Ask question

Ask question

All categories

Katyanochek1 [597]

3 years ago

9

sitting on the dock of the bay wasting time with my sister. i get bored and push her off the 2 m dock. How fast is she moving wh

en she belly flops into the water? (And more importantly how badly is she going to hurt me when she catches me?)

1 answer:

MrMuchimi3 years ago

3 0

Answer:

The data that we have is:

Initial height = 2m

Initial vertical speed: As the sister is only pushed, we can assume that we do not have initial speed in the vertical axis.

Then at the beginning, we only have potential energy, that (using the water as the zero in our axis) can be written as:

U = m*g*h

where m is the mass of your sister, g is the gravitational acceleration, and h is the initial height:

h = 2m

g = 9.8m/s^2

Now, when your sister is about to touch the water, all the potential energy has be transformed into kinetic energy:

K = (m/2)*v^2

Then we have, at that moment:

K = U

(m/2)*v^2 = m*g*h

We want to solve this for v, that is the velocity.

v = √(2*g*h) = √(2*9.8m/s^2*2m) = 6.26 m/s

So the speed at which your sister hits the water is 6.26 m/s

You might be interested in

A 13.5 μF capacitor is connected to a power supply that keeps a constant potential difference of 22.0 V across the plates. A pie

-BARSIC- [3]

a) $3.27\cdot 10^{-3} J$

b) $11.60\cdot 10^{-3} J$

c) $8.33\cdot 10^{-3} J$

Explanation:

a)

The energy stored in a capacitor is given by

$U=\frac{1}{2}CV^2$

where

C is the capacitance of the capacitor

V is the potential difference across the plates of the capacitor

For the capacitor in this problem, before insering the dielectric, we have:

$C=13.5 \mu F = 13.5\cdot 10^{-6}F$ is its capacitance

V = 22.0 V is the potential difference across it

Therefore, the initial energy stored in the capacitor is:

$U=\frac{1}{2}(13.5\cdot 10^{-6})(22.0)^2=3.27\cdot 10^{-3} J$

b)

After the dielectric is inserted into the plates, the capacitance of the capacitor changes according to:

$C'=kC$

where

k = 3.55 is the dielectric constant of the material

C is the initial capacitance of the capacitor

Therefore, the energy stored now in the capacitor is:

$U'=\frac{1}{2}C'V^2=\frac{1}{2}kCV^2$

where:

$C=13.5\cdot 10^{-6}F$ is the initial capacitance

V = 22.0 V is the potential difference across the plate

Substituting, we find:

$U'=\frac{1}{2}(3.55)(13.5\cdot 10^{-6})(22.0)^2=11.60\cdot 10^{-3} J$

C)

The initial energy stored in the capacitor, before the dielectric is inserted, is

$U=3.27\cdot 10^{-3} J$

The final energy stored in the capacitor, after the dielectric is inserted, is

$U'=11.60\cdot 10^{-3} J$

Therefore, the change in energy of the capacitor during the insertion is:

$\Delta U=11.60\cdot 10^{-3}-3.27\cdot 10^{-3}=8.33\cdot 10^{-3} J$

So, the energy of the capacitor has increased by $8.33\cdot 10^{-3} J$

8 0

3 years ago

Which trial’s cart has the greatest momentum at the bottom of the ramp?

Licemer1 [7]

The momentum of each cart is given by:
$p=mv$
where
m is the mass of the cart
v is its velocity (at the bottom of the ramp)

To answer the problem, let's calculate the momentum of each of the 4 carts:
1) $p=(200 kg)(6.5 m/s)=1300 kg m/s$
2) $p=(220 kg)(5.0 m/s)=1100 kg m/s$
3) $p=(240 kg)(6.4 m/s)= 1536 kg m/s$
4) $p=(260 kg)(4.8 m/s)=1248 kg m/s$

Therefore, the cart with greatest momentum is cart 3, so the right answer is
<span>- trial 3, because this trial has a large mass and a large velocity</span>

8 0

3 years ago

An object displaces 652ml of water. What is the volume of the object

Eva8 [605]

If the object is completely submerged at that time, then the volume of
the full object is 652 cm³.

If only part of the object is in the water, then the volume of the part that's
in the water is 652 cm³, but we have no idea what its full volume is.

8 0

3 years ago

If the velocity of a pitched ball has a magnitude of 47.0 m/s and the batted ball's velocity is 50.5 m/s in the opposite directi

Naya [18.7K]

Answer:

The magnitude of change in momentum of the ball is $97.5 m$ and impulse is also $97.5 m$

Explanation:

Given:

Velocity of a pitched ball $v _{i} = 47$ $\frac{m}{s}$

Velocity of ball after impact $v_{f} = -50.5$ $\frac{m}{s}$

From the formula of change in momentum,

$\Delta P = m (v_{f} -v_{i} )$

Here mass is not given in question,

Mass of ball is $m$

Change in momentum is given by,

$\Delta P = m (-50.5 -47)$

$\Delta P = -97.5 m$

Magnitude of change in momentum is

$\Delta P = 97.5 m$

And impulse is given by

$J = \Delta P$

$J = -97.5 m$

So impulse and

Therefore, the magnitude of change in momentum of the ball is $97.5 m$ and impulse is also $-97.5 m$

6 0

3 years ago

Which of the following is a category of mechanical wave?

givi [52]

Answer:

a

because the mechanical wave is when it goes over and over again

8 0

3 years ago

Read 2 more answers