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6 days ago

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Two identical conducting spheres, A and B, sit atop insulating stands. When they are touched, 1.51 × 1013 electrons flow from sp

here B to sphere A. If the total net charge on the spheres is +3.68 μC, what was the initial charge on sphere B?

1 answer:

Sav [2.2K]6 days ago

6 0

A = -0.576 μC B = 4.256 μC Suppose we consider a single electron charge. The total charge transferred from B to A is: Let A and B represent the initial charges of spheres A and B. Given that the overall charge amounts to 3.68μC, we can establish the following equation. After they come into contact, 2.416μC flows from B to A until their charges are equal, leading us to another equation. By combining both equations, we arrive at the solution.

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A spin bike has a flywheel in two parts—a 12.5 kg disk with radius 0.23 m, and a 7.0 kg ring with mass concentrated at the outer

Keith_Richards [2256]

Response:

Clarification:

Provided

weight of disk $m=12.5 kg$

diameter of disc $R=0.23 m$

weight of ring $m_r=7 kg$

Force $F=9.7 N$

$N=180 rpm$

$\omega =\frac{2\pi N}{60}$

$\omega =6\pi rad/s$

Overall moment of inertia

=Disc's moment of inertia +Ring's Moment of Inertia

$=0.5\cdot 12.5\times 0.23^2+7\times 0.23^2$

$=13.25\times 0.23^2=0.7009 kg-m^2$

At this point, Torque is $T=F\times R=I\cdot \alpha$

$9.7\times 0.23=0.7\times \alpha$

$\alpha =3.18 rad/s^2$

Utilizing $\omega _f=\omega +\alpha t$

$\omega _f=0$ in this scenario

$0=6\pi -3.18\times t$

$t=\frac{6\pi }{3.18}$

$t=5.92 s$

7 0

1 day ago

Determine the final state and temperature of 100 g of water originally at 25.0°c after 50.0 kj of heat have been added to it.

inna [2205]

The heat required to raise the temperature of a substance by $\Delta T$ is represented by
$Q=m C_p \Delta T$
where m stands for the mass of the substance and $C_p$ indicates the specific heat of the substance. In this situation, we possess $m=100~g=0.1~Kg$ and $C_p=4.19~KJ/(Kg K)$ , the specific heat of water.
Consequently, we can ascertain the temperature rise $\Delta T$ :
$\Delta T = \frac{Q}{m C_p}= \frac{50~KJ}{0.1~Kg cdot 4.19~KJ/(Kg K)}=119~K =119^{\circ}C$
Initially, the water's temperature was $25^{\circ}C$ , so the end temperature should be
$T_f = 25^{\circ}C+119^{\circ}C=144^{\circ}C$
Thus, the water is expected to be vapor by now.

However, to give a more accurate statement, during the liquid to vapor transition, the heat added to the system is used to break molecular bonds instead of raising the system's temperature. The heat necessary for the phase change from liquid to vapor is expressed as
$Q=m C_L=0.1~Kg \cdot 2265~KJ/Kg=226.5~KJ$
where $C_L$ denotes the latent heat of vaporization for water.
Nevertheless, the initial heat input of 50 KJ is less than this requirement, indicating there isn't sufficient heat to finish the liquid-vapor transition. Therefore, the water will remain in the liquid-vapor change phase at a temperature of $100^{\circ}C$ (the temperature at which the phase change begins)

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25 days ago

A 5.8 × 104-watt elevator motor can lift a total weight of 2.1 × 104 newtons with a maximum constant speed of

Keith_Richards [2256]

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1 month ago

Violet light (λ = 400 nm) passing through a diffraction grating for which the slit spacing is 6.0 μm forms a pattern on a screen

Maru [2355]

Answer:

The third-order dark fringe

Explanation:

y = Distance from central bright fringe = 204 mm

λ = Wavelength = 400 nm

L = Distance from screen to source = 1 m

d = Slit spacing = 6 μm

$tan\theta =\frac{y}{L}\\\Rightarrow tan\theta =\frac{204}{1000}=0.2012^{\circ}$

$dsin\theta=m\lambda\\\Rightarrow m=\frac{dsin\theta}{\lambda}\\\Rightarrow m=\frac{6\times 10^{-6}sin0.2012}{400\times 10^{-9}}=2.9982\approx 3$

The order of the fringe is 3

Thus, it’s identified as a dark fringe.

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18 days ago

Below you are given data about a wave in three different substances.

inna [2205]

1) The wave's period remains constant across different media

2) The wave's velocity varies depending on the medium it travels through

3) As a wave transitions between media, its speed, direction, and wavelength can change, while its frequency stays unchanged

Clarification:

1)

The period of a wave signifies the duration it takes for one full oscillation.

The wave's period is the inverse of its frequency:

$T=\frac{1}{f}$

where

T denotes the period

f is the frequency

The provided table illustrates that the frequency remains consistent across the three media; hence, the period is unchanged as it solely relies on frequency. We can compute it as we know that

f = 350 Hz

thus the period equals

$T=\frac{1}{350}=2.86\cdot 10^{-3} s = 2.86 ms$

2)

The velocity of a wave can be derived from the wave equation:

$v=f \lambda$

where

f indicates the frequency

$\lambda$ is the wavelength

<pin the="" first="" medium="">

$f=350 Hz, \lambda = 0.75 m$ , resulting in a speed of

$v_1 = (350)(0.75)=262.5 m/s$

In the second medium,

$f=350 Hz, \lambda = 0.70 m$ , leading to a speed of

$v_2 = (350)(0.70)=245 m/s$

In the third medium,

$f=350 Hz, \lambda = 0.65 m$ , showing a speed of

$v_3 = (350)(0.65)=227.5 m/s$

As a result, we conclude that the wave's speed varies with the medium.

3)

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16 days ago