All categories

1 year ago

PpA hollow plastic ball is projected into the air.

Physics

1 answer:

LenKa [72]1 year ago

4 0

a > g , is the best shows the magnitude a and the direction of the ball's acceleration at time.

<h3>What is acceleration?</h3>

Acceleration is the rate at which the speed and direction of a moving object change over time. A point or object travelling in a straight path is accelerated if it accelerates or decelerates. Even if the speed is constant, motion on a circle is accelerated because the direction is always changing.

Given data

The velocity of the ball is: V

The angle of velocity with the horizontal is (θ) = 45

The acceleration of the ball is: a

The acceleration due to gravity is: g

When the hollow ball is projected into the air, air resistance will act on the ball that will slow down the ball's motion. Therefore the velocity direction is opposite to the reducing force. The reducing force relates to the opposite to the acceleration of the hollow ball.

F(delaying) ∝ -VF(delaying) ∝ m×a(r)×F(delaying) ∝ a(r)

Here m is the mass of the hollow ball.

The force of gravity on the ball which is acting downward given as:

F(gravity) ∝ g

So, a > g

Correct option: (B)

To know more about acceleration refer to:

brainly.com/question/460763

#SPJ1

You might be interested in

Please help me

melamori03 [73]

<h2>Question:</h2>

An automobile is driving uphill. Which form of energy is not involved in this process?

<h2>Choosing:</h2>

electromagnetic

potential

kinetic

chemical

<h2>Answer:</h2>

<u>Electromagnetic</u><u> </u>

<h3><u>#READINGHELPSWITHLEARNING</u><u> </u></h3><h3><u>#CARRYONLEARNING</u><u> </u></h3><h3><u>#STUDYWELL</u><u> </u></h3>

7 0

3 years ago

(physical science) could someone please help me out with this lab? if i’m being honest i did the lab but i lost all of my work :

djverab [1.8K]

Explanation:

hdhhrhhrhehhshsujwuuwuwuwwh

6 0

3 years ago

Example of first, second and third law of motion.

statuscvo [17]

Examples of Newton's three law of motion.

First law of motion: A rocket being launched up in the atmosphere.

Second law of motion:while riding a bicycle, a bicycle acts as a mass and our legs pushing on the pedals of the bicycle is the force.

Third law of motion:when we jump off from the boat,the boat moves backward.

Hope,it will helpyouu!

4 0

3 years ago

Mark creates a graphic organizer to review his notes about electrical force. Which labels belong in the regions marked X and Y?

sveta [45]

Answer:

The correct answer is A

Explanation:

The question requires as well the attached image, so please see that below.

Coulomb's Law.

The electrical force can be understood by remembering Coulomb's Law, that describes the electrostatic force between two charged particles. If the particles have charges $q_1$ and $q_2$ , are separated by a distance r and are at rest relative to each other, then its electrostatic force magnitude on particle 1 due particle 2 is given by:

$|F|=k \cfrac{q_1 q_2}{r^2}$

Thus if we decrease the distance by half we have

$r_1 =\cfrac r2$

So we get

$|F|=k \cfrac{q_1 q_2}{r_1^2}$

Replacing we get

$|F|=k \cfrac{q_1 q_2}{(r/2)^2}\\|F|=k \cfrac{q_1 q_2}{r^2/4}$

We can then multiply both numerator and denominator by 4 to get

$|F|=k \cfrac{4q_1 q_2}{r^2}$

So we have

$|F|=4 \left(k \cfrac{q_1 q_2}{r^2}\right)$

Thus if we decrease the distance by half we get four times the force.

Then we can replace the second condition

$q_{2new} =2q_2$

So we get

$|F|=k \cfrac{q_1 q_{2new}}{r_1^2}$

which give us

$|F|=k \cfrac{q_1 2q_2}{r_1^2}\\|F|=2\left(k \cfrac{q_1 q_2}{r_1^2}\right)$

Thus doubling one of the charges doubles the force.

So the answer is A.

8 0

3 years ago

Read 2 more answers

A 40 kg skier starts at the top of a 12-meter high slope at the bottom she is traveling 10 m/s how much energy does she lose fri

Iteru [2.4K]

Potential energy at top:
PE = mgh
PE = 40 x 9.81 x 12
P.E = 4,708.8 J

Kinetic energy at bottom:
KE = 1/2 mv²
KE = 1/2 x 40 x 10²
K.E = 2,000 J

P.E = K.E + Frictional losses
Frictional losses = 4708 - 2000
Frictional losses = 2708 J

The answer is D.

7 0

3 years ago

Read 2 more answers