Select all that apply.

A thin, metallic spherical shell of radius 0.227 m has a total charge of 6.03 × 10 − 6 C placed on it.

KATRIN_1 [288]

Answer:

Explanation:

Given

radius $r=0.227 m$

Charge on surface $Q=6.03\times 10^{-6} C$

Point Charge inside sphere $q=1.15\times 10^{-6} C$

Electric Field at $r=0.735 m$

Treating Surface charge as Point charge and applying Gauss law

$E_{total}A=\frac{q_{enclosed}}{\epsilon _0}$

where A=surface area up to distance r

$E_{total}=\frac{Q+q}{4\pi r^2}$

$E_{total}=\frac{6.03\times 10^{-6}+1.15\times 10^{-6}}{4\pi (0.735)^2\times 8.85\times 10^{-12}}$

$E_{total}=1.194\times 10^{5} N/C$

3 0

3 years ago

How does the total amount of energy change as the spring goes up and down?

NeX [460]

The total amount of energy always stays the same because of the law of conservation of energy, meaning that there is always the same amount of energy, it just gets turned into different forms like potential energy, kinetic energy, sound, thermal, etc.

7 0

3 years ago

You are trying to overhear a juicy conversation, but from your distance of 25.0 m , it sounds like only an average whisper of 20

spayn [35]

Answer:So You Decide To Move Closer To Give The Conversation A Sound Level Of 80.0dB Instead. ... You are trying to overhear a juicy conversation, but from your distance of 24.0m , it sounds like only an average whisper of 40.0dB .

Explanation:

7 0

3 years ago

If two particles have equal kinetic energies, are their momenta necessarily equal? explain.

Mandarinka [93]

Answer:

No the given statement is not necessarily true.

Explanation:

We know that the kinetic energy of a particle of mass 'm' moving with velocity 'v' is given by

$K.E=\frac{1}{2}mv^{2}$

Similarly the momentum is given by $m\times v$

For 2 particles with masses $m_{1},m_{2}$ and moving with velocities $v_{1},v_{2}$ respectively the respective kinetic energies is given by

$K.E_{1}=\frac{1}{2}m_{1}v_{1}^{2}$

$K.E_{2}=\frac{1}{2}m_{2}v_{2}^{2}$

Similarly For 2 particles with masses $m_{1},m_{2}$ and moving with velocities $v_{1},v_{2}$ respectively the respective momenta are given by

$p_{1}=m_{1}\times v_{1}$

$p_{2}=m_{2}\times v_{2}$

Now since it is given that the two kinetic energies are equal thus we have

$\frac{1}{2}m_{1}v_{1}^{2}=\frac{1}{2}m_{2}v_{2}^{2}\\\\(m_{1}v_{1})\times v_{1}=(m_{2}v_{2})\times v_{2}\\\\p_{1}\times v_{1}=p_{2}\times v_{2}\\\\\therefore \frac{p_{1}}{p_{2}}=\frac{v_{2}}{v_{1}}............(i)$

Thus we infer that the moumenta are not equal since the ratio on right of 'i' is not 1 , and can be 1 only if the velocities of the 2 particles are equal which becomes a special case and not a general case.

5 0

3 years ago

A 1150 kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 7.69 m before contacting the beam,

Natasha_Volkova [10]

Answer:

the average force 11226 N

Explanation:

Let's analyze the problem we are asked for the average force, during the crash, we can find this from the impulse-momentum equation, but this equation needs the speeds and times of the crash that we could look for by kinematics.

Let's start looking for the stack speeds, it has a free fall, from rest (Vo=0)

Vf² = Vo² - 2gY

Vf² = 0 - 2 9.8 7.69 = 150.7

Vf = 12.3 m / s

This is the speed that the battery likes when it touches the beam. They also give us the distance it travels before stopping, let's calculate the time

Vf = Vo - g t

0 = Vo - g t

t = Vo / g

t = 12.3 / 9.8

t = 1.26 s

This is the time to stop

Now let's use the equation that relates the impulse to the amount of movement

I = Δp

F t = pf-po

The amount of final movement is zero because the system stops

F = - po / t

F = - mv / t

F = - 1150 12.3 / 1.26

F = -11226 N

This is the average force exerted by the stack on the vean

7 0

3 years ago