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2 years ago

The light waves have that particular interference from question #1 because they are emitted _____.

Physics

2 answers:

Elis [28]2 years ago

7 0

I am attaching the rest of your question so it makes sense,
<span>
Since lasers are made from stacking light waves that add together into a larger wave due to CONSTRUCTIVE INTERFERENCE.
</span>
Then, <span>light waves have that constructive interference (from question #1) because they are emitted IN PHASE with each other.

This means that they arrive at the same point of space with the same characteristics and their effects do not cancel each other, but the opposite, their intensity increases.</span>

Verdich [7]2 years ago

4 0

Answer;

- In phase

Explanation;

Constructive interference occurs when the maxima of two waves add together (the two waves are in phase), so that the amplitude of the resulting wave is equal to the sum of the individual amplitudes.

Constructive interference occurs when waves come together so that they are in phase with each other. This means that their oscillations at a given point are in the same direction, the resulting amplitude at that point being much larger than the amplitude of an individual wave.

Therefore; For two waves of equal amplitude interfering constructively, the resulting amplitude is twice as large as the amplitude of an individual wave.

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In a fast-pitch softball game the pitcher is impressive to watch, as she delivers a pitch by rapidly whirling her arm around so

Pepsi [2]

Answer:

(a) 181.05 m/s²

(b) 13.2°

Explanation:

Given:

Radius of the circle (R) = 0.610 m

Angular acceleration (α) = 67.6 rad/s²

Angular speed (ω) = 17.0 rad/s

(a)

Radial acceleration of the ball is given as:

$a_r=\omega^2R$

Plug in the given values and solve for $a_r$ . This gives,

$a_r=(17.0\ rad/s)^2\times (0.610\ m)\\\\a_r=289\times 0.610\ m/s^2\\\\a_r=176.29\ m/s^2$

Now, tangential acceleration is given by the formula:

$a_t=R\alpha$

Plug in the given values and solve for $a_t$ . This gives,

$a_t=(0.610\ m)(67.6\ rad/s^2)\\\\a_t=41.236\ m/s^2$

Now, the magnitude of total acceleration is given as the square root of the sum of the squares of tangential and centripetal accelerations. Therefore,

$a_{Total}=\sqrt{(a_r)^2+(a_t)^2}$

Plug in the given values and solve for total acceleration, $a_{Total}$ . This gives,

$a_{Total}=\sqrt{(176.29)^2+(41.236)^2}\\\\a_{Total}=181.05\ m/s^2$

Therefore, the magnitude of total acceleration is 181.05 m/s².

(b)

Angle of total acceleration relative to radial direction is given by the formula:

$\theta=\tan^{-1}(\frac{a_t}{a_r})\\\\\theta=\tan^{-1}(\frac{41.236}{176.29})\\\\\theta=13.2\°$

Therefore, the total acceleration makes an angle of 13.2° relative to radial direction.

4 0

2 years ago

A frictionless pendulum clock on the surface of the earth has a period of 1.00 s. On a distant planet, the length of the pendulu

Sergio [31]

Answer:

The gravitational acceleration on the planet is slightly less than g.

Explanation:

The period of a pendulum is given by:

$T=2\pi \sqrt{\frac{L}{g}}$

where

L is the length of the pendulum

g is the acceleration due to gravity

The formula can also be rewritten as

$g=(\frac{2\pi}{T})^2 L$ (1)

In this problem, we have a pendulum which has a period of T=1.00 s on Earth. The length of the same pendulum must be shortened on the distant planet to have the same period of T'=1.00 s: this means that the length of the pendulum on the distant planet, L', is shorter than the length of the pendulum on Earth, L

$L'$

By looking at formula (1), we see that g (the gravitational acceleration) is directly proportional to L. therefore, if L is shortened on the distant planet, it means that also the value of g is lower than on Earth:

so, the correct answer is

The gravitational acceleration on the planet is slightly less than g.

7 0

2 years ago

A girl swings on a playground swing in such a way that at her highest point she is 4.1 m from the ground, while at her lowest po

Umnica [9.8K]

Answer:

V1 =8.1 m/s

Explanation:

height at highest point (h2) = 4.1 m

height at lowest point (h1) = 0.8 m

acceleration due to gravity (g) = 9.8 m/s^{2}

from conservation of energy, the total energy at the lowest point will be the same as the total energy at the highest point. therefore

mgh1 + $0.5mV1^{2}$ = mgh2 + $0.5mV2^{2}$

where

speed at highest point = V2

speed at lowest point = V1

mass of the girl and swing = m

at the highest point, the speed is minimum (V1 = 0)

at the lowest point the speed is maximum (V2 is the maximum speed)

therefore the equation becomes mgh1 + $0.5mV1^{2}$ = mgh2

m(gh1 + $0.5V1^{2}$ ) = m(gh2)

gh1 + $0.5V1^{2}$ = gh2

V1 = $\sqrt{\frac{gh2 - gh1}{0.5}}$

now we can substitute all required values into the equation above.

V1 = $\sqrt{\frac{(9.8x4.1) - (9.8x0.8)}{0.5}}$

V1 = $\sqrt{\frac{32.34}{0.5}}$

V1 =8.1 m/s

8 0

2 years ago

A sphere has surface area 1.25 m2, emissivity 1.0, and temperature 100.0°C. What is the rate at which it radiates heat into empt

Yuri [45]

The total power emitted by an object via radiation is:
$P=A\epsilon \sigma T^4$
where:
A is the surface of the object (in our problem, $A=1.25 m^2$
$\epsilon$ is the emissivity of the object (in our problem, $\epsilon=1$ )
$\sigma = 5.67 \cdot 10^{-8} W/(m^2 K^4)$ is the Stefan-Boltzmann constant
T is the absolute temperature of the object, which in our case is $T=100^{\circ} C=373 K$

Substituting these values, we find the power emitted by radiation:
$P=(1.25 m^2)(1.0)(5.67 \cdot 10^{-8}W/(m^2K^4)})(373 K)^4=1371 W = 1.4 kW$
So, the correct answer is D.

6 0

2 years ago

A rod 14.0 cm long is uniformly charged and has a total charge of -22.0 μC. Determine the magnitude and direction of the net ele

lesya692 [45]

Explanation:

It is given that,

Length of the rod, l = 14 cm = 0.14 m

Charge on the rod, $q=-22\ \mu C=-22\times 10^{-6}\ C$

We need to find the magnitude and direction of the net electric field produced by the charged rod at a point 36.0 cm to the right of its center along the axis of the rod, z = 36 cm = 0.36 m

Electric field at the axis of the rod is given by :

$E=\dfrac{\lambda}{2\pi \epsilon_o z}$