An electric iron of resistance 20Ω takes a current of 5A. Calculate the heat developed in 30seconds?
1 answer:
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Answer
given,
frequency from Police car= 1240 Hz
frequency of sound after return = 1275 Hz
Calculating the speed of the car = ?
Using Doppler's effect formula
Frequency received by the other car
..........(1)
u is the speed of sound = 340 m/s
v is the speed of the car
Frequency of the police car received
now, inserting the value of equation (1)
1.02822(340 - v) = 340 + v
2.02822 v = 340 x 0.028822
2.02822 v = 9.799
v = 4.83 m/s
hence, the speed of the car is equal to v = 4.83 m/s
Corrected and Formatted Question:
A wave on a string is described by y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m) + t/(0.20 s))], where x is in m and t in s.
(a) In what direction is this wave traveling?
(b) What are the wave speed, frequency, and wavelength?
(c) At t = 0.50 , what is the displacement of the string at x = 0.20 m?
Answer:
The wave is travelling in the negative x direction
The wave speed = 12.0m/s
The frequency = 5Hz
The wavelength = 2.4m
The displacement at t = 0.50s and x = 0.20m is -0.029m
Explanation:
The general wave equation is given by;
y(x, t) = y cos (2(x/λ) - 2
ft) --------------------------------(i)
Where;
y(x, t) is the displacement of the wave at position x and a given time t
y = amplitude of the wave
f = frequency of the wave
λ = wavelength of the wave
Given;
y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m) + t/(0.20 s))] ------------------(ii)
Which can be re-written as;
y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m)) + 2π(t/(0.20 s))] -------------(iii)
Comparing equations (i) and (iii) we have that;
=> 2π(x/(2.4 m) = 2π(x/λ)
=> λ = 2.4m
Therefore the wavelength of the wave is 2.4m
Also, still comparing the two equations;
=> 2π(t/(0.20 s) = 2πft
=> f = 1 / 0.20
=> f = 5Hz
Therefore the frequency of the wave is 5Hz
To get the wave speed (v), it is given by;
v = f x λ
Where f = 5Hz and λ = 2.4m
=> v = 5 x 2.4
=> v = 12.0m/s
Therefore, the speed of the wave is 12.0m/s
At t = 0.50s and x = 0.20m;
The displacement, y(x,t) of the string wave is given by
y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m) + t/(0.20 s))]
<em>Convert the amplitude of 3.0cm to m</em>
=> 3.0cm = 0.03m
<em>Substitute this back into the equation</em>
=> y(x, t) = (0.03m) × cos[2π(x/(2.4 m) + t/(0.20 s))]
<em>Substitute the values of t and x into the equation above;</em>
=> y(x, t) = (0.03m) × cos[2π((0.20)/(2.4 m) + 0.50/(0.20 s))]
<em>Carefully solve the equation</em>
=> y(x, t) = (0.03m) × cos[2π((0.20)/(2.4 m)) + 2π(0.50/(0.20 s))]
=> y(x, t) = (0.03m) × cos[0.08π + 5π]
=> y(x, t) = (0.03m) × cos[5.08π]
=> y(x, t) = (0.03m) × cos[15.96]
=> y(x, t) = (0.03m) × cos[15.96]
=> y(x, t) = (0.03m) × -0.9684
=> y(x, t) = 0.029m
Therefore the displacement at those points is -0.029m
Also, the sign of the displacement shows that the direction of the wave is in the negative x direction.
To solve the problem it is necessary to consider the concepts related to Potential Energy and Kinetic Energy.
Potential Energy because of a planet would be given by the equation,
Where,
G = Gravitational Universal Constant
M = Mass of Ocean
M = Mass of Moon
r = Radius
From the data given we can calculate the mass of the ocean water through the relationship of density and volume, then,
It is necessary to define the two radii, when the ocean is far from the moon and when it is facing.
When it is far away, it will be the total diameter from the center of the earth to the center of the moon.
When it's near, it will be the distance from the center of the earth to the center of the moon minus the radius,
PART A) Potential energy when the ocean is at its furthest point to the moon,
PART B) Potential energy when the ocean is at its closest point to the moon
PART C) The maximum speed. This can be calculated through the conservation of energy, where,