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pochemuha
3 years ago
14

There are some points on a standing

Physics
2 answers:
bija089 [108]3 years ago
7 0

Answer:

Nodes.

Explanation:

Nodes are a point that are on a standing wave that never move.

kipiarov [429]3 years ago
4 0

Answer:

C. Nodes

Explanation:

There are some points on a standing  wave that never move. These points are called Nodes.

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When a p-n-p transistor is operated in saturation region, then its ___________<br><br>​
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Base-emitter and Base-collector junctions are forward biased

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• Describe the main ideas of physical science
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Branch of science that studies matter and its motion through space and time, along with related concepts such as energy and force. Describing the nature, measuring and quantifying of bodies and their motion, and dynamics.
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2. A bus drove 8 meters East, then turned to drive 8 meters North, then 2 meters
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2 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
3 years ago
A circular loop of wire of cross-sectional area 0.12 m2 consists of 200 turns, each carrying 0.50 A. It is placed in a magnetic
defon

Answer:

0.52 Nm

Explanation:

A = 0.12 m^2, N = 200, i = 0.5 A, B = 0.050 T

Angle between the plane of loop and magnetic field = 30 Degree

Angle between the normal of loop and the magnetic field = 90 - 30 = 60 degree

θ = 60°

Torque = N i A B Sinθ

Torque = 200 x 0.5 x 0.12 x 0.050 x Sin 60

Torque = 0.52 Nm

4 0
3 years ago
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