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lisov135 [29]
3 years ago
12

What is the frequency of a wave

Physics
2 answers:
MAVERICK [17]3 years ago
6 0
The frequency of a wave is the number of waves that passes through a point in a certain time. The less waves that pass in a period of time the lower the frequency of the wave. The more waves that pass in a period of time the higher the frequency of the wave. When measuring wave length the time period used is usually one second.
Xelga [282]3 years ago
4 0

Explanation:

The frequency of a wave is defined as the number of vibrations or oscillations per unit time. It is also defined as the reciprocal of the time period. Its SI unit is hertz or s⁻¹.

Mathematically, it can be written as :

\nu=\dfrac{1}{T}

T is the time period of the wave.

The relation between the frequency and the wavelength is inverse. Also,

\nu=\dfrac{v}{\lambda}

The wave having higher frequency have higher energy and vice versa.

Hence, this is the required solution.                              

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I need help on 2/3 please asap thx ♥️
ch4aika [34]

<u>questions 2</u>

F=m

therefore

a=F/m

a=408/68

=6m/s^2.

5 0
3 years ago
What is the net force exerted by these two charges on a third charge q3 = 49.5 nC placed between q1 and q2 at x3 = -1.170 m ? Yo
insens350 [35]

Full Question

Consider two point charges located on the x axis: one charge, Q1 = -12.0 nC , is located at x1 = -1.705m ; the second charge, Q2 = 36.5 nC, is at the origin (x=0.0000).

What is the net force exerted by these two charges on a third charge q3 = 49.5 nC placed between q1 and q2 at x3 = -1.170 m ? Your answer may be positive or negative, depending on the direction of the force.

Answer:

6.79E6 N

Explanation:

Given

Q1 is negative and to the left of Q3 the force will be to the left

Q2 is positive and to the right of Q3 the the force will also be to the left

Net Force is calculated as:

Using Coulomb's law

Coulomb's law: F = kqQ / r²

the constant k = 8.99 x 10^9 N m2 / C2

F = -kQ1*Q3/(r1)² -kQ2*Q3/(r2)²)

F = -kQ3(Q1/(r1)² + Q2/(r2)²)

Where

Q1 = -12nC = -12 * 10^-9C

Q2 = 36.5nC = 36.5 * 10^-9C

Q3 = 49.5nC = 49.5 * 10^-9C

x1 = -1.705m

x2 = x = 0

x3 = -1.170m

r1 = x3 - x1

r1 = -1.170 - -1.705

r1 = -1.170 + 1.705

r1 = 0.535

r1² = 0.286225

r2 = x3

r2 = -1.170

r2² = -1.170²

r2² = 1.3689

So,

F = (-8.99 * 10^9)(49.5 *10^-9) [-12 * 10^-9/0.286225 + 36.5 * 10^-9/1.3689]

F = -445.005 (−4.192505895711E−8 + 2.6663744612462E−8)

F = -445.005 * −1.5261314344648E−8

F = -(8.99 * 10^9) * (49.5 * 10^-9) * [ (-12 * 10^-9) /(-1.770 - -1.705)² + (36.5 * 10^-9)/(-1.170)²]

F = -445.005( −0.000002813572941777538)

F = 0.00000679136118994008324

F = 6.79E6 N

3 0
3 years ago
I'm walking 1.6m/s to 7-11 and it started to rain so I sped up to 2.7m/s in 1.2
olga nikolaevna [1]

Answer:

Explanation:

a = \frac{v_f-v_0}{t} which is the final velocity minus the initial velocity in the numerator, and the change in time in the denominator.  For us:

a=\frac{2.7-1.6}{1.2} so

a = .92 m/s/s (NOT negative because you're speeding up)

5 0
3 years ago
A cat runs at 2 m/s for 3 s and then slows to a stop with an acceleration of 0.80 m/s2. What is
sp2606 [1]

Answer:

see that the correct one is B

Explanation:

To solve this exercise let us use the kinematic relations

           v² = v₀² - 2 a x

as they indicate that the car stops, therefore the final speed is yield v = 0

          x = v₀² / 2a

let's calculate

          x = 2²/(2 0.8)

         x = 2.5 m / s²

When reviewing the answers we see that the correct one is B

7 0
2 years ago
find the potential energy of an aircraft weighing 10000 bs at 5000 ft true altitude and 125 kts true air speed
Cloud [144]

Answer:

U=5*10^7ft.Ib

Explanation:

From the question we are told that

Weight  W= 10000bs

Altitude H=5000ft

Speed    V=125kts\\1kts=0.514m/s\\V=125*0.514=>64.25m/s

Generally the equation for Potential energy ids mathematically given as

Potential\ Energy\ U=mgh

U=Wh

U=10000*5000

U=5*10^7ft.Ib

6 0
3 years ago
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