Supposing the runner is condensed to a point and moves upward at 2.2 m/s.
It takes a time = 2.2/g = 2.2/9.8 = 0.22 seconds to increase to max height.
Now looking at this condition in opposite - that is the runner is at max height and drops back to earth in 0.22 s (symmetry of this kind of motion).
From what height does any object take 0.22 s to fall to earth (supposing there is no air friction)?
d = 1/2gt²= (0.5)(9.8)(0.22)²= 0.24 m

7 0

3 years ago

William Tell shoots an apple from his son's head. The speed of the 102-g arrow just before it strikes the apple is 26.7 m/s, and

xxMikexx [17]

To develop the problem, we require the values concerning the conservation of momentum, specifically as given for collisions.

By definition the conservation of momentum tells us that, $m_1V_1+m_2V_2 = (m1+m2)V_f$

To find the speed at which the arrow impacts the apple we turn to the equation of time, in which,

$t= \sqrt{\frac{2h}{g}}$

The linear velocity of an object is given by

$V=\frac{X}{t}$

Replacing the equation of time we have to,

$V_f = \frac{X}{t}\\V_f =\frac{X}{\sqrt{\frac{2h}{g}}}\\V_f = \frac{6.9}{\sqrt{\frac{2(1.85)}{9.8}}}\\V_f = 11.23m/s$

Velocity two is neglected since there is no velocity of said target before the collision, thus,

$m_1V_1 = (m1+m2)V_f$

Clearing for m_2

$m_2 = \frac{m_1V_1}{V_f}-m_1\\m_2 = \frac{(0.102)(26.7)}{11.23}-0.102\\m_2 = 0.1405KG= 140.5g$

8 0

3 years ago

A soap bubble appears red (λ = 633nm) at the point on its front surface nearest to the viewer. Assuming n = 1.35, what is the sm

alexira [117]

Answer:

The smallest film thickness is 117 nm.

Explanation:

Light interference on thin films can be constructive or destructive. Constructive interference is dependent on the film thickness and the refractive index of the medium.

For the first interference (surface nearest to viewer), the minimum thickness can be expressed as:

$2t_{min} = \frac{wavelenth}{2n}$