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marshall27 [118]
3 years ago
11

Which best describes the motion of air particles when a transverse wave passes through them?

Physics
1 answer:
Minchanka [31]3 years ago
3 0
C.
The particles move perpendicular to the direction of the wave.
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Identify the amplitude in the wave image below.*<br> F. G H J
alina1380 [7]
Answer:F
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6 0
2 years ago
What is the strength of the electric field ep 0.90 mm from a proton?
dem82 [27]

The electric field strength is inversely related to the square of the distance.so  the strength of the electric field is given by

E=\frac{1}{4\pi \epsilon _{0}  } \frac{q}{r^{2} } = \frac{k q}{r^{2} }

Here,  \frac{1}{4\pi \epsilon _{0}  } = k  is constant depend upon medium and its value is 9.0 \times10^{9} \ N m^2/C^2 and q is charge and  r is the distance.

Given  r = 0.90 mm = 9.0 \times 10^{-4} m and we know the charge of proton, q = 1.6\times 10^{-19} \ C.

Therefore,

E=\frac{9.0 \times10^{9} \ N m^2/C^2 \times1.6\times 10^{-19} \ C  }{(9.0 \times 10^{-4} m)^2} = 0.177 \times 10^{-2} \ N/C \\\\  E= 1.77 \times 10^{-3}  N/C

4 0
3 years ago
Read the scenario. A car travels 25 m/s forward for 10 s. Which option accurately identifies the measurements within the scenari
Phantasy [73]

Explanation:

It is given that,

A car travels 25 m/s forward for 10 s.

Solution,

For a vector, a quantity must have both magnitude as well as the direction. For a scalar, a quantity have only the magnitude. In this case, the car moves in forward direction.  This is the only difference between the vector and the scalar.

Out of given option,s the correct option is (c) "The measurement 25 m/s is the only vector quantity because it is a measurement of speed".

5 0
2 years ago
Assume a rectangular strip of a material with an electron density of n=5.8x1020 cm-3. The strip is 8 mm wide and 0.8 mm thick an
vampirchik [111]

Answer: I = 111.69 pA

Explanation: The hall effect is all about the fact that when a semiconductor is placed perpendicularly to a magnetic field, a voltage is generated which could be measured at right angle to the current path. This voltage is known as the hall voltage.

The hall voltage of a semiconductor sensor is given below as

V = I×B/qnd

Where V = hall voltage = 1.5mV =1.5/1000=0.0015V

I = current =?,

n= concentration of charge (electron density) = 5.8×10^20cm^-3 = 5.8×10^20/(100)³ = 5.8×10^14 m^-3

q = magnitude of an electronic charge=1.609×10^-19c

B = strength of magnetic field = 5T

d = thickness of sensor = 0.8mm = 0.0008m

By slotting in the parameters, we have that

0.0015 = I × 5/5.8×10^14 × 1.609×10^-19×0.0008

0.0015 = I×5/7.446×10^-8

I = (0.0015 × 7.446×10^-8)/5

I = 111.69*10^(-12)

I = 111.69 pA

3 0
2 years ago
The electric field in a region of space increases from 0 0 to 3450 N/C 3450 N/C in 4.40 s. 4.40 s. What is the magnitude of the
slavikrds [6]

Answer: B = 1380T

Explanation: please find the attached file for the solution

7 0
3 years ago
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