Ask question

Ask question

All categories

3 years ago

6

A baseball player hits a baseball with a bat. The mass of the ball is 0.05 kilograms. The ball accelerates at 2,000 meters per s

econd squared.
What is the net force, to the nearest newton, that accelerates the ball?

1 answer:

AlexFokin [52]3 years ago

8 0

Answer:

550

Explanation:

You might be interested in

2. With regard to the pH scale, a solution with a pH

Alexandra [31]

Answer:

ph of 10 b your welcome

Explanation:

8 0

4 years ago

A 3.1-kilogram gun initially at rest is free to

34kurt

The conservation of momentum states that the total momentum in a system is constant if there is no external force acting on the system. The total momentum in the gun bullet system is 0 so it must stay that way.

The momentum of the bullet is mv = 0.015*500=7.5
The momentum of the gun must be the same to keep the total momentum of the system equal to zero, so we know that p = 7.5 for the gun.
Substituting this in we get:

7.5=3.1x
x=7.5/3.1
x=2.42

So the speed of the gun is 2.4m/s.

8 0

3 years ago

Read 2 more answers

What is a property of “normal force”? a. It always points perpendicular to the contact surface. b. It always points parallel to

OleMash [197]

Answer:

a. It always points perpendicular to the contact surface.

Explanation:

"Normal" means perpendicular. Normal forces are always perpendicular to the contact surface.

6 0

3 years ago

A vertical cylindrical tank 10 ft in diameter, has an inflow line of 0.3 ft inside diameter and an outflow line of 0.4 ft inside

neonofarm [45]

Answer:

$\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}$ , level is rising.

Explanation:

Since liquid water is a incompresible fluid, density can be eliminated of the equation of Mass Conservation, which is simplified as follows:

$\dot V_{in} - \dot V_{out} = \frac{dV_{tank}}{dt}$

$\frac{\pi}{4}\cdot D_{in}^2 \cdot v_{in}-\frac{\pi}{4}\cdot D_{out}^2 \cdot v_{out}= \frac{\pi}{4}\cdot D_{tank}^{2} \cdot \frac{dh}{dt} \\D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out} = D_{tank}^{2} \cdot \frac{dh}{dt} \\\frac{dh}{dt} = \frac{D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out}}{D_{tank}^{2}}$

By replacing all known variables:

$\frac{dh}{dt} = \frac{(0.3 ft)^{2}\cdot (5 \frac{ft}{s} ) - (0.4 ft)^{2} \cdot (2 \frac{ft}{s} )}{(10 ft)^{2}}\\\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}$

The positive sign of the rate of change of the tank level indicates a rising behaviour.

6 0

4 years ago

A ship's propeller of diameter 3 m makes 10.6 revolutions in 30s. What is the angular velocity of the propeller?

ycow [4]

Answer:

The angular velocity of the propeller is 2.22 rad/s.

Explanation:

The angular velocity (ω) of the propeller is:

$\omega = \frac{\Delta \theta}{\Delta t}$

Where:

θ: is the angular displacement = 10.6 revolutions

t: is the time = 30 s

$\omega = \frac{\Delta \theta}{\Delta t} = \frac{10.6 rev*\frac{2\pi rad}{1 rev}}{30 s} = 2.22 rad/s$

Therefore, the angular velocity of the propeller is 2.22 rad/s.

I hope it helps you!

5 0

3 years ago