PHYSICS COLLEGE

What is the speed of an electron with a de Broglie wavelength of 0.20 nm?

Answers

Answer 1

Answer:

Answer:

$v = 3.65 * 10^6 m/s$

Explanation:

Given

de Broglie wavelength = 0.20nm

Required

Determine the speed (v)

The speed is calculated using the following formula;

$L = (h)/(mv)$

Where

$L = Wavelength = 0.2nm$

$h = Planck's\ constant = 6.63 * 10^(-34)Js$

$m = Mass\ of\ Electron = 9.11 * 10^(-31) kg$

Substitute these values in the above formula

$0.2nm = (6.63 * 10^(-34))/(9.11 * 10^(-31) * v)$

-----------------------------------------------------

Convert 0.2nm to metre (m)

$0.2nm = 0.2 * 10^(-9)m$

-----------------------------------------------------

$0.2 * 10^(-9) = (6.63 * 10^(-34))/(9.11 * 10^(-31) * v)$

Multiply both sides by v

$v * 0.2 * 10^(-9) = (6.63 * 10^(-34))/(9.11 * 10^(-31) * v) * v$

$v * 0.2 * 10^(-9) = (6.63 * 10^(-34))/(9.11 * 10^(-31) )$

$v * 0.2 * 10^(-9) = (0.73 * 10^(-34))/(10^(-31) )$

$v * 0.2 * 10^(-9) = 0.73 * 10^(-34) * 10^(31)$

Apply law of indices

$v * 0.2 * 10^(-9) = 0.73 * 10^(-34 + 31)$

$v * 0.2 * 10^(-9) = 0.73 * 10^(-3)$

Divide both sides by $0.2 * 10^(-9)m$

$v = (0.73 * 10^(-3) )/( 0.2 * 10^(-9) )$

$v = (3.65 * 10^(-3) )/(10^(-9) )$

Apply law of indices

$v = 3.65 * 10^(-3) * 10^9$

$v = 3.65 * 10^(-3+9)$

$v = 3.65 * 10^6$

Hence;

The velocity is

$v = 3.65 * 10^6 m/s$

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A wagon is pulled at a speed of 0.40 m/s by a horse exerting 1800 Newtons of horizontal Force. how much work was done by the horse

Answers

The amount of work done per second by the horse exerting a force of 1800 N on a wagon moving with a speed of 0.4 m/s is 720 J/s.

What is power?

Power is the workdone by a body in one second.

To calculate the work done by the horse in one seconds, we use the formula below

Formula:

P = Fv................ Equation 1

Where:

P = work done on the horse in one second
F = Force of the horse
v = Velocity of the wagon

From the question,

Given:

F = 1800 N
v = 0.4 m/s

Substitute these values into equation 1

P = 1800×0.4
P = 720 J/s

Hence, the amount of work done per second by the horse is 720 J/s.

Learn more about power here: brainly.com/question/25864308

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Complete question: A wagon is pulled at a speed of 0.40 m/s by a horse exerting 1800 Newtons of horizontal Force. how much work was done by the horse per second.

Light from a lamp is shining on a surface. How can you increase the intensity of the light on the surface? Light from a lamp is shining on a surface. How can you increase the intensity of the light on the surface? A. Use a lens to focus the power into a smaller area. B. Increase the power output of the lamp. C. Either A or B.

Answers

Answer:

the correct option is C

Explanation:

The intensity of a lamp depends on the power of the lamp that is provided by the current flowing over it, therefore the intensity would increase if we raise the current.

Another way to increase the intensity is to decrease the area with a focusing lens, as the intensity is power over area, decreasing the area increases the power.

When we see the possibilities we see that the correct option is C

Two charged particles attract each other with a force of magnitude F acting on each. If the charge of one is doubled and the distance separating the particles is also doubled, the force acting on each of the two particles has magnitude (a) F/2,
(b) F/4,
(c) F,
(d) 2F,
(e) 4F,
(f) None of the above.

Answers

Answer:

F/2

Explanation:

In the first case, the two charges are Q1 and Q2 and the distance between them is r. K is the Coulomb's constant

Hence;

F= KQ1Q2/r^2 ------(1)

Where the charge on Q1 is doubled and the distance separating the charges is also doubled;

F= K2Q1 Q2/(2r)^2

F2= 2KQ1Q2/4r^2 ----(2)

F2= F/2

Comparing (1) and (2)

The magnitude of force acting on each of the two particles is;

F= F/2

A turntable with a rotational inertia of 0.0120 kg∙m2 rotates freely at 2.00 rad/s. A circular disk of mass 200 g and radius 30.0 cm, and initially not rotating, slips down a spindle and lands on the turntable. (a) Find the new angular velocity. (b) What is the change in kinetic energy?

Answers

To solve this problem it is necessary to apply the related concepts to the moment of inertia in a disk, the conservation of angular momentum and the kinematic energy equations for rotational movement.

PART A) By definition we know that the moment of inertia of a disk is given by the equation

$I = (1)/(2) MR^2$

Where

M = Mass of the disk

R = Radius

Replacing with our values we have

$I = (1)/(2) (0.2)(0.3)^2$

$I = 9*10^(-3)kg\cdot m^2$

The initial angular momentum then will be given as

$I = I_1 \omega_1$

$I = 0.012*2$

$I = 0.024kg\cdot m^2/s$

Therefore the total moment of inertia of the table and the disc will be

$I_2 = 9*10^(-3)+0.012$

$I_2 = 0.021kg\cdot m^2$

The angular velocity at the end point will be given through the conservation of the angular momentum for which it is understood that the proportion of inertia and angular velocity must be preserved. So

$I_1 \omega_1 = I_2\omega_2$

$(0.012)(2)=(1.08*10^(-4))\omega_2$

$\omega_2 = (0.012*2)/(0.021)$

$\omega_2 = 1.15rad/s$

Therefore the new angular velocity is 1.15rad/s

PART B) Through the conservation of rotational kinetic energy we can identify that its total change is subject to

$\Delta KE = (1)/(2)I_1\omega_1^2-(1)/(2)I_2\omega^2$

$\Delta KE = (1)/(2)(I_1\omega_1^2-I_2\omega^2)$

$\Delta KE = (1)/(2)(0.024*2^2-0.021*1.15^2)$

$\Delta KE = 0.034J$

Therefore the change in kinetic energy is 0.034J

How does light move?

Answers

Answer:

Light travels as a wave. But unlike sound waves or water waves, it does not need any matter or material to carry its energy along. This means that light can travel through a vacuum—a completely airless space. It speeds through the vacuum of space at 186,400 miles (300,000 km) per second.

Explanation:

Hope this helps :))

NASA scientists suggest using rotating cylindrical spacecraft to replicate gravity while in a weightless environment. Consider such a spacecraft that has a diameter of d = 148 m. What is the speed v, in meters per second, the spacecraft must rotate at its outer edge to replicate the force of gravity on earth?