A student wants to model the physical weathering of rock. Which activity could the student perform?

a ball is dropped from rest at a height of 45.0 m above the ground. ignore the effects of air resistance. What is the speed of c

NemiM [27]

So, the final velocity of the ball when it is 10.0 m above the ground approximately <u>26.2 m/s</u>.

<h3>Introduction</h3>

Hi ! In this question, I will help you. This question uses the principle of final velocity in free fall. Free fall occurs only when an object is dropped (without initial velocity), so the falling object is only affected by the presence of gravity. In general, the final velocity in free fall can be expressed by this equation :

$\boxed{\sf{\bold{v = \sqrt{2 \times g \times h}}}}$

With the following condition :

v = final velocity (m/s)
h = height or any other displacement at vertical line (m)
g = acceleration of the gravity (m/s²)

<h3>Problem Solving</h3>

We know that :

$\sf{h_1}$ = initial height = 45.0 m
$\sf{h_2}$ = final height = 10.0 m
g = acceleration of the gravity = 9.8 m/s²

Note :

At this point 10 m above the ground, the object can still complete its movement up to exactly 0 m above the ground.

What was asked :

v = final velocity = ... m/s

Step by Step

$\sf{v = \sqrt{2 \times g \times \Delta h}}$

$\sf{v = \sqrt{2 \times g \times (h_1 - h_2)}}$

$\sf{v = \sqrt{2 \times 9.8 \times (45 - 10)}}$

$\sf{v = \sqrt{19.6 \times 35}}$

$\sf{v = \sqrt{686}}$

$\boxed{\sf{v \approx 26.2 \: m/s}}$

<h3>Conclusion</h3>

So, the final velocity of the ball when it is 10.0 m above the ground approximately 26.2 m/s.

The relationship between acceleration and the change in velocity and time in free fall brainly.com/question/26486625

5 0

2 years ago

A pitcher releases a fastball that moves toward home plate.

Vsevolod [243]

The two additional forces that act on the ball as it travels between the pitcher and the home plate are air resistance and gravity.

<h3>What are the forces that affect object in motion;</h3>

Air resistance: this is the force that oppose the motion of an object in air due to frictional force
Gravity: this is the force due to weight of the object and acts downwards.

The two additional forces that act on the ball as it travels between the pitcher and the home plate include:

Air resistance and
Gravitational force

<h3>How the forces affect the motion of the ball</h3>

Air resistance oppose the motion of the ball as it travels in air.
Gravity is the force due to weight of the ball and acts downwards.

Learn more about forces on object in motion here: brainly.com/question/10454047

5 0

3 years ago

What is the velocity of the buggy? <br><br> At 20 seconds the buggy will have a position of ____ ?

lbvjy [14]

At 20 seconds it will be 12.6 because at 10 seconds it was as at approximately 6.3 so we times it by 2 to get the 20s

6 0

3 years ago

What is the magnitude of the free-fall acceleration at a point that is a distance 2R above the surface of the Earth, where R is

ss7ja [257]

Answer:

g' = g/9 = 1.09 m/s²

Explanation:

The magnitude of free fall acceleration at the surface of earth is given by the following formula:

g = GM/R² ----- equation 1

where,

g = free fall acceleration

G = Universal Gravitational Constant

M = Mass of Earth

R = Distance between the center of earth and the object

So, in our case,

R = R + 2 R = 3 R

Therefore,

g' = GM/(3R)²

g' = (1/9) GM/R²

using equation 1:

g' = g/9

g' = (9.8 m/s)/9

3 0

3 years ago

A clock is designed that uses a mass on the end of a spring as a timing mechanism. If the oscillation time needed is exactly one

user100 [1]

Answer:

The value is $k = 51.34 \ N/ m$

Explanation:

From the question we are told that

The mass is $m = 1.3 \ kg$

The needed oscillation time is $T = 1 \ s$

Generally the spring constant is mathematically represented as

$k = \frac{4 \pi^2 * m }{ T^2}$

=> $k = \frac{4* 3.142^2 * 1.3 }{ 1^2}$

=> $k = 51.34 \ N/ m$

5 0

3 years ago