Beer producers and bread bakers depend on types of yeast to respirate anaerobically to make their products.

A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 3.05 m/s. If the roof is pitched at

vekshin1

1) Time in the air: 0.78 s

The motion of the ball is a projectile motion, which consists of two independent motions:

- A horizontal motion with constant horizontal velocity

- A vertical motion with constant downward acceleration of

$g=-9.8 m/s^2$ (acceleration of gravity)

The initial vertical velocity is

$u_y = u sin \theta = (3.05)(sin(-40^{\circ}))=-1.96 m/s$

where the negative sign means the direction is downward.

The vertical position of the ball is given by

$y(t) = h + u_y t + \frac{1}{2}gt^2$

where

h = 4.50 m is the initial heigth of the ball when it starts falling down

The ball reaches the ground when y = 0, so we have:

$0 = 4.50 -1.96t-4.9t^2$

This is a second-order equation; solving for t, we get

t = -1.18 s

t = 0.78 s

We discard the negative solution since it has no physical meaning, so we can say that the ball spent 0.78 s in the air.

2) Horizontal distance: 1.83 m

For this second part of the problem, we just have to consider the horizontal motion of the ball.

As we said previously, the motion of the ball along the horizontal direction is a uniform motion with constant velocity, which is given by

$v_x = u cos \theta = (3.05)(cos (-40.0^{\circ}))=2.34 m/s$

where u = 3.05 m/s is the initial speed and $\theta$ the angle of projection.

For a uniform motion, we can use the following relationship between distance covered and velocity:

$d=v_x t$

and substituting t = 0.78 s, we find the total distance travelled along the horizontal direction by the ball before reaching the ground:

$d=(2.34)(0.78)=1.83 m$

7 0

3 years ago

What is Aristotle famous for?

Mila [183]

Answer:

History

Explanation:

Im. Sorry. Im. Not. Much. Help

8 0

3 years ago

A car is traveling at 16 m/s^2

Leya [2.2K]

Really? wow that's pretty cool

4 0

3 years ago

What is the force in N of an object that has a mass of 7 kilograms and an acceleration of 4 m/s/s

VMariaS [17]

Newton's 2nd law of motion: Force = (mass) x (acceleration)

If you want to move a 7-kg object with an acceleration of 4 m/s²,
then you will need to push it with (7 x 4) = 28 newtons of force.

4 0

3 years ago

Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore f

tatiyna

Answer:

a) h=3.16 m, b) v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

Em₀ = U = mg h

final point. Lowest point

$Em_{f}$ = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

K = ½ m $v_{cm }^{2}$ + ½ $I_{cm}$ w²

angular and linear speed are related

v = w r

w = v / r

K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

Em₀ = Em_{f}

mg h = ½ v_{cm }^{2} (m + I_{cm} / r²) (1)

h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

I_{cm} = ⅔ m r²

we substitute

h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

h = ½ v_{cm }^{2}/g 5/3

h = 5/6 v_{cm }^{2} / g

let's calculate

h = 5/6 6.1 2 / 9.8

h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

I_{cm} = ½ m r²

we substitute

v_{cm } = √ [2gh / (1 + ½)]

v_{cm } = √(4/3 gh)

let's calculate

v_{cm } = √ (4/3 9.8 3.16)

v_{cm }^ = 6.43 m / s

4 0

3 years ago