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GarryVolchara [31]

3 years ago

8

There is a known potential difference between two charged plates of 12000 Volts. An object with a charge of 6.5 x 10-6 C charge

and a mass of 0.02 kg is placed next to the positive plate. How fast will it be traveling when it gets to the negative plate

1 answer:

s344n2d4d5 [400]3 years ago

7 0

Answer:

1.97 m/s.

Explanation:

From the question,

Using the law of conservation of energy,

The energy stored in the charged plate = Kinetic energy of the mass

1/2(qV) = 1/2mv².......................... Equation 1

Where q = charge, V = voltage, m = mass, v = velocity.

make v the subject of the equation

v = √(qV/m)......................... Equation 2

Given: q = 6.5×10⁻⁶ C, V = 12000 Volts, m = 0.02 kg

Substitute these values into equation 2

v = √(6.5×10⁻⁶×12000 /0.02)

v = √3.9

v = 1.97 m/s.

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A 1-kilogram mass is attached to a spring whose constant is 21 N/m, and the entire system is then submerged in a liquid that imp

amm1812

Answer:

the required value is $x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

Explanation:

Given that,

mass, m = 1kg

spring constant k = 21N/M

damping force = $-\beta\frac{dx}{dt} = \frac{-10dx}{dt}$

$\beta = 10$

By Newtons second law ,

The diffrential equation of motion with damping is given by

$m\frac{d^2x}{dt^2} = -kx-\beta\frac{dx}{dt}$

substitute the value of m =1kg, k = 21N/M, and $\beta = 10$

$1\frac{d^2x}{dt^2} = -21x=10\frac{dx}{dt}$

$\frac{d^2x}{dt^2} + 10\frac{dx}{dt} + 21x = 0$

suppose the equation of the form $x =e^m^t$ ,

and the auxilliary equation is given by

$m^2 + 10m + 21 = 0\\\\m^2 + 7m+3m+21=0\\\\m(m+7)+3(m+7)=0\\\\(m+7)(m+3)=0\\\\m=-7\\m=-3$

The general solution for the above differential equation is

$x(t) =C_1e^{-3t}+C_2e^{-7t}$

Derivate with respect to t

$x'(t)=-3C_1e^{-3t}-7C_2e^{-7t}$

(a)

since time is 0 then mass is one meter below

so x(0) = 1

Also it start from rest , that implies , velocity is 0 and time is 0

$x'(0) = 0$

substitute the initial condition

$C_1 +C_2 = 1$

$-3C_1-7C_2=0$

Solve the above equation to get C₁ and C₂

$C_1 =\frac{7}{4} and C_2 = -\frac{3}{4}$

substitute for C₁ and C₂ in general solution

$x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

Thus the required value is $x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

3 0

3 years ago

Read 2 more answers

I need help!!!!<br> Please and thank u!

alexandr1967 [171]

Answer:

False

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Explanation:

5 0

3 years ago

A train travels 77 kilometers in 1 hour, and the 66 kilometers in 1 hour. What is the average speed?

Kipish [7]

Average speed = (total distance covered) / (time to cover the distance)

Total distance = (77km + 66km) = 143 kilometers

Time to cover the distance = 2 hours

Average speed = (143 km) / (2 hours) = 71.5 km per hour

6 0

3 years ago

The work function for tungsten metal is 4.52eV a. What is the cutoff (threshold) wavelength for tungsten? b. What is the maximum

Tanya [424]

Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V

Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:

h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.

In order to calculate the cutoff wavelength we have to consider that Ek=0

in this case h*ν=W

(h*c)/λ=4.52 eV

λ= (h*c)/4.52 eV

λ= (1240 eV*nm)/(4.52 eV)=274.34 nm

From this h*ν = Ek+W; we can calculate the kinetic energy for a radiation wavelength of 198 nm

then we have

(h*c)/(λ)-W= Ek

Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV

Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this acts to slow down the ejected electrons from the catode.

5 0

2 years ago

An alpha particle (Î±), which is the same as a helium-4 nucleus, is momentarily at rest in a region of space occupied by an elec

vaieri [72.5K]

Answer:

Speed = 575 m/s

Mechanical energy is conserved in electrostatic, magnetic and gravitational forces.

Explanation:

Given :

Potential difference, U = $-3.45 \times 10^{-3} \ V$

Mass of the alpha particle, $m_{\alpha} = 6.68 \times 10^{-27} \ kg$

Charge of the alpha particle is, $q_{\alpha} = 3.20 \times 10^{-19} \ C$

So the potential difference for the alpha particle when it is accelerated through the potential difference is

$U=\Delta Vq_{\alpha}$

And the kinetic energy gained by the alpha particle is

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

From the law of conservation of energy, we get

$K.E. = U$

$\frac{1}{2}m_{\alpha}v_{\alpha}^2 = \Delta V q_{\alpha}$

$v_{\alpha} = \sqrt{\frac{2 \Delta V q_{\alpha}}{m_{\alpha}}}$

$v_{\alpha} = \sqrt{\frac{2(3.45 \times 10^{-3 })(3.2 \times 10^{-19})}{6.68 \times 10^{-27}}}$

$v_{\alpha} \approx 575 \ m/s$

The mechanical energy is conserved in the presence of the following conservative forces :

-- electrostatic forces

-- magnetic forces

-- gravitational forces

5 0

3 years ago