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2 years ago

A 5.0 kg book is lying on a 0.25 meter high table.

Physics

1 answer:

Wittaler [7]2 years ago

3 0

Answer:

Below in the picture:-

I hope it helps.

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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

Explain how you can work out the

Ne4ueva [31]

Answer:

The speed of light changes as it moves between media. This causes refraction. Angles of refraction can be calculated using known speeds or wavelengths. Beyond the critical angle, light is reflected.

8 0

3 years ago

See the diagram above. Which of the following is the best prediction of what would happen if you increased the distance in secti

Inessa [10]

By the definition of wavelength, the answer is the letter D, the wavelength would decrease.

We can see in the diagram a wave motion.

A wave has some characteristics:

Has an amplitude, the distance from 0 to the crest (highest point in the y-direction, point (3) in the figure) it would see in the figure as (2)
Has wavelength, the distance between the crests.
Has a trough, the lowest point in the y-direction.

Now, if we increase the distance of the crests, by the definition shown above, we will increase the wavelength.

Therefore, the answer is letter D, the wavelength would increase.

You can learn more about wave motion here:

brainly.com/question/22763521

7 0

3 years ago

While finding the spring constant, if X1 = 12 cm, X2 = 15 cm, and hanging mass = 22 grams, the value of spring constant K would

Pavel [41]

Answer:

If x₁=12 cm then k=1.7985 N/m

If x₂=15 cm then k=1.4388 N/m

Explanation:

Hanging mass= 22 g=0.022 kg

Acceleration due to gravity g=9.81 m/s²

If x₁=displacement= 12 cm=0.12 m

k= spring constant

$F=ma\\\Rightarrow F=0.022\times 9.81\\\Rightarrow F=0.21582\ N$

$\text {For spring}\\F=kx\\\Rightarrow 0.21582=k\times 0.012\\\Rightarrow k=1.7985\ N/m\\$

∴k = 1.7985 N/m

If x₂=15 cm=0.15 m

Force of the hanging mass is same however the spring constant will change

$\text {For spring}\\F=kx\\\Rightarrow 0.21582=k\times 0.015\\\Rightarrow k=1.4388\ N/m\\$

∴k = 1.4388 N/m

As the mass is not changing the spring constant has to change. That means that here there are two spring one with k=1.7985 N/m and the other with k= 1.4388 N/m

4 0

3 years ago

A vertical spring stretches 4.0 cm when a 12-g object is hung from it. The object is replaced with a block of mass 28 g that osc

marin [14]

Answer:

Time period of the osculation will be 0.0671 sec

Explanation:

It is given a vertical spring is stretched by 4 cm

So change in length of the spring x = 4 cm = 0.04 m

Mass which is hung from it m = 12 gram = 0.012 kg

Sprig force will be equal to weight of the mass

So $kx=mg$