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2 years ago

Which equation could be rearranged to calculate the frequency of a wave?

Physics

1 answer:

777dan777 [17]2 years ago

7 0

Answer:

wavelength = speed/frequency

Explanation:

Required

Determine which of the options can be used to calculate frequency

The relationship between wavelength, speed and frequency is as follows;

$Frequency = \frac{Wave\ Speed}{Wave\ Length}$ ---- Equation 1

When option (1), (2) and (4) are rearranged, they do not result in the above formula; only option (3) does

Checking option (3)

$Wave\ Length = \frac{Speed}{Frequency}$

Multiply both sides by Frequency

$Wave\ Length * Frequency = \frac{Speed}{Frequency} * Frequency$

$Wave\ Length * Frequency = Speed$

Divide both sides by Wave Length

$\frac{Wave\ Length * Frequency}{Wave\ Length} = \frac{Speed}{Wave\ Length }$

$Frequency = \frac{Speed}{Wave\ Length }$ --- Equation 2

<em>Comparing equation 1 and 2; both equations are the same.</em>

<em>Hence, option (3) answers the question</em>

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A 10. g cube of copper at a temperature T1 is placed in an insulated cup containing 10. g of water at a temperature T2. If T1 &g

Anna35 [415]

Answer:

a. The temperature of the copper changed more than the temperature of the water.

Explanation:

Because we're only considering the isolated system cube-water, the heat of the system should be constant, that implies the heat the cube loses is equal the heat the water gains (because by zero law of thermodynamics heat (Q) flows from hot body to cold body until reach thermal equilibrium and T1>T2). So:

$Q_{cube}=Q_{water}$ (1)

But Q is related with mass (m), specific heat (c) and changes in temperature ( $\varDelta T$ )in the next way:

$Q=cm\varDelta T$ (2)

Using (2) on (1):

$c_{cooper}*m_{cooper}*\varDelta T_{cooper}=c_{water}*m_{waterer}*\varDelta T_{water}$

$(10g)(0.385 \frac{J}{g\,C})(\varDelta T_{cooper})=(10g)(4.186 \frac{J}{g\,C})(\varDelta T_{water})$

$(0.385 \frac{J}{g\,C})(\varDelta T_{cooper})=(4.186 \frac{J}{g\,C})(\varDelta T_{water})$

Because we have an equality and 0.385 < 4.186 then $\varDelta T_{cooper}>\varDelta T_{waterer}$ to conserve the equality

4 0

2 years ago

When you are standing on Earth, orbiting the Sun, and looking at a broken cell phone on the ground, there are gravitational pull

Mandarinka [93]

Answer:

The answer is "Option b, c, and a".

Explanation:

Here that the earth pulls on the phone, as it will accelerate towards Earth when we drop it.

We now understand the effects of gravity:

$F \propto M\\\\F\propto \frac{1}{r^2}\\\\or\\\\F \propto \frac{M}{r^2}\\\\Sun (\frac{M}{r^2}) = \frac{10^{28}}{(10^9)^2} = 10^{10}$

The force of the sun is, therefore, $10^{10}$ times greater and the proper sequence, therefore, option steps are:

b. Pull-on phone from earth

c. Pull-on phone from sun

a. Pull phone from you

5 0

1 year ago

What is the rate of heat transfer by radiation, with an unclothed person standing in a dark room whose ambient temperature is 22

SIZIF [17.4K]

Answer:

$5.45\times 10^{-4}$ W

Explanation:

$T_{r}$ = Temperature of the room = 22.0 °C = 22 + 273 = 295 K

$T_{s}$ = Temperature of the skin = 33.0 °C = 33 + 273 = 306 K

$A$ = Surface area = 1.50 m²

$\epsilon$ = emissivity = 0.97

$\sigma$ = Stefan's constant = 5.67 x 10⁻⁸ Wm⁻² K⁻⁴

Rate of heat transfer is given as

$R = \epsilon \sigma A (T_{s}^{2} - T_{r}^{2})$

$R = (0.97)(5.67\times 10^{-8}) (1.50) ((306)^{2} - (295)^{2})$

$R = 5.45\times 10^{-4}$ W

3 0

2 years ago

A test car and its driver, with a combined mass of 600 kg, are moving along a straight,horizontal track when a malfunction cause

ANEK [815]

Answer:

The two of the following measurements, when taken together, would allow engineers to find the total mechanical energy dissipated during the skid

B. The contact area of each tire with the track.

C. The co-efficent of static friction between the tires and the track.

D. The co-efficent of static friction between the tires and the track.

Explanation:

4 0

1 year ago

Write the equivalent formulas for velocity, acceleration, and force using the relationships covered for UCM, Newton’s Laws, and

yKpoI14uk [10]

Answer:

The newton’s second law is F=ma

The Gravitational force is $F=\dfrac{Gm_{1}m_{2}}{r^2}$

Explanation:

Given that,

The equivalent formulas for velocity, acceleration, and force using the relationships covered for UCM, Newton’s Laws, and Gravitation.

We know that,

Velocity :

The velocity is equal to the rate of position of the object.

$v=\dfrac{dx}{dt}$ ....(I)

Acceleration :

The acceleration is equal to the rate of velocity of the object.

$a=\dfrac{dv}{dt}$ ....(II)

Newton’s second Laws

The force is equal to the change in momentum.

In mathematically,

$F=\dfrac{d(p)}{dt}$

Put the value of p

$F=\dfrac{d(mv)}{dt}$

$F=m\dfrac{dv}{dt}$

Put the value from equation (II)

$F=ma$

This is newton’s second laws.

Gravitational force :

The force is equal to the product of mass of objects and divided by square of distance.

In mathematically,

$F=\dfrac{Gm_{1}m_{2}}{r^2}$

Where, m₁₂ = mass of first object

m= mass of second object

r = distance between both objects

Hence, The newton’s second law is F=ma

The Gravitational force is $F=\dfrac{Gm_{1}m_{2}}{r^2}$

3 0

2 years ago