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2 years ago

If a seismic wave has a period of 0.0202s, find the frequency of the wave.

Physics

1 answer:

maw [93]2 years ago

7 0

Answer:

49.5 Hz.

Explanation:

From the question given above, the following data were obtained:

Period (T) = 0.0202 s

Frequency (f) =?

The frequency and period of a wave are related according to the following equation:

Frequency (f) = 1 / period (T)

f = 1/T

With the above formula, we can obtain the frequency of the wave as follow:

Period (T) = 0.0202 s

Frequency (f) =?

f = 1/T

f = 1/0.0202

f = 49.5 Hz

Therefore the frequency of the wave is 49.5 Hz.

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If you add 50.0 mL of water to a beaker containing 22 mL of water, then how much water do you have in the beaker?

saw5 [17]

Answer:

72 mL

Explanation:

if you add 50.0 mL of water to a beaker containing 22 mL of water, then the total in the beaker is 50.0 + 22 = 72 mL

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2 years ago

A vehicle of mass 100kg has a kinetic energy of 5000 J at an instant. The velocity at that instant is

snow_lady [41]

Answer:

$\bf\pink{10\:m/s}$

Explanation:

<h2><u>Given</u> :-</h2>

$\sf\red{Mass \ of \ vehicle \ (m) = 100 \ kg}$
$\sf\orange{Kinetic \ energy \ (K.E.) = 5000 \ J}$

$\sf\green{Velocity\: of\: the \:vehicle \:at \:the \:instant}$

<h2><u>Formula to be used</u> :-</h2>

$\sf\blue{K.E. = \dfrac{1}{2} mv^{2}}$

Where,

K.E. = Kinetic energy possessed by the body
M = Mass of the body
V = Velocity of the body

<h2><u>Solution</u> :-</h2>

$\to\:\:\sf\red{K.E. = \dfrac{1}{2}mv^{2}}$

$\to\:\:\sf\orange{5000 = \dfrac{1}{2} \times 100 \times v^{2}}$

$\to\:\:\sf\green{5000 = 50 \times v^{2}}$

$\to\:\:\sf\blue{\dfrac{5000}{50} = v^{2}}$

$\to\:\:\sf\purple{100 = v^{2}}$

$\to\:\:\sf\red{\sqrt{100} = v}$

$\to\:\:\sf\orange{ 10 = v}$

$\to\:\:\bf\pink{v = 10\:m/s}$

Velocity of the vehicle at the instant is $\bf\green{10\:m/s}$

7 0

2 years ago

Read 2 more answers

A hollow conducting spherical shell has radii of 0.80 m and 1.20 m, The radial component of the electric field at a point that i

mars1129 [50]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The electric field at that point is $E = 7500 \ N/C$

Explanation:

From the question we are told that

The radius of the inner circle is $r_i = 0.80 \ m$

The radius of the outer circle is $r_o = 1.20 \ m$

The charge on the spherical shell $q_n = -500nC = -500*10^{-9} \ C$

The magnitude of the point charge at the center is $q_c = + 300 nC = + 300 * 10^{-9} \ C$

The position we are considering is x = 0.60 m from the center

Generally the electric field at the distance x = 0.60 m from the center is mathematically represented as

$E = \frac{k * q_c }{x^2}$

substituting values

$E = \frac{k * q_c }{x^2}$

where k is the coulomb constant with value $k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.$

substituting values

$E = \frac{9*10^9 * 300 *10^{-9}}{0.6^2}$

$E = 7500 \ N/C$

7 0

2 years ago

A 1900kg car starts from rest and drives around a flat 65-m-diameter circular track. The forward force provided by the car's dri

s344n2d4d5 [400]

Answer:

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Explanation:

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2 years ago

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Makovka662 [10]

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2 years ago