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3 years ago

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Which of the following is true about green light? a) It has a greater frequency than red light and blue light. b) It has a small

er frequency than red light and blue light. c) It has a greater frequency than red light but a smaller frequency than blue light. d) It has a smaller frequency than red light but a larger frequency than blue light.

1 answer:

lesantik [10]3 years ago

4 0

Answer:

c) It has a greater frequency than red light but a smaller frequency than blue light.

Explanation:

According to the relation:

c = frequency × Wavelength

The higher the frequency, the lower the value of wavelength

The order of wavelength is:

Violet < Indigo < Blue < Green < Yellow < Orange < Red

Stated above, frequency is inversely proportional to the wavelength. Thus, the order of wavelength is:

Violet > Indigo > Blue > Green > Yellow > Orange > Red

Thus,

<u>Green light has lower frequency than blue light and higher than red light.</u>

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3. Christina is on her college softball team, and she is practicing swinging the bat. Her coach wants her to work on the speed a

inessss [21]

Answer:

The average angular acceleration is $\alpha =125.487 rad /s^2$

Explanation:

From the question we are told that

From the question we are told that

The length of the bat is $l = 0.85m$ \

The initial linear velocity is $u = 0 m/s$

The time is $t = 0.15s$

The velocity at t is $v = 16 m/s$

Generally average angular acceleration is mathematically represented as

$\alpha = \frac{w_f - w_o}{t}$

Where $w_f$ is the finial angular velocity which is mathematically evaluated as

$w_f = \frac{v}{l}$

$w_f = \frac{16}{0.85}$

$= 18.823 rad/s$

and $w_o$ is the initial angular velocity which is zero since initial linear velocity is zero

So

$\alpha = \frac{18.823 - 0}{0.15}$

$\alpha =125.487 rad /s^2$

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4 years ago

What are some ways to conserve electricity

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3 years ago

Read 2 more answers

A spring with force constant of 59 N/m is compressed by 1.3 cm in a hockey game machine. The compressed spring is used to accele

Furkat [3]

Answer:

The puck moves a vertical height of 2.6 cm before stopping

Explanation:

As the puck is accelerated by the spring, the kinetic energy of the puck equals the elastic potential energy of the spring.

So, 1/2mv² = 1/2kx² where m = mass of puck = 39.2 g = 0.0392 g, v = velocity of puck, k = spring constant = 59 N/m and x = compression of spring = 1.3 cm = 0.013 cm.

Now, since the puck has an initial velocity, v before it slides up the inclined surface, its loss in kinetic energy equals its gain in potential energy before it stops. So

1/2mv² = mgh where h = vertical height puck moves and g = acceleration due to gravity = 9.8 m/s².

Substituting the kinetic energy of the puck for the potential energy of the spring, we have

1/2kx² = mgh

h = kx²/2mg

= 59 N/m × (0.013 m)²/(0.0392 kg × 9.8 m/s²)

= 0.009971 Nm/0.38416 N

= 0.0259 m

= 2.59 cm

≅ 2.6 cm

So the puck moves a vertical height of 2.6 cm before stopping

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3 years ago

A force of 150N at an angle of 60 degree to the horizontal to pull a box through a distance of 50m calculate the work done

Alexxandr [17]

Force=150N
Angle=60°
Displacement=50m

$\boxed{\sf W=Fscos\Theta}$

$\\ \sf\longmapsto W=150(50)cos 60$

$\\ \sf\longmapsto W=7500\times \dfrac{1}{2}$

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3 years ago

A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7

aleksandrvk [35]

(a) 1440.5 Hz

The general formula for the Doppler effect is

$f'=(\frac{v+v_r}{v+v_s})f$

where

f is the original frequency

f is the apparent frequency

$v$ is the velocity of the wave

$v_r$ is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

$v_s$ is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

$v_s = v_A = -28.7 m/s$ (surface A is the source, which is moving towards the receiver)

$v_r = +62.2 m/s$ (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

$f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz$

(b) 0.232 m

The wavelength of a wave is given by

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

where

v is the speed of the wave

f is the frequency

In the reflector frame,

f = 1440.5 Hz

So the wavelength is

$\lambda=\frac{334 m/s}{1440.5 Hz}=0.232 m$

(c) 1481.2 Hz

Again, we can use the same formula

$f'=(\frac{v+v_r}{v+v_s})f$

In the source frame (= on surface A), we have

$v_s = v_B = -62.2 m/s$ (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

$v_r = +28.7 m/s$ (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

$f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz$

(d) 0.225 m

The wavelength of the wave is given by

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

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

$\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m$

8 0

3 years ago