Every object in the universe exerts a force on every other object. This force is called

A lead ball is dropped into a lake from a diving board 5.0 m above the water. After entering the water, it sinks to the bottom w

nirvana33 [79]

Answer:

$|D_{depth} |=19.697m$

Explanation:

To find Depth D of lake we must need to find the time taken to hit the water.So we use equation of simple motion as:

Δx=vit+(1/2)at²

$x_{f}-x_{i}=v_{i}t+(1/2)at^{2}\\ -5.0m=(o)t+(1/2)(-9.8m/s^{2} )t^{2}\\ -4.9t^{2}=-5.0\\ t^{2}=5/4.9\\t=\sqrt{1.02} \\t=1.01s$

As we have find the time taken now we need to find the final velocity vf from below equation as

$v_{f}=v_{i}+at\\v_{f}=0+(-9.8m/s^{2} )(1.01s) \\v_{f}=-9.898m/s$

So the depth of lake is given by:

first we need to find total time as

t=3.0-1.01 =1.99 s

$|D_{depth} |=|vt|\\|D_{depth} |=|(-9.898m/s)(1.99s)|\\|D_{depth} |=19.697m$

6 0

3 years ago

PLEASE HELPPP Count the number of atoms of each element in this molecule: 2H2K(OH2)2

yulyashka [42]

Answer:

H = 6 atoms

K= 2 atoms

O = 2 atoms

7 0

3 years ago

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What is the magnitude of the momentum of a 3400 kg airplane traveling at a speed of 450 miles per hour?

Vitek1552 [10]

Answer:

The magnitude of momentum of the airplane is $6.83\times 10^5\ kg-m/s$ .

Explanation:

Given that,

Mass of the airplane, m = 3400 kg

Speed of the airplane, v = 450 miles per hour

Since, 1 mile per hour = 0.44704 m/s

v = 201.16 m/s

We need to find the magnitude of momentum of the airplane. It is given by the product of mas and velocity such that,

$p=mv$

$p=3400\ kg\times 201.16 \ m/s$

$p=683944\ kg-m/s$

or

$p=6.83\times 10^5\ kg-m/s$

So, the magnitude of momentum of the airplane is $6.83\times 10^5\ kg-m/s$ . Hence, this is the required solution.

7 0

3 years ago

Read 2 more answers

A referee will toss up the ball between to opponents.what is this called

OlgaM077 [116]

This happens in basketball. It is known as "jump ball".

5 0

3 years ago

2. Below what depth would a submarine have to submerge so that it would not be swayed by surface waves with a wavelength of 24 m

Travka [436]

<u>Answer:</u> Below 12m of depth, the submarine has to submerge so that it would not be swayed by surface waves

<u>Explanation:</u>

To avoid the surface waves, a submarine has to submerge below the wave base. It is the position below which the motion of the waves is negligible.

This wave base is equal to half of the wavelength. The equation becomes:

Wave base = $\frac{\text{Wavelength}}{2}$

We are given:

Wavelength = 24 m

Putting values in above equation, we get:

Wave base = $\frac{24m}{2}=12m$

Hence, below 12m of depth, the submarine has to submerge so that it would not be swayed by surface waves

4 0

2 years ago