rightarrow \dfrac{m_1}{m_2}u_1+u_2=\left(\dfrac{m_1}{m_2}+1\right)v\\\\\Rightarrow 0+1=0.25\dfrac{m_1}{m_2}+0.25\\\\\Rightarrow \dfrac{m_1}{m_2}=\dfrac{0.75}{0.25}\\\\\Rightarrow \dfrac{m_1}{m_2}=3$

6 0

3 years ago

A ball is thrown vertically down from the edge of a cliff with a speed of 4 m/s, how high is the cliff, if it took 12 s for the

Vesna [10]

Answer:

The height of the cliff from which the ball was dropped from is 224.4m.

\overline{v}={\frac{\Delta x}{\Delta t}}

Given the data in the question;

Initial velocity of the ball;

Time taken by the ball to reach the ground;

Distance or Height of the cliff from which the ball was thrown from;

To get the height of the Cliff, we use the Second Equation of Motion:

Where s is the distance or height, is the initial velocity, t is the time and a is the acceleration. Since the ball was thrown down from a certain height (cliff), its is now under the influence of gravity. acceleration due to gravity;

Hence, the equation becomes

We substitute the given values into the equation

Therefore, the height of the cliff from which the ball was dropped from is 224.4m

Explanation:

3 0

2 years ago

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