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solniwko [45]
3 years ago
5

Dimensional formula of Amplitude ?​

Physics
1 answer:
jeka57 [31]3 years ago
5 0

Answer:

The dimensional formula of the amplitude of a wave is given by,

[ M⁰L¹T⁰ ]

Where,

M = Mass

L = Length

T = Time

Derivation

The amplitude of wave = Maximum displacement

Since, the dimensional formula of displacement = [ M⁰L¹T⁰ ]

Therefore, the amplitude of a wave is dimensionally represented as [ M⁰L¹T⁰ ]

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Garrick rubs an inflated balloon against his hair. He then touches the balloon against a non-conducting wall.
Sauron [17]

Answer:

Figure A

Explanation:

At first, the inflated balloon is rubbed against the hair.

In this situation, the balloon is charged by friction: because of the friction between the surface of the balllon and the hair, electrons are transferred from the hair to the surface of the balloon.

As a result, when the balloon is detached from the hair, it will have an excess of negative charge (due to the acquired electrons).

Then, the balloon is placed in contact with the non-conducting wall.

The non-conducting wall is initially neutral (equal number of positive and negative charges).

Because the wall is made of a non-conducting material (=isolant), the charges cannot move easily through it. Therefore, even though the charges on the wall feel a force due to the presence of the electrons in the balloon, they will not redistribute along the wall.

Therefore, the charges on the wall will remain equally distributed, as shown in figure A.

7 0
4 years ago
Two golf balls are hit from the same point on a flat field. Both are hit at an angle of 55 degree above the horizontal. Ball 2 h
juin [17]

Answer:

d_2 = 4d_1

Explanation:

The range or horizontal distance covered by a projectile projected with a velocity U at an angel of θ to the horizontal is given by

R = U²sin2θ/g

Let the range or horizontal distance of ball 1 with initial velocity U projected at an angle θ = 55° be

d_1 = U²sin2θ/g

Let the range or horizontal distance of ball 2 with initial velocity V = 2U projected at an angle θ = 55° be

d_2 = V²sin2θ/g

= (2U)²sin2θ/g

= 4U²sin2θ/g

= 4d_1   (since d_1 = U²sin2θ/g)

So, the ball 2 lands a distance d_2 = 4d_1 from the initial point.

4 0
3 years ago
Suppose a ray of light traveling in a material with an index of refraction n a reaches an interface with a material having an in
KATRIN_1 [288]

Answer: C and D

Explanation: One of the first rule for total internal reflection to occur is that the ray must move from a dense to a less dense medium, hence refractive index of medium a must be greater than that of b.

When a ray moves from a dense to a less dense medium, the refracted ray moves away from the normal thus increasing the size of the angle of refraction (total internal refraction occurs when the angle of refraction is 90° and the angle of incidence at this point is known as the critical angle), hence the angle of incidence must be greater than the critical angle.

These points verifies option C and D

5 0
3 years ago
The rocket now has a thruster that malfunctions and is now pushing the rocket in the wrong direction. What is the new net force
Lana71 [14]

Answer:

Daddy Dr Dr Dr Dr

Explanation:

3 0
3 years ago
Read 2 more answers
1. A 1.5 kg ball moves in a circle that is 0.5 m in radius at a speed of 5.1 m/s.
kolezko [41]

Answer: a= 52.02 m/s²

Fc= 78.03 N

Explanation: Solution attached:

3 0
3 years ago
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