S= 1/2(V f + V I )t solve for V f <br><br> show me how

You wash the dishes. when you let the water down the drain, where is the water flowing the fastest?

Goshia [24]

The water flows the fastest at the center.

This is due to the friction between the water and the sink/drain. The outer parts of the water that are in contact with the kitchen sink/drain experience a greater frictional force and slow down. Their slowing down also affects the layers of water molecules adjacent to them, thereby reducing their speed as well. The layer of water molecules least affected by the friction with the sink/drain is at the center, and so they are the ones that move the fastest.

3 0

4 years ago

A celebrated Mark Twain story has motivated contestants in the Calaveras County Jumping Frog Jubilee, where frog jumps as long a

vladimir1956 [14]

Answer: x= 4.761 m/s

t=0.786 sec

Explanation: In a projectile motion (or 2D motion), the object is launched with an initial angle and an initial velocity

The components of the velocity are

<em>The magnitude, which is the speed, and the direction in which the motion is happening.</em>

Similarly the displacement has the components

The last formula is valid only if the object is launched at ground level, as our frog does.

There are two times where the value of y is zero, when t=0 (at launching time) and when it lands back from the air. We need to find that time t by making y=0

Dividing by t (assuming t different from zero)

Then we find the total flight as

Replacing this time in the formula of x

We can solve for

Knowing that x=2.20 m and °

We now compute t

6 0

3 years ago

What is the outermost layer of the sun? photosphere corona core radiative zone chromosphere convective zone

murzikaleks [220]

The answer is Corona

7 0

3 years ago

Read 2 more answers

A wheel is rotating about an axis that is in the z direction The angular velocity ωz is 6.00 rad s at t 0 increases linearly wit

Amanda [17]

A) $+1.67 rad/s^2$

The angular acceleration of the wheel is given by

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

where

$\omega_i = -6.00 rad/s$ is the initial angular velocity of the wheel (initially clockwise, so with a negative sign)

$\omega_f = 4.00 rad/s$ is the final angular velocity (anticlockwise, so with a positive sign)

$\Delta t= 6.00 s - 0=6.00 s$ is the time interval

Substituting into the equation, we find the angular acceleration:

$\alpha = \frac{4.00 rad/s - (-6.00 rad/s)}{6.00 s}=+1.67 rad/s^2$

And the acceleration is positive since the angular velocity increases steadily from a negative value to a positive value.

B) 3.6 s

The time interval during which the angular velocity is increasing is the time interval between the instant $t_1$ where the angular velocity becomes positive (so, $\omega_i=0$ ) and the time corresponding to the final instant $t_2 = 6.0 s$ , where $\omega_f = +6.00 rad/s$ . We can find this time interval by using

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

And solving for $\Delta t$ we find

$\Delta t = \frac{\omega_f - \omega_i}{\alpha}=\frac{+6.00 rad/s-0}{+1.67 rad/s^2}=3.6 s$

C) 2.4 s

The time interval during which the angular velocity is idecreasing is the time interval between the initial instant $t_1=0$ when $\omega_i=-4.00 rad/s$ ) and the time corresponding to the instant in which the velovity becomes positive $t_2$ , when $\omega_f = 0 rad/s$ . We can find this time interval by using