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3 years ago

What happens to kinetic energy when you increase the mass?

Physics

1 answer:

Rus_ich [418]3 years ago

5 0

Increasing mass increases kinetic energy. This can be seen in the equation KE = 1/2 (m) (v)^2

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What is the magnitude of the Box's Acceleration?

mojhsa [17]

The Box's Acceleration : g sin θ

<h3>Further explanation </h3>

Newton's 2nd law explains that the acceleration produced by the resultant force on an object is proportional and in line with the resultant force and inversely proportional to the mass of the object

∑F = m. a

F = force, N

m = mass = kg

a = acceleration due to gravity, m / s²

We plot the forces acting on the block (picture attached) according to the y-axis and the x-axis.

Because the motion of the block is in the same direction as the x-axis, ignoring the friction force with the inclined plane, then

$\tt \sum F_x=m.a\\\\W.sin\theta=m.a\\\\mgsin\theta=m.a\\\\a=gsin\thet\theta$

4 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

A car passes point “A” and then 120 meters later. It’s velocity was measured 21 m/s. If it’s acceleration was constant at 0.853

Norma-Jean [14]

Recall that

${v_f}^2-{v_i}^2=2a\Delta x$

where $v_i$ and $v_f$ are the initial and final velocities, respecitvely; $a$ is the acceleration; and $\Delta x$ is the change in position.

So we have

$\left(21\dfrac{\rm m}{\rm s}\right)^2-{v_i}^2=2\left(0.853\dfrac{\rm m}{\mathrm s^2}\right)(120\,\mathrm m)$

$\implies v_i\approx\boxed{15.4\dfrac{\rm m}{\rm s}}$

(Normally, this equation has two solutions, but we omit the negative one because the car is moving in one direction.)

7 0

3 years ago

Name the 2 types of tissue that form your skin?

QveST [7]

Answer:

Epithelial tissue and Muscle tissue

Explanation:

7 0

3 years ago

A football player runs 20 meters North of a football field, and then 15 meters East. The total motion lasted 15 seconds. What wa

vlada-n [284]

Given:
1st run: 20 meters North
2nd run: 15 meters East
time: 15 seconds

Average speed = total distance covered / total time taken
Ave. Speed = (20m + 15m) / 15s
Ave. Speed = 35m / 15s
Ave. Speed = 2 1/3 meters per second

4 0

3 years ago