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hjlf
3 years ago
6

Consider the following three concentric systems two thick shells and a solid sphere all conductors The radii in the increasing o

rder are a b c d and e The small sphere is given an excess charge of 3 C and the smaller shell is given an excess charge of 7 C The larger shell is electrically neutral The system quickly comes to electrostatic equilibrium state a Note that there are 5 conducting surfaces What are the electric charges values and signs on the each of them Are these charges distributed uniformly
Physics
1 answer:
Margarita [4]3 years ago
5 0

Answer:

Explanation:

From the given question, the small sphere was provided with an excess charge of +3 C, while the smaller shell was given an excess of -7 C, it should be -7 C and not 7 C.

So, in light of that, to determine the electric charges values & signs on each of them, we have:

on a = +3 C

on b = -7 C

on c = -7 C

on d = +3 C

on e = -7 C

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A student uses a spring loaded launcher to launch a marble vertically in the air. The mass of the marble is
GarryVolchara [31]

Answer:

Part a)

When spring compressed by 2 cm

H = 1.47 m

Part b)

When spring is compressed by 4 cm

H = 5.94 m

Explanation:

Part a)

As we know that the spring is compressed and released

so here spring potential energy is converted into gravitational potential energy at its maximum height

So we will have

\frac{1}{2}kx^2 = mg(H + x)

0.5(220)(0.02)^2 = 0.003(9.81)(H + 0.02)

so we have

H = 1.47 m

Part b)

Similarly when spring is compressed by 4 cm

then we have

\frac{1}{2}kx^2 = mg(H + x)

0.5(220)(0.04)^2 = 0.003(9.81)(H + 0.04)

so we have

H = 5.94 m

8 0
3 years ago
Consider a spring mass system (mass m1, spring constant k) with period T1. Now consider a spring mass system with the same sprin
tatuchka [14]

Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of m_2 would be four times that of m_1.

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2.

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is T, then the angular velocity of the SHM would be

\displaystyle \omega = \frac{2\pi}{T}.

Assume that the mass starts with a zero displacement and a positive velocity. If A represent the amplitude of the SHM, then the displacement of the mass at time t would be:

\mathbf{x}(t) = A\sin(\omega\cdot t).

The velocity of the mass at time t would be:

\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t).

The acceleration of the mass at time t would be:

\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t).

Let m represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time t would be:

\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t),

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant k will be equal to:

\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}.

Since \displaystyle \omega = \frac{2\pi}{T}, it can be concluded that:

\begin{aligned} k &= m \, \omega^2 = m \left(\frac{2\pi}{T}\right)^2\end{aligned}.

For the first mass m_1, if the time period is T_1, then the spring constant would be:

\displaystyle k = m_1\, \left(\frac{2\pi}{T_1}\right)^2.

Similarly, for the second mass m_2, if the time period is T_2, then the spring constant would be:

\displaystyle k = m_2\, \left(\frac{2\pi}{T_2}\right)^2.

Since the two springs are the same, the two spring constants should be equal to each other. That is:

\displaystyle m_1\, \left(\frac{2\pi}{T_1}\right)^2 = k = m_2\, \left(\frac{2\pi}{T_2}\right)^2.

Simplify to obtain:

\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2.

6 0
3 years ago
Calculate the displacement of a motorcycle that accelerates forward from rest to 7.0 m/s in 3.5 s.
Harrizon [31]

Answer:

12.25 meters

Explanation:

s=1/2(v+u)t

s= displacement

v= final velocity

u= initial velocity

t= time

7m/s+0m/s divide by 2= 3.5 m/s velocity  Times 3.5 seconds=  12.25 meters

3 0
3 years ago
Why does too much carbon in the air causen a problem?
Sonbull [250]
It causes global warming
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3 years ago
A laser shines violet light with a frequency of 703 thz onto the metal with a threshold frequency of 295 thz, what is the kineti
hram777 [196]

The kinetic energy of the ejected electrons will be 2782.5 ×10⁻²² eV.

<h3>What is threshold frequency?</h3>

The threshold frequency of incoming radiation is the lowest frequency at which photoelectric emission or electron emission is impossible.

The threshold frequency is the light frequency that causes an electron to dislodge and emit from the metal's surface.

From the photoelectric effect, the equation obtained as;

\rm  hv=hv_0+KE\\\\ KE=h(v-v_0)\\\\ KE=6.625 \times 10^{-34}(702-282)Hz \\\\ KE= 6.625 \times 10^{-34}\times 420 \times 10^{12} \ Hz \\\\ KE=2782 .5 \times 10^{-22} \ eV

Hence,the kinetic energy of the ejected electrons will be 2782.5 ×10⁻²² eV.

To learn more about the threshold frequency, refer to the link;

brainly.com/question/2499414

#SPJ1

3 0
2 years ago
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