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Karolina [17]
4 years ago
13

"If we increase the amplitude of transverse waves on a string, what happens to the speed of a wave along the string?"

Physics
1 answer:
Gnesinka [82]4 years ago
7 0

Answer:

It is increased

Explanation:

If the amplitude of a transverse wave in a string is increased, then also, the speed of that wave along the string also increases.

In transverse waves, the particles of the medium does move vertically, m either up or down and also moves at right angles to the direction of the wave. It should be noted also, that the wave amplitude of a transverse wave is defined as the difference in height between the crest(high) and the resting position(low).

The highest point particles of the medium reach is often tagged as the crest, while the lowest point is tagged as the resting position. The higher the height of the crests are, the greater the amplitude of the wave will be.

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A soccer goal is 2.44 m high. A player kicks the ball at a distance 13 m from the goal at an angle of 40°, and the ball just hit
Vera_Pavlovna [14]

Let <em>v</em> denote the initial speed of the ball. The ball's position at time <em>t</em> is given by the vector

\mathbf r(t)=v\cos40^\circ\,t\,\mathbf i+\left(v\sin40^\circ\,t-\dfrac g2t^2\right)\,\mathbf j

where <em>g</em> is the acceleration due to gravity with magnitude 9.80 m/s^2.

The ball reaches the goal 13 m away at time <em>t</em> such that

10\,\mathrm m=v\cos40^\circ t\implies t=\dfrac{10\,\mathrm m}{v\cos40^\circ}

at which point it attains a height of 2.44 m, so that

2.44\,\mathrm m=v\sin40^\circ\left(\dfrac{10\,\mathrm m}{v\cos40^\circ}\right)-\dfrac g2\left(\dfrac{10\,\mathrm m}{v\cos40^\circ}\right)^2

2.44\,\mathrm m=(10\,\mathrm m)\tan40^\circ-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)\left(\dfrac{100\,\mathrm m^2}{v^2\cos^240^\circ}\right)

\implies\boxed{v\approx3.75\dfrac{\rm m}{\rm s}}

6 0
4 years ago
Establish the dimension of density​
mario62 [17]
Dimensional Formula of Density
The dimensional formula of density is given by,

[M1 L-3 T0]
Where,

M = Mass
L = Length
T = Time
Derivation

Density (ρ) = Mass × Volume-1 . . . . . (1)

The dimensional formula of volume = [M0 L3 T0] . . . . (2)

And, the dimensional formula of mass = [M1 L0 T0] . . . . (3)

On substituting equation (2) and (3) in equation (1) we get,

Density = Mass × Volume-1

Or, (ρ) = [M1 L0 T0] × [M0 L3 T0]-1 = [M1 L-3 T0]

Therefore, density is dimensionally represented as [M1 L-3 T0].

Hope i helped!
5 0
4 years ago
Read 2 more answers
A5 kg frisbee is thrown from rest to a final speed of 12 m/s. What is the impulse of the frisbee?
lesya692 [45]

Answer:

60kgm/s

Explanation:

Given parameters:

Mass of frisbee  = 5kg

Final speed  = 12m/s

Unknown:

Impulse of the frisbee  = ?

Solution:

The impulse of the frisbee is the same as the change in momentum.

It is given as:

 Impulse  = mass (Final velocity  -  Initial velocity)

 Impulse  = 5(12  - 0)  = 60kgm/s

3 0
3 years ago
When a wire with a current is placed in a magnetic field, electrical energy is transformed into mechanical energy
babymother [125]

Answer:

TRUE

Explanation:

When a wire with a current is placed in a magnetic field, electrical energy is transformed into mechanical energy. The transformation happens when the magnetic field produced by the current causes the wire to move.

8 0
3 years ago
To make ice, a freezer that is a reverse Carnot engine extracts 37 kJ as heat at -17°C during each cycle, with coefficient of pe
Dvinal [7]

Answer:

a) 6.4 kJ

b) 43.4 kJ

Explanation:

a)

Q_{a} = Heat absorbed = 37 kJ

β  = Coefficient of performance = 5.8

W = Work done

Heat absorbed is given as

Q_{a} = β W

37 = (5.8) W

W = 6.4 kJ

b)

Q  = work per cycle required

Q  = Q_{a} + W

Q  = 37 + 6.4

Q  = 43.4 kJ

5 0
4 years ago
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