Which law of motion explains what happens during a ride on the bumper cars ?

A wave travels at a constant speed. How does the frequency change if the wavelength increases by a factor of 2?

NARA [144]

Answer : The frequency decreases by a factor of 2.

Explanation :

Given that the wave travels at a constant speed. The speed of the wave is given as :

$v=\nu\times \lambda$

Where

υ is the frequency of the wave

and λ is the wavelength of the wave.

In this case, the speed is constant. So, the relation between the frequency and the wavelength is inverse.

$\nu\propto \dfrac{1}{\lambda}$

If the wavelength increases by a factor of 2, its frequency will decrease by a factor of 2.

Hence, the correct option is (A) " The frequency decreases by a factor of 2 ".

3 0

3 years ago

Read 2 more answers

An fm radio station broadcasts at 98. 6 mhz. What is the wavelength of the radiowaves?.

victus00 [196]

The wavelength of the radio waves is 3.04 cm.

<h3>Calculation:</h3>

λf = c

λ = c/f

where,

λ = wavelength

c = speed of light

f = frequency

Given,

f = 98.6 MHz = 98.6 × 10⁶

c = 3 × 10⁸

To find,

λ =?

Put the values in the formula,

λ = c/f

λ = 3 × 10⁸/98.6 × 10⁶

= 0.0304 × 10² m

= 3.04 cm

Therefore, the wavelength of the radio waves is 3.04 cm.

Learn more about the calculation of wavelength here:

brainly.com/question/8422432

#SPJ4

3 0

2 years ago

A series circuit has a capacitor of 0.25 × 10−6 F, a resistor of 5 × 103 Ω, and an inductor of 1 H. The initial charge on the ca

svetoff [14.1K]

Answer:

$q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

Explanation:

Given that $L=1H, R=5000\Omega, \ C=0.25\times10^{-6}F, \ \ E(t)=12V$ , we use Kirchhoff's 2nd Law to determine the sum of voltage drop as:

$E(t)=\sum{Voltage \ Drop}\\\\L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\\\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+\frac{1}{0.25\times10^{-6}}q=12\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+4000000q=12\\\\m^2+5000m+4000000=0\\\\(m+4000)(m+1000)=0\\\\m=-4000 \ or \ m=-1000\\\\q_c=c_1e^{-4000t}+c_2e^{-1000t}$

#To find the particular solution:

$Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

Hence the charge at any time, t is $q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

6 0

3 years ago

A force acts on a body of mass 13 kg initially at rest. The force

PtichkaEL [24]

Answer:

Force that acted on the body was F = 13 N

Explanation:

If once accelerated, the body covers 60 meters in 6 seconds, then its velocity is 60/6 m/s = 10 m/s

When the force was acting (for 10 seconds) the object accelerated from rest (initial velocity vi = 0) to 10 m/s (its final velocity). therefore we can use the kinematic equation for the velocity in an accelerated motion given by:

$v_f=v_i+a*t$

which in our case becomes;

$10\,m/s=0+a*(10\,s)$

and we can solve for the acceleration as:

a = 10/10 m/s^2 = 1 m/s^2

Therefore the force acting on the body, based on Newton's 2nd Law expression: F = m * a is:

F = 13 kg * 1 m/s^2 = 13 N

4 0

3 years ago

Which states of matter have particles that move independently of one another with very little attraction

Crazy boy [7]

The state of matter that the particles move independently of one another with very little attraction is, I believe, gas

6 0

3 years ago