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4 years ago

9

Consider a single slit diffraction pattern for awavelength of 589 nm porjected ona screen that is 100m from a slit of width 0.25

mm.How far from the center of the pattern are the center of first and second dark fringes?

1 answer:

3241004551 [841]4 years ago

6 0

Answer:

c

Explanation:

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A circular loop (radius of 20 cm) is in a uniform magnetic field of 0.15 T. What angle(s) between the normal to the plane of the

Wittaler [7]

Answer:

The angle other than 90 degrees would result in magnetic flux of non zero magnitude.

Explanation:

The magnetic flux through a given closed area is equal to $\underset{B}{\rightarrow}.\underset{A}{\rightarrow}$ .

Where $\underset{B}{\rightarrow}$ is the magnitude of magnetic flux through loop .

$\underset{A}{\rightarrow}$ is the area vector of the given loop .

We know that $\underset{B}{\rightarrow}.\underset{A}{\rightarrow}$ = $\left | A \right |\left | B \right |\cos \theta$ ,

Where $\theta$ is the angle made by magnetic field with area vector.

Area vector of a closed loop is always normal to the plane of the loop .

So from above equation we can deduce that

$\theta = \cos^{-1}\frac{\phi }{\left | A \right |\left | B \right |}$

$\phi$ is the magnitude of magnetic flux which is non zero when $\theta$ not equal to 90 degrees.

8 0

3 years ago

A 10.0-µF capacitor is charged so that the potential difference between its plates is 10.0 V. A 5.0-µF capacitor is similarly ch

IceJOKER [234]

Answer:

Explanation:

Given that,

First Capacitor is 10 µF

C_1 = 10 µF

Potential difference is

V_1 = 10 V.

The charge on the plate is

q_1 = C_1 × V_1 = 10 × 10^-6 × 10 = 100µC

q_1 = 100 µC

A second capacitor is 5 µF

C_2 = 5 µF

Potential difference is

V_2 = 5V.

Then, the charge on the capacitor 2 is.

q_2 = C_2 × V_2

q_2 = 5µF × 5 = 25 µC

Then, the average capacitance is

q = (q_1 + q_2) / 2

q = (25 + 100) / 2

q = 62.5µC

B. The two capacitor are connected together, then the equivalent capacitance is

Ceq = C_1 + C_2.

Ceq = 10 µF + 5 µF.

Ceq = 15 µF.

The average voltage is

V = (V_1 + V_2) / 2

V = (10 + 5)/2

V = 15 / 2 = 7.5V

Energy dissipated is

U = ½Ceq•V²

U = ½ × 15 × 10^-6 × 7.5²

U = 4.22 × 10^-4 J

U = 422 × 10^-6

U = 422 µJ

6 0

4 years ago

Find the resultant of the following vector

Mashutka [201]

You can use the Pythagorean theorem to calculate the hypotenuse, since the resultant is the hypotenuse of the triangle.

$r^{2} = x^{2} + y^{2}\\r^{2} = 12^{2} + 16^{2}\\r^{2} = 144 + 256\\r^{2} = 400\\r = \sqrt{400}\\r = 20N$

8 0

3 years ago

If the half-life of a 2.0 gram sample of a radionuclide is 15 hours, then the half-life of a 1.0 gram sample of the same radionu

Lerok [7]

7.5 hours or 450 minutes. 15/2=7.5

5 0

3 years ago

A girl launches a toy rocket from the ground. The engine experiences an average thrust of 5.26 N. The mass of the engine plus fu

Triss [41]

Answer: The answer for A is - v = 786.93 m/s

The answer for B is - v = 122.40 m/s

Explanation:

a) To find the average exhaust speed (v) of the engine we can use the following equation:

F = vΔm

Where:

F: is the thrust by the engine = 5.26 N

Δm: is the mass of the fuel = 12.7 g

Δt: is the time of the burning of fuel = 1.90 s

v = F×ΔT/ΔT

b) To calculate the final velocity of the rocket we need to find the acceleration.

The acceleration (a) can be calculated as follows:

a = F/M

In the above equation, m is an average between the mass of the engine plus the rocket case mass and the mass of the engine plus the rocket case minus the fuel mass:

m = (m_{engine} + m_{rocket}) + (m_{engine} + m_{rocket} - m_{fuel})}{2} = {2*m_{engine} + 2*m_{rocket} - m_{fuel}}{2} = 2*25.0 g + 2*63.0 g - 12.7 g}{2} = 81.65 g

Now, the acceleration is:

a = 5.26 N/81.65-t 10^³kg} = 64.42 m*s^²

Finally, the final velocity of the rocket can be calculated using the following kinematic equation:

v= v_{0} + at = 0 + 64.42 m*s^{-2}*1.90 s = 122.40 m/s

7 0

3 years ago