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3 years ago

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Suppose a radar gun sends out radio waves with a frequency of 2,000,000.0 Hz. The waves bounce off a moving car and return with

a frequency of 2,000,000.2 Hz. If the speed of light is 300,000,000 m/s, what is the velocity of the car

1 answer:

Igoryamba3 years ago

6 0

Answer:

300,000,030m/s

Explanation:

Using the relationship between frequency and speed of a wave;

F ∝ V

F = kV

k = F/V

F1/V1 = F2/V2 = k

Let F1 be the frequency of the radio wave = 2,000,000.0Hz

V1 be the speed of light = 300,000,000m/s

F2 = frequency produced by the car = 2,000,000.2Hz

V2 be the velocity of the moving car

Substituting this values in the equation above;

2,000,000/300,000,000 = 2,000,000.2/V2

Cross multiplying

2,000,000V2 = 300,000,000×2,000,000.2

2V2 = 300×2,000,000.2

V2 = 150×2,000,000.2

V2 = 300,000,030m/s

The velocity of the car is 300,000,030m/s

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A 6,000N is applied to a formula one car that weighs 500kg. What is the car's acceleration?

Vesna [10]

Answer:

<h2>12 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

f is the force

m is the mass

From the question

f = 6000 N

m = 500 kg

We have

$a = \frac{6000}{500} = \frac{60}{5} = 12 \\$

We have the final answer as

<h3>12 m/s²</h3>

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2 years ago

An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is

Andreas93 [3]

Answer:

$v=1.08\times 10^7\ m/s$

Explanation:

Initial speed of the electron, u = 0

The charge per unit area of each plate, $\dfrac{Q}{A}=1.69\times 10^{-7}\ C/m^2$

Separation between the plates, $d=1.75\times 10^{-2}\ m$

An electron is released from rest, u = 0

Using equation of kinematics,

$v^2-u^2=2ad$ ..........(1)

Acceleration of the electron in electric field, $a=\dfrac{qE}{m}$ ............(2)

Electric field, $E=\dfrac{\sigma}{\epsilon_o}$ ............(3)

From equation (1), (2) and (3) :

$v=\sqrt{\dfrac{2q\sigma d}{m\epsilon_o}}$

$v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 1.69\times 10^{-7}\times 1.75\times 10^{-2}}{9.1\times 10^{-31}\times 8.85\times 10^{-12}}}$

v = 10840393.1799 m/s

or

$v=1.08\times 10^7\ m/s$

So, the electron is moving with a speed of $1.08\times 10^7\ m/s$ before it reaches the positive plate. Hence, this is the required solution.

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3 years ago

a wooden block has a mass of 1.2 kg, a specific heat of 710, and is at a temperature of 25*C. what is the block's final temperat

mash [69]

The final temperature of the block is $27.5^{\circ} C$

Explanation:

The amount of thermal energy Q supplied to a substance is related to the increase in temperature of the substance, $\Delta T$ , according to the equation

$Q=mC_s \Delta T$

where:

m is the mass of the substance

$C_s$ is the specific heat capacity of the substance

In this problem, we have:

m = 1.2 kg is the mass of the block

$Q = 2,130 J$ is the amount of energy supplied to the block

$C_s = 710 J/kg^{\circ}C$ is the specific heat capacity of the block

Solving for $\Delta T$ , we find the increase in temperature:

$\Delta T = \frac{Q}{m C_s}=\frac{2130}{(1.2)(710)}=2.5^{\circ}C$

And since the initial temperature was

$T_i = 25^{\circ}C$

The final temperature will be

$T_f = T_i + \Delta T = 25+2.5=27.5^{\circ} C$

Learn more about specific heat capacity:

brainly.com/question/3032746

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