1. A bag dropped from a helicopter falls with an acceleration of 9.8m/s2. What is its velocity after 5s?

Physics

1 answer:

Tanzania [10]2 years ago

5 0

Answer:

The answer is 49m/s.

Explanation:

Given,

acceleration (a) =9.8m/s^2

initial velocity (u)=0 (as it was in rest condition in helicopter)

time (t) =5 s

now, final velocity (v) after 5s =?

we have,

$a = \frac{v - u}{t}$

$or \: 9.8 = \frac{v - 0}{5}$

or, v = 49m/s

Therefore, the velocity after 5s is 49m/s.

Hope it helps..

You might be interested in

Select all THREE of the correct response about the diagram of an atom.

STALIN [3.7K]

Answer: HI

Explanation:

5 0

2 years ago

A carnival Ferris wheel has a 15-m radius and completes five turns about its horizontal axis every minute. What is the accelerat

Tanzania [10]

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.

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