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Oksanka [162]
2 years ago
14

What is the angle of reflection of the incident angle is 35°? 35 25° 55 90

Physics
1 answer:
Anestetic [448]2 years ago
4 0

Answer:

i think its the third one or second one

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a class of seventh grade students conducted various scientific investigations throughout the year. which statement is the best e
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c

Explanation:it is

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3 years ago
Complete the sentences to describe the difference between speed and velocity.
xeze [42]

Answer:

velocity =displacement/time

and speed =distance/time

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2 years ago
Which is a sub-atomic particle?
Natalka [10]
A particle that is smaller than an atom or a cluster of particles.
4 0
3 years ago
An electron enters the gap between the plates of a capacitor at the center of the gap traveling parallel to theplates at 2.0 x 1
Svetlanka [38]

Answer:

How far will the electron travel beforehitting a plate is 248.125mm

Explanation:

Applying Gauss' law:

Electric Field E = Charge density/epsilon nought

Where charge density=1.0 x 10^-6C/m2 & epsilon nought= 8.85× 10^-12

Therefore E = 1.0 x 10^-6/8.85× 10^-12

E= 1.13×10^5N/C

Force on electron F=qE

Where q=charge of electron=1.6×10^-19C

Therefore F=1.6×10^-19×1.13×10^5

F=1.808×10^-14N

Acceleration on electron a = Force/Mass

Where Mass of electron = 9.10938356 × 10^-31

Therefore a= 1.808×10^-14 /9.11 × 10-31

a= 1.985×10^16m/s^2

Time spent between plate = Distance/Speed

From the question: Distance=1cm=0.01m and speed = 2×10^6m/s^2

Therefore Time = 0.01/2×10^6

Time =5×10^-9s

How far the electron would travel S =ut+ at^2/2 where u=0

S= 1.985×10^16×(5×10^-9)^2/2

S=24.8125×10^-2m

S=248.125mm

4 0
3 years ago
An airplane traveling at one third the speed of sound (i.e., 114 m/s) emits a sound of frequency 3.72 kHz. At what frequency doe
nlexa [21]

Answer:

f'=5.58kHz

Explanation:

This is an example of the Doppler effect, the formula is:

f'=\frac{(v+v_o)}{(v+v_s)}f

Where f is the actual frequency, f' is the observed frequency, v is the velocity of the sound waves, v_o the velocity of the observer (which is negative if the observer is moving away from the source)  and v_s the velocity of the source  (which is negative if is moving towards the observer). For this problem:

f=3.72kHz\\v=342m/s\\v_o=0m/s\\v_s=-114m/s

f'=\frac{(342+0)}{(342-114)}3.72\times10^3\\f'=\frac{342}{228}3.72\times10^3\\f'=(1.5)3.72\times10^3\\f'=5580Hz=5.58kHz

5 0
2 years ago
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