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d1i1m1o1n [39]
2 years ago
10

In a transverse wave the particles Name

Physics
1 answer:
7nadin3 [17]2 years ago
3 0

Answer:

The direction a wave propagates is perpendicular to the direction it oscillates for transverse waves. A wave does not move mass in the direction of propagation; it transfers energy.

Explanation:

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6

Explanation:

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Read 2 more answers
What is the new acceleration (in m/s/s)?
Delicious77 [7]

The new acceleration is 108 m/s^2

Explanation:

We can answer this problem by applying Newton's second law, which states that:

F=ma

where

F is the net force on an object

m is the mass of the object

a is its acceleration

The equation can be rewritten as

a=\frac{F}{m}

In this problem, the initial acceleration is

a=18.0 m/s^2

Later:

- The net force is tripled: F'=3F

- The mass is halved: m'=\frac{m}{2}

Therefore, the new acceleration is:

a'=\frac{F'}{m'}=\frac{3F}{m/2}=6\frac{F}{m}=6a

which means that the new acceleration is 6 times the original acceleration, therefore

a'=6(18)=108 m/s^2

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8 0
2 years ago
A fixed 11.2-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.10 s , the field is changed to
Karolina [17]

Answer:

The average induced emf in the coil is 0.0286 V

Explanation:

Given;

diameter of the wire, d = 11.2 cm = 0.112 m

initial magnetic field, B₁ = 0.53 T

final magnetic field, B₂ = 0.24 T

time of change in magnetic field, t = 0.1 s

The induced emf in the coil is calculated as;

E = A(dB)/dt

where;

A is area of the coil = πr²

r is the radius of the wire coil = 0.112m / 2 = 0.056 m

A = π(0.056)²

A = 0.00985 m²

E = -0.00985(B₂-B₁)/t

E = 0.00985(B₁-B₂)/t

E = 0.00985(0.53 - 0.24)/0.1

E = 0.00985 (0.29)/ 0.1

E = 0.0286 V

Therefore, the average induced emf in the coil is 0.0286 V

3 0
2 years ago
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