3 years ago

Please refer to the picture

Physics

1 answer:

Vinvika [58]3 years ago

3 0

I think is c not sure though

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. A grindstone increases in angular speed from 4.00 rad/s to 12.00 rad/s in 4.00 s. Through what angle does it turn during that

ZanzabumX [31]

Answer:

$\theta=32rad$

Explanation:

In an uniformly accelerated circular motion, the angle traveled by the object is given by:

$\theta=\frac{\omega_f+\omega_0}{2}t$

Here $\omega_f$ is the final angular speed, $\omega_0$ is the initial angular speed and t is the time of the motion. Replacing the given values:

$\theta=\frac{12\frac{rad}{s}+4\frac{rad}{s}}{2}(4s)\\\theta=32rad$

5 0

4 years ago

Fan object moves in uniform circular motion in a circle of radius R=200 meters, and the objectes 5.00 seconds to

Leviafan [203]

Answer:

The centripetal acceleration of the object is $31550.72\ m/s^2$ .

Explanation:

We have,

Radius of a circular path is 200 m

It takes 5 seconds to complete 10 revolutions. The angular velocity of the object is given by the rate of change of angular displacement per unit time :

$\omega=\dfrac{\Delta \theta}{\Delta t}\\\\\omega=\dfrac{2\pi \times 10}{5}\\\\\omega=12.56\ rad/s$

The centripetal acceleration of the object is given by :

$a=\omega^2 r\\\\a=(12.56)^2 \times 200\\\\a=31550.72\ m/s^2$

So, the centripetal acceleration of the object is $31550.72\ m/s^2$ .

6 0

3 years ago

Divers in acapulco, mexico, dive headfirst at 8 feet per second from the top of a cliff 87 feet above the pacific ocean. during

xxMikexx [17]

Depends on the wieght of his genitals.

8 0

3 years ago

I can not find the answer to the one after potassium ?

Zepler [3.9K]

Answer:

it’s halogen

Explanation:

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3 years ago

A busy chipmunk runs back and forth along a straight line of acorns that has been set out between his burrow and a nearby tree.

saveliy_v [14]

Answer: a = 1.32m/s2

Therefore, the average acceleration is 1.32m/s2

Explanation:

Acceleration is the rate of change in the velocity per time

a = change in velocity/time

a = ∆v/t

average acceleration a = (v2 -v1)/t. ....1

Given;

Final velocity v2 = 1.63m/s

Initial velocity v1 = -1.15ms

time taken t = 2.11s

Substituting into eqn 1

a = [1.63 - (-1.15)]/2.11

a = (1.63+1.15)/2.11

a = 2.78/2.11

a = 1.32m/s2

Therefore, the average acceleration is 1.32m/s2

6 0

4 years ago