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2 years ago

15

Which type of diffraction occurs when the point source and the screen are at finite distances from the obstacle forming the diff

raction pattern?
A. fraunhofer
B. fresnel
C. far-field
D. single slit

1 answer:

Paha777 [63]2 years ago

4 0

Fresnel diffraction

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An 8.0-ohm resistor and a 6.0-ohm resistor are connected in series with a battery. The potential difference across the 6.0-ohm r

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Answer:

a) 14 Ω

b) 2.0 A

c) 28 V

Explanation:

a) The total resistance of resistors in series is the sum:

R = R₁ + R₂

R = 8.0 Ω + 6.0 Ω

R = 14 Ω

b) The current in the 6.0 Ω resistor can be found with Ohm's law:

V = IR

12 V = I (6.0 Ω)

I = 2.0 A

c) Since the resistors are in series, they have the same current. So the total voltage is:

V = IR

V = (2.0 A) (14 Ω)

V = 28 V

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What is the mass m of the elevator? use g=10m/s2 for the magnitude of the acceleration of gravity.

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Your own car has a mass of 2000. Kg if your car produces a force of 5000 N how fast will it accelerate?

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F = m*a
5000 = 2000 * x
5000/2000 = x
2.5 = x
2.5m/s^2 = a

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3 years ago

A bowling ball has an initial momentum of +30 kg m/s and hits a stationary bowling pin. After the collision, the bowling ball le

dangina [55]

Momentum should be conserved. The momentum of both objects must balance with their initial and final momentum.

Let m1 and v1 be the mass and velocity of the bowling ball

And m2 and v2 be the mass and velocity of the bowling pin

(m1v1)i + (m2v2)i = (m1v1)f + (m2v2)f

30 kg m/s + (1.5 kg)(0 m/s) = 13kg m/s + 1.5v2f

V2f = 11.33 m/s

<span>So the momentum = 1.5 kg(11.33 m/s) = 17 kg m/s</span>

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2 years ago

A 1.20-m cylindrical rod of diameter 0.570 cm is connected to a power supply that maintains a constant potential difference of 1

nasty-shy [4]

(a) $1.72\cdot 10^{-5} \Omega m$

The resistance of the rod is given by:

$R=\rho \frac{L}{A}$ (1)

where

$\rho$ is the material resistivity

L = 1.20 m is the length of the rod

A is the cross-sectional area

The radius of the rod is half the diameter: $r=0.570 cm/2=0.285 cm=2.85\cdot 10^{-3} m$ , so the cross-sectional area is

$A=\pi r^2=\pi (2.85\cdot 10^{-3} m)^2=2.55\cdot 10^{-5} m^2$

The resistance at 20°C can be found by using Ohm's law. In fact, we know:

- The voltage at this temperature is V = 15.0 V

- The current at this temperature is I = 18.6 A

So, the resistance is

$R=\frac{V}{I}=\frac{15.0 V}{18.6 A}=0.81 \Omega$

And now we can re-arrange the eq.(1) to solve for the resistivity:

$\rho=\frac{RA}{L}=\frac{(0.81 \Omega)(2.55\cdot 10^{-5} m^2)}{1.20 m}=1.72\cdot 10^{-5} \Omega m$

(b) $8.57\cdot 10^{-4} /{\circ}C$

First of all, let's find the new resistance of the wire at 92.0°C. In this case, the current is

I = 17.5 A

So the resistance is

$R=\frac{V}{I}=\frac{15.0 V}{17.5 A}=0.86 \Omega$

The equation that gives the change in resistance as a function of the temperature is

$R(T)=R_0 (1+\alpha(T-T_0))$

where

$R(T)=0.86 \Omega$ is the resistance at the new temperature (92.0°C)

$R_0=0.81 \Omega$ is the resistance at the original temperature (20.0°C)

$\alpha$ is the temperature coefficient of resistivity

$T=92^{\circ}C$

$T_0 = 20^{\circ}$

Solving the formula for $\alpha$ , we find

$\alpha=\frac{\frac{R(T)}{R_0}-1}{T-T_0}=\frac{\frac{0.86 \Omega}{0.81 \Omega}-1}{92C-20C}=8.57\cdot 10^{-4} /{\circ}C$

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2 years ago