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4 years ago

I will give Brainliest to whoever can answer!!!!

Physics

1 answer:

Alexus [3.1K]4 years ago

4 0

Answer:

The nodes and anti nodes would reverse roles.

Explanation:

I believe it has to do with the path differences. If waves are in phase, then the path differences are such that the waves reach the screen with crests superimposing crests and troughs superimposing troughs. This happens when the periods of each wave are equal or the paths themselves differ by a whole number multiple of the wavelength (λ, 2λ, 3λ, ...).

Now make these waves out of phase. Then half of the waves will travel half a wavelength farther than the rest. So the path difference will be 0.5λ, 1.5λ, 2.5λ, ....

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Two parallel-plate capacitors have the same plate area, but the plate gap in capacitor 1 is twice as big as capacitor 2. If capa

-BARSIC- [3]

Answer:

Capacitance of the second capacitor = 2C

Explanation:

$\texttt{Capacitance, C}=\frac{\varepsilon_0A}{d}$

Where A is the area, d is the gap between plates and ε₀ is the dielectric constant.

Let C₁ be the capacitance of first capacitor with area A₁ and gap between plates d₁.

We have

$\texttt{Capacitance, C}_1=\frac{\varepsilon_0A_1}{d_1}=C$

Similarly for capacitor 2

$\texttt{Capacitance, C}_2=\frac{\varepsilon_0A_2}{d_2}=\frac{\varepsilon_0A_1}{\frac{d_1}{2}}=2\times \frac{\varepsilon_0A_1}{d_1}=2C$

Capacitance of the second capacitor = 2C

6 0

3 years ago

The intensity level of a "Super-Silent" power lawn mower at a distance of 1.0 m is 100 dB. You wake up one morning to find that

OLEGan [10]

Answer: The intensity level of sound in the bedroom is 80dB

Explanation:

Intensity of lawn mower at r=1m is 100dB

Beta1= 10dBlog(I1/Io)

100dB= 10dB log(I1/Io)

10^10= I1/Io

I1= Io(10^10)

10^12)×(10^10)= I1

I1=10^-2w/m^2

Intensity of lawn mower at r=20m

I2/I1=(r1/r2)^2 =(1/20)^2

I2= I1(1/400)

I2=2.5×10^-3W_m^2

Intensity of 4 lown mowers at 20m fro. Window

= 10dBlog(4I2/Io)

= 10^-4/10^-12

=80dB

6 0

3 years ago

The maximum Compton shift in wavelength occurs when a photon isscattered through 180^\circ .

vlabodo [156]

Answer: $90\°$

Explanation:

The Compton Shift $\Delta \lambda$ in wavelength when the photons are scattered is given by the following equation:

$\Delta \lambda=\lambda_{c}(1-cos\theta)$ (1)

Where:

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}c}$ , being $h$ the Planck constant, $m_{e}$ the mass of the electron and $c$ the speed of light in vacuum.

$\theta)$ the angle between incident phhoton and the scatered photon.

We are told the maximum Compton shift in wavelength occurs when a photon isscattered through $180\°$ :

$\Delta \lambda_{max}=\lambda_{c}(1-cos(180\°))$ (2)

$\Delta \lambda_{max}=\lambda_{c}(1-(-1))$

$\Delta \lambda_{max}=2\lambda_{c}$ (3)

Now, let's find the angle that will produce a fourth of this maximum value found in (3):

$\frac{1}{4}\Delta \lambda_{max}=\frac{1}{4}2\lambda_{c}(1-cos\theta)$ (4)

$\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}(1-cos\theta)$ (5)

If we want $\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}$ , $1-cos\theta$ must be equal to 1:

$1-cos\theta=1$ (6)

Finding $\theta$ :

$1-1=cos\theta$

$0=cos\theta$

$\theta=cos^{-1} (0)$

Finally:

$\theta=90\°$ This is the scattering angle that will produce $\frac{1}{4}\Delta \lambda_{max}$

7 0

3 years ago

The top of the pool table is 0.810 m from the floor. the placement of the tape is such that 0 m is aligned with the edge of the

8090 [49]

Compute first for the vertical motion, the formula is:

y = gt²/2

0.810 m = (9.81 m/s²)(t)²/2

t = 0.4064 s

whereas the horizontal motion is computed by:

x = (vx)t

4.65 m = (vx)(0.4064 s)

4.65 m/ 0.4064s = (vx)

(vx) = 11.44 m / s
So look for the final vertical speed.

(vy) = gt

(vy) = (9.81 m/s²)(0.4064 s)

(vy) = 3.99 m/s

speed with which it hit the ground:

v = sqrt[(vx)² + (vy)²]

v = sqrt[(11.44 m/s)² + (3.99 m/s)²]

v = 12.12 m / s

6 0

3 years ago

Respond to the following based on your reading (write your answers using 1–2 paragraphs). 1. Describe the universal gravitationa

yaroslaw [1]

Answer:

Explanation:

1. Discovered by Sir Isaac Newton, this law states that every object in the universe that has mass attracts every other object in the universe that has mass. The force is proportional to the product of the two masses and inversely proportional to the square of the distance between their centers. When applying this to a situation with two objects, the object with the smaller mass will do most of the moving because the other object has too much inertia to move any noticeable amount.

2. Without advanced technology like we have today, Ptolemy and Copernicus tried to best explain the model of the universe through observation. Ptolemy’s model came first and placed a stationary earth at the center of the model. Everything else moved in respect to earth. This was widely accepted since it seemed like earth wasn’t moving. Ptolemy stated that the planetary bodies moved around earth in circular paths. However, this wasn’t always witnessed through observation. He adjusted his model to state that some planets must be moving in secondary orbits.

Copernicus put a rotating earth in a sun-centered model. The rotation of earth was able to account for the rising and setting of stars. The orbital motion of the earth and moon also accounted for the motion of the sun and moon with respect to the stars. This was easier to understand but encountered scrutiny due to its differences from religious teachings.

One big difference between the approaches in the two is that Copernicus didn’t try to adjust his model to match what was going on; he used observations to develop the model. In addition, one common trend in science is that the simplest explanation is usually most accurate or closer to accurate. Copernicus’ model was more straightforward; Ptolemy’s was more complex.

3. Acceleration in a circle is toward the center of the circle, while velocity is always a straight line that's tangent to the circle. Thus, when the boy lets go of the rope, the centripetal force (acceleration) toward the center of the circle disappears. The ball then follows the straight path, tangent to the circle, and follows Path A.

7 0

3 years ago