A baseball is thrown horizontally with a velocity of 44 m/s. It travels a horizontal distance of 18m to the plate before it is caught. How long does the ball stay in the air?

A.0.41 sec
B.41 sec
C.4.1 sec
D.4 sec

Answers

Answer 1

Answer:

A horizontal baseball pitch is launched at 44 m/s. The ball will stay for 4.1 sec (approx) in the air. Hence, option C is correct.

What is Velocity?

The rate at which an object's position changes when observed from a specific point of view and when measured against a specific unit of time is known as its velocity.

Its SI unit is represented as m/s, and it is a vector quantity, it means that it has both magnitude and direction.

According to the question, the given values are :

Initial Velocity, u = 44 m/s,

Distance travelled, s = 18 m and,

Final velocity, v = 0.

Use equation of motion :

v = u + at

0 = 44 + (-9.8)t

t = 44 / 9.8

t = 4.3 (approx)

Hence, the time for which the ball stay in the air is 4.1 sec (approx).

To get more information about velocity :

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Answer 2

Answer:

Answer:

a 0.41

plug number into equation

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Ezra (m = 20.0 kg) has a tire swing and wants to swing as high as possible. He thinks that his best option is to run as fast as he can and jump onto the tire at full speed. The tire has a mass of 10.0 kg and hangs 3.50 m straight down from a tree branch. Ezra stands back 10.0 m and accelerates to a speed of 3.62 m/s before jumping onto the tire swing. (a) How fast are Ezra and the tire moving immediately after he jumps onto the swing? m/s (b) How high does the tire travel above its initial height?
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A local meteorologist reports the day’s weather. "Currently sunny outside, 34°F. Skies will become overcast later this afternoon, as temperatures drop to 25°F, with windy conditions out of the north at 10–15 miles per hour. Radar indicates 2–3 inches of snow expected to fall later tonight.” Which information is qualitative? These are non-numerical, descriptive data. These are numerical data that have been measured. “sunny” “25°F” “2–3 inches of snow” “10–15 miles per hour”

Answers

Answer:

sunny

Explanation:

took the test

Answer:

A.) Sunny

Explanation:

What is the speed of an ocean wave if it’s wavelength is 5.0 m and it’s frequency is 3/s?

Answers

Answer:

15 m/s

Explanation:

We know that $v = f * d$ where f = frequency & d = wavelength .

So here.

Wavelength = 5 m

Frequency = 3 s⁻¹

Hence Speed = 5 * 3 = 15 m/s

Impulse is the______of the force and time of contact

Answers

Answer:

Product

Explanation:

Impulse is defined as the average force acting on an object times the time the force acts:

Impulse = F · Δt

A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end. The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it. The mass of the bullet is one-fourth the mass of the rod. What is thefinal angular speed of the rod?

Answers

The value of final angular speed of the uniform rod which rests on the frictionless horizontal surface is,

$\omega=(6v)/(19L)$

What is angular speed of a body?

The angular speed of a body is the rate by which the body changed its angle with respect to the time. It can be given as,

$\omega= (\Delta \theta)/(\Delta t)$

A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end.

The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it.

The mass of the bullet is one-fourth the mass of the rod. The diagram for the above condition is attached below.

In the attached image the angular momentum about the point A is constant just before and after the collision. Thus,

$L_i=L_f\nL_i=(m)/(4)*(vL)/(2)=I\omega\n(m)/(4)*(vL)/(2)=I\omega$

Put the value of inertia as,

$(m)/(4)*(vL)/(2)=\left((mL^2)/(3)+(mL^2)/(4*4)\right)\n(mvL)/(8)*(vL)/(2)=\left((19mL^2)/(48)\right)$

Solving it further we get,

$(v)/(8)=(19)/(48)L\omega\n\omega=(6v)/(19L)$

Hence, the value of final angular speed of the uniform rodwhich rests on the frictionless horizontal surface is,

$\omega=(6v)/(19L)$

Learn more about the angular speed here;

brainly.com/question/540174

Answer: a) 0.315 (V/L)

Explanation:

From Conservation of angular momentum, we know that

L1 = L2 ,

Therefore MV L/2 = ( Irod + Ib) x W

M/4 x V x L/2 = (M (L/2)^2 + 1/3xMxL^2) x W

M/8 X VL = (ML^2/16 + ML^2 /3 )

After elimination we have,

V/8 = 19/48 x L x W

W = 48/8 x V/19L = 6/19 x V/L

Therefore W = (0.136)X V/L

20.0 moles, 1840 g, of a nonvolatile solute, C 3H 8O 3 is added to a flask with an unknown amount of water and stirred. The solution is allowed to reach 90.0°C . The vapor pressure of pure water at this temperature is 528.8 mm Hg. The vapor pressure of the solution is 423.0 mm Hg. How many kg of water was present?

Answers

Answer:

0.144 kg of water

Explanation:

From Raoult's law,

Mole fraction of solvent = vapor pressure of solution ÷ vapor pressure of solvent = 423 mmHg ÷ 528.8 mmHg = 0.8

Let the moles of solvent (water) be y

Moles of solute (C3H8O3) = 2 mole

Total moles of solution = moles of solvent + moles of solute = (y + 2) mol

Mole fraction of solvent = moles of solvent/total moles of solution

0.8 = y/(y + 2)

y = 0.8(y + 2)

y = 0.8y + 1.6

y - 0.8y = 1.6

0.2y = 1.6

y = 1.6/0.2 = 8

Moles of solvent (water) = 8 mol

Mass of water = moles of water × MW = 8 mol × 18 g/mol = 144 g = 144/1000 = 0.144 kg

A ball with a mass of 4 kg is initially traveling at 2 m/s and has a 5 N force applied for 3 s. What is the initial momentum of the ball?

Answers

Answer:

The initial momentum of the ball is 8 kg-m/s.

Explanation:

Given that,

Mass of the ball is 4 kg

Initial speed of the ball is 2 m/s

Force applied to the ball is 5 N for 3 seconds

It is required to find the initial momentum of the ball. Initial momentum means that the product of mass and initial velocity of the ball. It is given as :

$p_i=mu\n\np_i=4\ kg* 2\ m/s\n\np_i=8\ kg-m/s$