What is the force needed to throw a ball 7 meters when 1778 J of work is done?

Look at the image to answer the question correctly.

r-ruslan [8.4K]

Answer:

1- b: 2- a : 3- c : 4- d

Explanation:

it starts 2 move away from strting point, then no motion, then moves toward the start, the slows up.

3 0

3 years ago

Which term defines the energy of motion?

Korvikt [17]

D kinetic energy✨
nnnnnnnnnnnnbbbhhbbbbbbb

4 0

3 years ago

A 6.7kg object moves with a velocity of 8m/s. What's its kinetic energy?

Snezhnost [94]

Given parameters:

Mass of object = 6.7kg

Velocity = 8m/s

Unknown parameter:

Kinetic energy = ?

Energy is defined as the ability to do work. There are two forms of energy;

Kinetic and potential energy.

Kinetic energy is the energy due to the motion of a body. Whereas, potential energy is the energy due to the position of a body usually at rest.

Kinetic energy is mathematically expressed as;

Kinetic energy = $\frac{1}{2} m v^{2}$

where m is the mass of the body

v is the velocity of the body

Since we have been given both mass and velocity, input the parameter to solve for the unknown;

Kinetic energy = $\frac{1}{2}$ x 6.7 x 8² = 214.4‬J

So the kinetic energy of the body is 214.4‬J

3 0

3 years ago

When a carpenter shuts off his circular saw, the 10.0 inch diameter blade slows from 4250 rpm to 0.00 in 4.00 s. (a) What is the

MaRussiya [10]

Answer:

(a) $\alpha=-111.26rad/s$

(b) $s=4450.6in$

(c) $8.66in$

Explanation:

First change the units of the velocity, using these equivalents $1rev=2\pi rad$ and $1 min =60s$

$4250rpm(\frac{2\pi rad}{1rev})(\frac{1 min}{60 s} )=445.06rad/s$

The angular acceleration $\alpha$ the time rate of change of the angular speed $\omega$ according to:

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

$\Delta \omega=\omega_i-\omega_f$

Where $\omega_i$ is the original velocity, in the case the velocity before starting the deceleration, and $\omega_f$ is the final velocity, equal to zero because it has stopped.

$\alpha=\frac{\Delta \omega}{\Delta t} =\frac{\omega_i-\omega_f}{4}\frac{0-445.06}{4} =\frac{-445.06}{4} =-111.26rad/s$

b) To find the distance traveled in radians use the formula:

$\theta = \omega_i t + \frac{1}{2} \alpha t^2$

$\theta = 445.06 (4) + \frac{1}{2}(-111.26) (4)^2=1780.24-890.12=890.12rad$

To change this result to inches, solve the angular displacement $\theta$ for the distance traveled $s$ ( $r$ is the radius).

$\theta=\frac{s}{r} \\s=\theta r$

$s=890.12(5)=4450.6in$

c) The displacement is the difference between the original position and the final. But in every complete rotation of the rim, the point returns to its original position. so is needed to know how many rotations did the point in the 890.16 rad of distant traveled:

$\frac{890.12}{2\pi}=141.6667$

The real difference is in the 0.6667 (or 2/3) of the rotation. To find the distance between these positions imagine a triangle formed with the center of the blade (point C), the initial position (point A) and the final position (point B). The angle $\gamma=\frac{2\pi}{3}=\frac{360^o}{3}=120$ is between the two sides known. Using the theorem of the cosine we can find the missing side of the the triangle(which is also the net displacement):