Why is the sign of the emf of the second peak opposite to the sign of the first peak?

4 of

Maksim231197 [3]

Answer:

all of the above????

Explanation:

my teacher told me.

8 0

3 years ago

Neil Armstrong, the first human being to step on the moon, left a foot print that is essentially unchanged nearly 50 years later

Likurg_2 [28]

Since the boot-print was left there nearly 50 years ago, there has been very little wind and very little rain in that area, and plus, there have been very few people or other animals walking around in that spot to disturb it.

6 0

3 years ago

An N-slit system has slit separation d and slit width a. Plane waves with intensity I and wavelength O are incident normally on

Leokris [45]

Answer:

E. d and O

Explanation:

"Light passing through a single slit forms a diffraction pattern somewhat different from those formed by double slits or diffraction gratings".

According to Huygens’s principle, "for each element of the wavefront in the slit emits wavelets. These are like rays that start out in phase and head in all directions. (Each ray is perpendicular to the wavefront of a wavelet.) Assuming the screen is very far away compared with the size of the slit, rays heading toward a common destination are nearly parallel".

The destructive interference for a single slit is given by:

$d sin \theta = m\lambda , m=1,-1,2,-2,3,...$

Where

d is the slit width

$\lambda=O$ is the light's wavelength

$\theta$ is the angle relative to the original direction of the light

m is the order od the minimum

I represent the intensity

When the intensity and the wavelength are incident normally the angular as we can see on the expression above the angular separation just depends of the distance d and the wavelength O.

7 0

2 years ago

A bowling ball with a mass of 7.10 kg has a velocity of 5 m/s. what is the bowling ball's momentum?

melomori [17]

P = mv

P momentum
m mass
v velocity

4 0

2 years ago

A basketball player makes a jump shot. The 0.599 kg ball is released at a height of 2.18 m above the floor with a speed of 7.05

7nadin3 [17]

Answer:

$W_{drag} = 4.223\,J$

Explanation:

The situation can be described by the Principle of Energy Conservation and the Work-Energy Theorem:

$U_{g,A}+K_{A} = U_{g,B} + K_{B} + W_{drag}$

The work done on the ball due to drag is:

$W_{drag} = (U_{g,A}-U_{g,B})+(K_{A}-K_{B})$

$W_{drag} = m\cdot g\cdot (h_{A}-h_{B})+ \frac{1}{2}\cdot m \cdot (v_{A}^{2}-v_{B}^{2})$

$W_{drag} = (0.599\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.18\,m-3.10\,m)+\frac{1}{2}\cdot (0.599\,kg)\cdot [(7.05\,\frac{m}{s} )^{2}-(4.19\,\frac{m}{s} )^{2}]$

$W_{drag} = 4.223\,J$

7 0

3 years ago

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