Which Gas Law explains the relationship between temperature and volume?

If a Ferris wheel has a 15-m radius and completes five turns about its horizontal axis every minute then the acceleration of a passenger at his lowest point during the ride is 4.11 $\text{m/s}^{2}$ .

Calculation:

Step-1:

It is given that the radius of the Ferris wheel is r=15 m, and the angular speed of the wheel is $\omega$ =5rev/min.

It is required to find the angular acceleration of a passenger at his lowest point during the ride.

The formula required to calculate the angular acceleration is, $a=\omega ^2 r$ .

Step-2:

Now substituting the given values into the equation to get the value of the angular acceleration.

$\begin{aligned}a &=\omega^{2} r \\&=(5 \mathrm{rev} / \mathrm{min})^{2}(15 \mathrm{~m}) \\&=\left(5 \mathrm{rev} / \mathrm{min} \times \frac{\left(\frac{2 \pi}{60}\right) \mathrm{rad} / \mathrm{sec}}{1 \mathrm{rev} / \mathrm{min}}\right)^{2}(15 \mathrm{~m}) \\&=(0.2741 \times 15) \mathrm{m} / \mathrm{s}^{2} \\&=4.11 \mathrm{~m} / \mathrm{s}^{2}\end{aligned}$

The acceleration is towards upwards that means towards the center of the wheel.

Learn more about the angular acceleration:

brainly.com/question/1592013

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3 0

1 year ago

The launch velocity of a toy car launcher is determined to be 5 m/s. If the car is to be launched from a height of 0.5 m, where

lora16 [44]

Answer:

1.6 m

Explanation:

Given that the launch velocity of a toy car launcher is determined to be 5 m/s. If the car is to be launched from a height of 0.5 m.

The time for landing should be calculated by using the second equation of motion formula

h = Ut + 1/2gt^2

Let U = 0

0.5 = 1/2 × 9.8 × t^2

0.5 = 4.9t^2

t^2 = 0.5 / 4.9

t^2 = 0.102

t = 0.32 s

The target should be placed so that the toy car lands on it at:

Distance = 5 × 0.32

distance = 1.597 m

Distance = 1.6 m

Therefore, the target should be placed so that the toy car lands on it 1.6 metres away.

7 0

3 years ago