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3 years ago

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A wave has a period of 2 seconds and a wavelength of 4 meters. Calculate its frequency and speed. Note: Recall that the frequenc

y of a wave equals 1/period & the period of a wave equals 1/frequency.

1 answer:

Angelina_Jolie [31]3 years ago

4 0

Answer:

frequency = 0.5 Hz and speed = 2 m/s

Explanation:

Given that,

The period of a wave, T = 2 s

Wavelength, $\lambda=4\ m$

If f be the frequency. So,

f = 1/T

$f=\dfrac{1}{2}\\\\f=0.5\ Hz$

Speed of a wave is given by :

$v=f\lambda\\\\v=0.5\times 4\\\\v=2\ m/s$

So, the frequency of the wave is 0.5 Hz and speed is 2 m/s.

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Answer:

the moment of force is 25 Nm.

Explanation:

Given;

weight of the book, F = 50 N

distance through which his hand was stretched, d = 50 cm = 0.5 m

The moment of force is calculated as;

moment of force = F x r

moment of force = 50 N x 0.5 m

moment of force = 25 Nm

Therefore, the moment of force is 25 Nm.

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Describe using examples how objects can be at rest and in motion simultaneously

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An object can be at rest and still be in motion because the earth is always in motion.

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A closed box is filled with dry ice at a temperature of -87.1°C, while the outside temperature is 17.6°C. The box is cubical, me

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Answer:

K=24.17 x 10⁻² J s⁻¹c⁻¹m⁻¹

Explanation:

Rate of flow of heat through a material is given by the following expression

$\frac{Q}{t} =\frac{KA\delta T}{d}$

where Q is amount of heat flowing in time t through area A and a medium of thickness d having two faces at temperature difference δT . K is thermal conductivity of the medium .

Here Q = 3.34 x 10⁶/6 , t = 24 x 60 x 60 = 86400 s , A = .332 X .332 = .0110224 m² , δT = 104.7

Put these values here

$\frac{3.34\times10^6}{6\times86400}= \frac{k\times.011224\times104.7}{4.41\times10^{-2}}$

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3 years ago

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Given vectors D (3.00 m, 315 degrees wrt x-axis) and E (4.50 m, 53.0 degrees wrt x-axis), find the resultant R= D + E. (a) Write

Eva8 [605]

Answer:

R = ( 4.831 m , 1.469 m )

Magnitude of R = 5.049 m

Direction of R relative to the x axis= 16°54'33'

Explanation:

Knowing the magnitude and directions relative to the x axis, we can find the Cartesian representation of the vectors using the formula

$\vec{A}= | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )$

where $| \vec{A} |$ its the magnitude and θ.

So, for our vectors, we will have:

$\vec{D}= 3.00 m \ ( \ cos(315) \ , \ sin (315) \ )$

$\vec{D}= ( 2.121 m , -2.121 m )$

and

$\vec{E}= 4.50 m \ ( \ cos(53.0) \ , \ sin (53.0) \ )$

$\vec{E}= ( 2.71 m , 3.59 m )$

Now, we can take the sum of the vectors

$\vec{R} = \vec{D} + \vec{E}$

$\vec{R} = ( 2.121 \ m , -2.121 \ m ) + ( 2.71 \ m , 3.59 \ m )$

$\vec{R} = ( 2.121 \ m + 2.71 \ m , -2.121 \ m + 3.59 \ m )$

$\vec{R} = ( 4.831 \ m , 1.469 \ m )$

This is R in Cartesian representation, now, to find the magnitude we can use the Pythagorean theorem

$|\vec{R}| = \sqrt{R_x^2 + R_y^2}$

$|\vec{R}| = \sqrt{(4.831 m)^2 + (1.469 m)^2}$

$|\vec{R}| = \sqrt{23.338 m^2 + 2.158 m^2}$

$|\vec{R}| = \sqrt{25.496 m^2}$

$|\vec{R}| = 5.049 m$

To find the direction, we can use

$\theta = arctan(\frac{R_y}{R_x})$

$\theta = arctan(\frac{1.469 \ m}{4.831 \ m})$

$\theta = arctan(0.304)$

$\theta = 16\°54'33''$

As we are in the first quadrant, this is relative to the x axis.

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3 years ago