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3 years ago

You throw a ball into the air. Which two forces cause the ball to gradually stop moving upward and then fall back to Earth?

Physics

2 answers:

Sindrei [870]3 years ago

4 0

Answer:

it should be A and B

Explanation:

because obviously gravity causes it to fall down and the second force acted upon it is Friction from the air and ball coming in contact with eachother. I hope this helps

Virty [35]3 years ago

3 0

Answer:

I'm pretty sure the answer is D

Explanation:

Honestly it's just a guess so let me know if it's right :3

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A person is riding a bicycle, and its wheels have an angular velocity of 10.7 rad/s. Then, the brakes are applied and the bike i

Scilla [17]

Answer:

(a) t = 22.9 s

(b) α= - 0.467 rad/s²

Explanation:

The uniformly accelerated circular movement, is a circular path movement in which the angular acceleration is constant.

We apply the equations of circular motion uniformly accelerated :

ωf ²= ω₀² + 2*α*θ Formula (1)

ωf= ω₀ + α*t Formula (2)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular speed ( rad/s)

ωf : final angular speed ( rad/s)

Data

θ = 19.5 revolutions : angular displacement of each wheel or angle that the wheel has rotated in a given time interval

ω₀= 10.7 rad/s : initial angular speed of the Wheel ( rad/s)

ωf = 0 : final angular speed of the Whee( rad/s)

Calculating of the angular acceleration (α )

We replace data in the fómula (1),considering that 1 revolution is equal to 2π radians :

ωf ²= ω₀² + 2*α*θ

(0 )²= (10.7)² + 2*α*(19.5*2*π )

0= 114.49 + (245.04)*α

-114.49 = (245.04)*α

α= (-114.49) /(245.04)

α= -114.49 /(245.04)

α= -0.467 rad/s²

Time does it take for the bike to come to rest

We replace data in the formula (2)

ωf = ω₀ + α*t

0 = 10.7 + -0.467*t

-10.7 = - 0.467*t we multiply by (-1) both sides of the equation :

10.7 = 0.467*t

t = 10.7 / 0.467

t = 22.9 s

3 0

3 years ago

A worker pushes horizontally on a large crate with a force of 200 N, and the crate is moved 3.5 m. How much work was done in Jou

sattari [20]

Work = (force) x (distance) =

(200 N) x (3.5 m) = <em>700 joules</em>

6 0

4 years ago

Claudia throws a baseball to her dog. Which free-body diagram shows the

chubhunter [2.5K]

Answer:

only the weight of the ball will act on the ball

Explanation: There is no contact force on the ball. Also there is no air resistance on the ball so the friction force on the ball due to air is not shown

6 0

3 years ago

A loaded ore car has a mass of 950 kg. and rolls on rails ofnegligible friction. It starts from rest ans is pulled up a mineshaf

stiks02 [169]

(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

$F - mg sin \theta = 0$

where

F is the force applied to pull the car

m = 950 kg is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=30.0^{\circ}$ is the angle of the incline

Solving for F,

$F=mg sin \theta = (950)(9.8)(sin 30.0^{\circ})=4655 N$

Now we know that the car is moving at constant velocity of

v = 2.20 m/s

So we can find the power done by the motor during the constant speed phase as

$P=Fv = (4655)(2.20)=10241 W$

(b) 10624 W

The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

$v=u+at$

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

$a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2$

So now the equation of the forces along the direction parallel to the incline is

$F - mg sin \theta = ma$

And solving for F, we find the maximum force applied by the motor:

$F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N$

The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:

$P=Fv=(4829)(2.20)=10624 W$

(c) $5.82\cdot 10^6 J$

Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.

The change in potential energy of the car is:

$\Delta U = mg \Delta h$