All categories

2 years ago

A football player kicks a 0.94 kg football with a force of 2.4 N.

Physics

2 answers:

Artyom0805 [142]2 years ago

6 0

2.6m/s²

Explanation:

Given parameters:

Mass of football = 0.94kg

Force on ball = 2.4N

Unknown:

The acceleration of the ball = ?

Solution:

According to Newton's second law of motion "the net force on a body is the product of its mass and acceleration".

Net force = mass x acceleration

Make acceleration the subject of the expression;

Acceleration = $\frac{net force}{mass}$

Acceleration = $\frac{2.4}{0.94}$ = 2.6m/s²

learn more:

Acceleration brainly.com/question/3820012

#learnwithBrainly

Juliette [100K]2 years ago

3 0

Answer:

The Answer is 2.6m/s(m/s)

Explanation:

Just took the test!

You might be interested in

PWEESE HELP ME WIT MY QUIZ

kaheart [24]

I believe it is acceleration

7 0

2 years ago

Read 2 more answers

8459299

sergeinik [125]

Answer:

Mechanical energy is the energy that is possessed by an object due to its motion or due to its position. Mechanical energy can be either kinetic energy (energy of motion) or potential energy

3 0

2 years ago

A record player turntable initially rotating at 3313 rev/min is braked to a stop at a constant rotational acceleration. The turn

Rus_ich [418]

Answer:

(A) It will take 22 sec to come in rest

(b) Work done for coming in rest will be 0.2131 J

Explanation:

We have given the player turntable initially rotating at speed of $33\frac{1}{3}rpm=33.333rpm=\frac{2\times 3.14\times 33.333}{60}=3.49rad/sec$

Now speed is reduced by 75 %

So final speed $\frac{3.49\times 75}{100}=2.6175rad/sec$

Time t = 5.5 sec

From first equation of motion we know that '

$\alpha =\frac{\omega -\omega _0}{t}=\frac{2.6175-3.49}{4}=-0.158rad/sec^2$

(a) Now final velocity $\omega =0rad/sec$

So time t to come in rest $t=\frac{0-3.49}{-0.158}=22sec$

(b) The work done in coming rest is given by

$\frac{1}{2}I\left ( \omega ^2-\omega _0^2 \right )=\frac{1}{2}\times 0.035\times (0^2-3.49^2)=0.2131J$

4 0

2 years ago

A ball is dropped from a building of height h. Assume the ball starts from rest and that air friction can be ignored. Derive an

agasfer [191]

Answer:

$t=\sqrt{h/g}$

Explanation:

We use the kinematics equation to solve this question:

$y(t)=y_{o}+v_{o}t+1/2*a*t^{2}$

$v_{o}=0$ because the ball is dropped

$a=-g$ the acceleration is the gravity, negative because it points downwards

$y_{o}=h$ initial height

$y(t)=h/2$ final height

So:

$h/2=h-1/2*g*t^{2}$

$t=\sqrt{h/g}$

8 0

2 years ago

Objects 1 and 2 attract each other with a electrostatic force of 36.0 units. If the charge of Object 1 is doubled AND the distan

melamori03 [73]

Answer:

8 units

Explanation:

F=(k*q1*q2)/(r^2)

K is a constant, q1 is charge of 1, q2 is charge of 2, r is distance between the two.

6 0

2 years ago