10. What is the acceleration due to gravity at a location

A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r

atroni [7]

(a) $2.79 rev/s^2$

The angular acceleration can be calculated by using the following equation:

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

where:

$\omega_f = 20.0 rev/s$ is the final angular speed

$\omega_i = 11.0 rev/s$ is the initial angular speed

$\alpha$ is the angular acceleration

$\theta=50.0 rev$ is the number of revolutions made by the disk while accelerating

Solving the equation for $\alpha$ , we find

$\alpha=\frac{\omega_f^2-\omega_i^2}{2d}=\frac{(20.0 rev/s)^2-(11.0 rev/s)^2}{2(50.0 rev)}=2.79 rev/s^2$

(b) 3.23 s

The time needed to complete the 50.0 revolutions can be found by using the equation:

$\alpha = \frac{\omega_f-\omega_i}{t}$

where

$\omega_f = 20.0 rev/s$ is the final angular speed

$\omega_i = 11.0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

t is the time

Solving for t, we find

$t=\frac{\omega_f-\omega_i}{\alpha}=\frac{20.0 rev/s-11.0 rev/s}{2.79 rev/s^2}=3.23 s$

(c) 3.94 s

Assuming the disk always kept the same acceleration, then the time required to reach the 11.0 rev/s angular speed can be found again by using

$\alpha = \frac{\omega_f-\omega_i}{t}$

where

$\omega_f = 11.0 rev/s$ is the final angular speed

$\omega_i = 0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

t is the time

Solving for t, we find

$t=\frac{\omega_f-\omega_i}{\alpha}=\frac{11.0 rev/s-0 rev/s}{2.79 rev/s^2}=3.94 s$

(d) 21.7 revolutions

The number of revolutions made by the disk to reach the 11.0 rev/s angular speed can be found by using

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

where:

$\omega_f = 11.0 rev/s$ is the final angular speed

$\omega_i = 0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

$\theta=?$ is the number of revolutions made by the disk while accelerating

Solving the equation for $\theta$ , we find

$\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{(11.0 rev/s)^2-0^2}{2(2.79 rev/s^2)}=21.7 rev$

4 0

3 years ago

A swimmer, capable of swimming at a speed of 1.0 m/s in still water (i.e., the swimmer can swim with a speed of 1.0 m/s relative

Vesnalui [34]

If the current takes him downstream we must find the resultant vector of the velocities: $V res= \sqrt{1^{2}+0.91^{2} } = \sqrt{1.8281}= 1.3520747$ Then if the river is 3000 m-wide the swimmer will have to pass:
1.3520747 · 300 = 4056.14 m t = 4056.14 m : 1 m/s
a ) It takes 4056.15 seconds ( 1 hour 7 minutes and 36 seconds ) to cross the river.
b ) 0.91 · 3000 = 2730 m
He will be 2730 m downstream.

5 0

3 years ago

The electric force between two charges A. increases with distance between the charges B. increases if either one of charges gets

beks73 [17]

Answer:

Option (A) , (b) and (d) are correct option

Explanation:

According to Coulomb's law electric force between two charges is given by

$F=\frac{1}{4\pi \epsilon _0}\frac{q_1q_2}{r^2}$

From the relation we can say that force is directly proportional to magnitude of charges and inversely proportional to distance between them '

So if we increase the distance then force will decrease

Increase if any of the charge get larger

If force is attractive then both the charge will be of different sign and is force is repulsive then both the charges of same sign

From above conclusion we can say that (a), (b) and (d) are correct option

6 0

3 years ago

Is there a definite end to our atmosphere?

Irina18 [472]

There is no definite end to earths atmosphere, but technically the border between the outer space and earth gets thinner as you move up from the earths surface. The Karman line is the closest definition there is which describes the end of the earth's atmosphere, it is 100 km above earth's sea level at approximately 1.56 % of total earth's radius. This describes the boundary between the outer space and the atmosphere.

7 0

3 years ago

The two properties of an electron that cannot be known exactly at the same time are the ____________________.

enyata [817]

Position and momentum.

This is Heisenberg's Uncertainty Principle:
Δx Δp ≥ h ÷ 4π, where Δx is the change in position, Δp is the change in momentum, and h is Planck's Constant.

6 0

3 years ago