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3 years ago

If the speed of a wave is 1500m/sec and its frequency is 200 Hz, what is its wavelength

Physics

1 answer:

Ray Of Light [21]3 years ago

4 0

Answer:

The wavelength of wave is 7.5 meter.

Given:

Speed of wave = 1500 $\frac{m}{s}$

Frequency of wave = 200 Hz

To find:

Wavelength of wave = ?

Formula used:

$\lambda = \frac{v}{n}$

Where $\lambda$ = wavelength of the wave

v = speed of wave

n = frequency of wave

Solution:

Wavelength of wave is given by,

$\lambda = \frac{v}{n}$

Where $\lambda$ = wavelength of the wave

v = speed of wave

n = frequency of wave

$\lambda = \frac{1500}{200}$

$\lambda$ = 7.5 m

The wavelength of wave is 7.5 meter.

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(33%) Problem 3: Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1100 kg and i

expeople1 [14]

Answer:

The final velocity of the cars is 8.38 m/s at an angle 39.1° south of west.

Explanation:

Given that,

Mass of first car = 1100 kg

Velocity of first car = 8.5 m/s

Mass of second car = 650 kg

Velocity of second car = 17.5 m/s

Suppose we need to find the final velocity of the cars and direction of the cars.

We need to calculate the velocity of the car in west direction

Using conservation of momentum in west direction

$m_{f}v_{f}+m_{s}v_{s}= (m_{f}+m_{s})v_{x}$

$v_{x}=\dfrac{m_{f}v_{f}+m_{s}v_{s}}{(m_{f}+m_{s})}$

Put the value into the formula

$v_{x}=\dfrac{1100\times0+650\times17.5}{1100+650}$

$v_{x}=6.5\ m/s$

We need to calculate the velocity of the car in south direction

Using conservation of momentum in south direction

$m_{f}v_{f}+m_{s}v_{s}= (m_{f}+m_{s})v_{y}$

$v_{y}=\dfrac{m_{f}v_{f}+m_{s}v_{s}}{(m_{f}+m_{s})}$

Put the value into the formula

$v_{y}=\dfrac{1100\times8.5+650\times0}{1100+650}$

$v_{y}=5.3\ m/s$

We need to calculate the final velocity of the cars

Using formula of velocity

$v_{eq}=\sqrt{(6.5)^2+(5.3)^2}$

$v_{eq}=8.38\ m/s$

We need to calculate the direction

Using formula of direction

$\tan\theta=\dfrac{v_{y}}{v_{x}}$

Put the value into the formula

$\tan\theta=\dfrac{5.3}{6.5}$

$\theta=\tan^{-1}(\dfrac{5.3}{6.5})$

$\theta=39.1^{\circ}$

Hence, The final velocity of the cars is 8.38 m/s at an angle 39.1° south of west.

8 0

3 years ago

A 2kg mass is moving at 3m/s. What is its kinetic energy?

Illusion [34]

<h2><u>KINETIC ENERGY</u></h2>

<h3>Problem:</h3>

» A 2kg mass is moving at 3m/s. What is its kinetic energy?

<h3>Answer:</h3>

$\color{hotpink} \bold{9 \: J} \\$

— — — — — — — — — —

<h3>Formula:</h3>

To calculate the velocity of a kinetic energy, we can use formula

$\underline{ \boxed{ \tt KE = \frac{1}{2} m{v}^{2} } }$

where,

v is the velocity in m/s
KE is the kinetic energy in J (joules)
m is the mass in kg

— — —

Based on the problem, the givens are:

KE (Kinetic energy) = ? (unknown)
m (mass) = 2 kg
v (velocity) = 3 m/s

<h3>Solution:</h3>

To get the velocity, substitute the givens in the formula above then solve.

$\: \: \tt KE = \frac{1}{2} m{v}^{2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt \: KE = \frac{1}{2} \times 2 \times {(3)}^{2} \\ \tt \: \: \: \: \: \: \: \: \: \: \: \: \: \: KE = \frac{1}{2} \times(2\times 9) \\ \tt KE = \underline{ \boxed{ \blue{ \tt9 \: J}}}$