Describe what happens to the magnitude of the net electrostatic force on the electron as the electron

A water balloon is dropped out of a fifth floor window and hits an unsuspecting person below 4.0 seconds later. The window is ap

andrew11 [14]

Answer:

80 meters high

Explanation:

The velocity of the balloon would be g*t (I won't calculate, but will us this later)

We know that the kinetic energy at the bottom equals the potential at the top.

KE = PE

1/2 * m * v^2 = m * g * h

1/2 * m * (g * t)^2 = m * g * h (substitution)

1/2 * m * g^2 * t^2 = m * g * h

1/2 * g * t^2 = h (simplification by dividing the commons between both sides)

h = 1/2 * 9.81 * 4^2

h = 78.48 m (roughly 80 m)

4 0

3 years ago

The radder clocked sonic going 300 miles per hour how long will it take him to go 48.1 miles from Miami to Florida

lys-0071 [83]

Answer:

9.62 minutes 0r 0.16 of an hour

Explanation:

Speed = distance/time

300mph = 48.1 m/t

xt

300t = 48.1

÷300

t = 48.1/300

t = 0.16033333333 hr

0.16033333333 x 60 = 9.62 minutes

60 minutes in an hour

9.62/60= 0.16033333333 hr

So, around 10 minutes.

Hope this helps!

8 0

2 years ago

An electric field of 3x10^16 n/c is needed to create a spark in the air. If the charged particles in the field are separated by

meriva

The answer of this problem is positive 3

6 0

3 years ago

Will an object with more mass roll faster down a hill?

inna [77]

Over time, yes. It will over time gain more momentum

5 0

3 years ago

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

So

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

3 0

3 years ago