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shutvik [7]
3 years ago
9

Which equations could be used as is, or rearranged to calculate for frequency of a wave? Check all that apply.

Physics
2 answers:
amm18123 years ago
7 0
-- Equations  #2  and  #6  are both the same equation,
and are both correct.

-- If you divide each side by  'wavelength', you get Equation #4,
which is also correct.

-- If you divide each side by  'frequency', you get Equation #3,
which is also correct. 
With some work, you can rearrange this one and use it to calculate
frequency.

Summary:

-- Equations #2, #3, #4, and #6 are all correct statements,
and can be used to find frequency.

-- Equations #1 and #5 are incorrect statements.
alexira [117]3 years ago
3 0

The correct answers to the question are 2,3,4 and 6 options respectively.

EXPLANATION:

Let us consider a wave which is moving with a speed v in a medium .

Let f and \lambda is the frequency and wavelength of the wave.

The relation between frequency,wavelength and speed is given as -

                       speed = =\ frequency\times wavelength

                                   = f\times \lambda

The wavelength of the wave is calculated as -

                             wavelength=\ \frac{speed}{frequency}

Similarly frequency of the wave can be calculated as-

                            frequency=\ \frac{speed}{wavelength}

The option 1 is a wrong relation. So, we can not calculate the wavelength through this equation.

The option 2 is a right relation. By rearranging it, we can calculate the frequency of the wave.

The option 3 is a correct relation. So, we can calculate the wavelength through it.

The option 4 is also a correct relation. By rearranging it, we can calculate the wavelength.

The option 5 is a wrong relation.

The option 6 is also a right relation. By rearranging it, we can calculate the wavelength of the wave.

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What must the charge (sign and magnitude) of a 1.45-g particle be for it to remain stationary when placed in a downward-directed
tekilochka [14]

Answer:

charge will be equal to 2.03\times 10^{-5}C  

Explanation:

We have given mass of the particle m = 1.45 gram = 0.00145 kg

Acceleration due to gravity g=9.8m/sec^2

Electric field E = 700 N/C

Electric force will be equal to F=qE, here q is charge and E is electric field

For particle to be stationary this force must be equal to force due to gravity , that is mg force

So qE = mg

q=\frac{mg}{E}=\frac{0.00145\times 9.8}{700}=2.03\times 10^{-5}C

So charge will be equal to 2.03\times 10^{-5}C

8 0
3 years ago
What is the kinetic energy of a 74.0 kg skydiver falling at a terminal velocity of 52.0m/s?
skad [1K]

Answer:

100,048

Explanation:

K.E = 1/2 m (v)^2

K.E = 1^/2 * 74 * (52)^2

K.E = 100,048J =100.048kJ

8 0
3 years ago
A single loop of wire with an area of 0.0780 m2 is in a uniform magnetic field that has an initial value of 3.80 T, is perpendic
Katarina [22]

Answer:

The induced current is 26.7 mA

Explanation:

Given;

area of the loop, A = 0.078 m²

initial magnetic field, B₁ = 3.8 T

change in the magnetic field strength, dB/dt = 0.24 T/s

The induced emf is calculated as;

emf = - \frac{d \phi}{dt} \\\\emf = -\frac{dB.A}{dt} \\\\emf = A (\frac{dB}{dt} )\\\\emf = 0.078(0.24)\\\\emf = 0.0187 \ V

The resistance of the loop = 0.7 Ω

The induced current is calculated as;

V = IR\\\\I = \frac{V}{R} = \frac{emf}{R} = \frac{0.0187}{0.7} = 0.0267 \ A = 26.7 \ mA

4 0
3 years ago
A rectangular loop of area A is placed in a region where the magnetic field is perpendicular to the plane of the loop. The magni
mina [271]

Answer:

Induced emf, \epsilon=-A\dfrac{B_{max}e^{-t/\tau}}{\tau}

Explanation:

The varying magnetic field with time t is given by according to equation as :

B=B_{max}e^{-t/\tau}

Where

B_{max}\ and\ t are constant

Let \epsilon is the emf induced in the loop as a function of time. We know that the rate of change of magnetic flux is equal to the induced emf as:

\epsilon=-\dfrac{d\phi}{dt}

\epsilon=-\dfrac{d(BA)}{dt}

\epsilon=-A\dfrac{d(B)}{dt}

\epsilon=-A\dfrac{d(B_{max}e^{-t/\tau})}{dt}

\epsilon=A\dfrac{B_{max}e^{-t/\tau}}{\tau}

So, the induced emf in the loop as a function of time is A\dfrac{B_{max}e^{-t/\tau}}{\tau}. Hence, this is the required solution.

7 0
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