WHAT- HOW WAS THIS NOT ACADEMIC? IT WAS A QUESTION ABOUT WHY BLACK HOLES ARE INVISIBLE LOL.

A green croquet ball of mass 0.50 kg is rolling at +12 m/s. It collides with a blue croquet ball that also

Svetach [21]

Answer:

a) 9.6 m/s

b) 11.7 m/s

c) 12 m/s

Explanation:

This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=m_{g} V_{o} + m_{b} U_{o}$ (2)

$p_{f}=m_{g} V_{f} + m_{b} U_{f}$ (3)

$m_{g}=0.5 kg$ is the mass of green ball

$m_{b}=0.5 kg$ is the mass of the blue ball

$V_{o}=12 m/s$ is the initial velocity of the green ball

$U_{o}=0 m/s$ is the initial velocity of the blue ball

$V_{f}$ is the final velocity of the green ball

$U_{f}$ is the final velocity of the blue ball

Substituting (2) and (3) in (1):

$m_{g} V_{o} + m_{b} U_{o}=m_{g} V_{f} + m_{b} U_{f}$ (4)

Isolating $U_{f}$ :

$U_{f}=\frac{m_{g} V_{o} - m_{g} V_{f}}{m_{b}}$ (5)

$U_{f}=\frac{m_{g} (V_{o} - V_{f})}{m_{b}}$ (6) This is the equation we will use for the next cases

Knowing this, let's begin with the answers:

a) In this case $V_{f}=2.4 m/s$ and we have to find $U_{f}$

$U_{f}=\frac{0.5 kg (12 m/s - 2.4 m/s)}{0.5 kg}$ (7)

$U_{f}=9.6 m/s$ (8)

b) In this case $V_{f}=0.3 m/s$ and we have to find $U_{f}$

$U_{f}=\frac{0.5 kg (12 m/s - 0.3 m/s)}{0.5 kg}$ (9)

$U_{f}=11.7 m/s$ (10)

c) In this case $V_{f}=0 m/s$ and we have to find $U_{f}$

$U_{f}=\frac{0.5 kg (12 m/s - 0 m/s)}{0.5 kg}$ (11)

$U_{f}=12 m/s$ (12)

4 0

3 years ago

A car is 2.0 km west of a traffic light at t = 0 and 5.0 km east of the light at t = 6.0 min. Assume the origin of the coordinat

egoroff_w [7]

Answer:

(a) Position Vectors V₁= -2î km, V₂=5î km

(b) Displacement Δx=7 km

Explanation:

Given data

Distance=2 km west at t=0

Distance=5 km east at t=6 min

Positive x is the east direction

To find

(a)Car position vector at given times

(b)Displacement between 0 to 6.0 min

Solution

For Part (a) car position vector at given times

At t=0 the distance=2 km west so conclude that x₁=-2 because it is in negative side So vector V₁

V₁= -2î km

At t=6.0 the distance=5 km east so conclude that x₂=5 because it is in positive side So vector V₂

V₂=5î km

For (b) displacement between 0 to 6.0 min

According to following mathematical law we can conclude that

Δx=x₂-x₁

Δx=5-(-2)km

Δx=7 km

6 0

3 years ago

What is a normal force?

Flauer [41]

Is the component perpendicular to the surface on contact of the contact force <span />

3 0

4 years ago

In a thunderstorm, electric charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can be considere

jolli1 [7]

1) 333.6 C

In order to have breakdown, the electric field at the surface of the cloud must be equal to the breakdown electric field:

$E=3.00\cdot 10^6 N/C$

The electric field strength at the surface of a charged sphere is given by

$E=\frac{1}{4\pi \epsilon_0} \frac{Q}{R^2}$

where

$\epsilon_0 = 8.85\cdot 10^{-12} C^2/(N^2 m^2)$ is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

Here we have a cloud of radius

$R = 1.00 km = 1000 m$

So we can re-arrange the previous equation in order to find the charge on the cloud:

$Q=4\pi \epsilon_0 ER^2=4\pi (8.85\cdot 10^{-12})(3.00\cdot 10^6)(1000)^2=333.6 C$

2) $2.08\cdot 10^{21}$ excess electrons

The total charge of the cloud must be (in magnitude)

Q = 333.3 C

We know that one electron carries a charge of

$e = 1.6 \cdot 10^{-19}C$

The total charge is just given by the charge of each electron multiplied by the number of excess electrons in the cloud:

$Q=Ne$

where

N is the number of excess electrons

Solving for N, we find:

$N=\frac{Q}{e}=\frac{333.3 C}{1.6\cdot 10^{-19} C}=2.08\cdot 10^{21}$

6 0

3 years ago

Why are fuse wires not provided in an electric circute containing an electric cell

Basile [38]

Hey there,

Electric cells we use usually carry limited voltage. Here's an example: 1.5V. Really, there is no chance of any shot circuit. that's why MCB and fuse wires are not used in their circuit .

:)

3 0

3 years ago