A screen is placed 60.0 cm from a single slit, which is illuminated with light of wavelength 690 nm. If the distance between the first and third minima in the diffraction pattern is 3.10 mm, what is the width of the slit?

While a power supply tester can be a useful tool for quickly checking voltage output, it might not reveal all the potential issues a faulty power supply can cause.

Even if a power supply tester shows that the voltage output of a power supply is within acceptable limits, it's still possible that the power supply may be faulty. Here's why:

1. Voltage Under Load: A power supply tester might only measure the voltage output under no load or very light load conditions.

A faulty power supply might provide the correct voltage at low loads but fail to deliver stable voltage under high loads, which could lead to system instability or crashes.

2. Voltage Ripple and Noise: Power supplies are expected to provide a stable and clean output voltage.

3. Short Circuits or Overloads: A power supply tester typically doesn't simulate the behavior of a real system.

4. Intermittent Issues: Faulty power supplies can exhibit intermittent issues. The power supply might work fine during the testing but fail when subjected to extended periods of operation or specific conditions.

5. Quality of Components: A power supply tester might not assess the quality of individual components within the power supply.

6. Compatibility Issues: Some power supplies might not be fully compatible with certain computer hardware. Even if the voltage seems fine, compatibility issues can still cause problems.

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Answer:

s = 6.25 10⁻²² m

Explanation:

Polarizability is the separation of electric charges in a structure, in the case of the atom it is the result of the separation of positive charges in the nucleus and the electrons in their orbits, macroscopically it is approximated by

                p = q s

               s = p / q

let's calculate

              s = 1 10⁻⁴⁰ / 1.6 10⁻¹⁹

              s = 0.625 10⁻²¹ m

              s = 6.25 10⁻²² m

We see that the result is much smaller than the size of the atom, therefore this simplistic model cannot be taken to an atomic scale.

Answer:

Heat flux = (598.3î + 204.3j) W/m²

a) Magnitude of the heat flux = 632.22 W/m²

b) Direction of the heat flux = 18.85°

Explanation:

- The correct question is the first image attached to this solution.

- The solution to this question is contained on the second and third images attached to this solution respectively.

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Answer:

1.2 s

Explanation:

Given:

v₀ = 8.0 m/s

v = -4.0 m/s

a = -10 m/s²

Find: t

v = at + v₀

(-4.0 m/s) = (-10 m/s²) t + (8.0 m/s)

t = 1.2 s

The question is incomplete. Here is the complete question.

Lightning bolts can carry currents up to approximately 20kA. We can model such a current as the equivalent of a very long, straight wire.

(a) If you were unfortunate enough to be 5.5m away from such a lightning bolt, how large a magnetic field would you experience?

(b) How does this field compare to one you would experience by being 5.5cm from a long, straight household current of 5A?

Answer: (a) B = 7.27 x 10⁻⁴ T

              (b) Approximately 40 times higher than a household one.

Explanation: Using Biot-Savart Law, the magnetic field in a straight, long wire is given by

where:

(permeability of free space) = T.m/A

(a) If lightning bolt is compared to a long and straight wire, then magnetic field is

B = 7.27 x 10⁻⁴ T

The magnitude of magnetic field in a lightning bolt is 7.27 x 10⁻⁴ T

(b) Magnetic field in a household wire will be

B = 1.82 x 10⁻⁵ T

Comparing fields:

≈ 40

The filed for a lightning bolt is approximately 40 times higher than for a household wire.

Answer:

69.69 g

Explanation:

Evaporation of water will take out latent heat of vaporization.  Let the mass of water be m and latent heat of vaporization of water be 2260000 J per kg

Heat taken up by evaporating water

= 2260000 x m J

Heat lost by body

= mass x specific heat of body x drop in temperature

60 x 3500 x .750  ( specific heat of human body is 3.5 kJ/kg.k)

= 157500 J

Heat loss = heat gain

2260000 m= 157500

m = .06969 kg

= 69.69 g

Final answer:

Approximately 78 grams of water would need to evaporate from a 60.0-kg person to lower their body temperature by 0.750ºC. This calculation is based on the principles of thermodynamic heat transfer and the specific body temperature, latent heat of water vaporization, and specific heat capacity of the human body.

Explanation:

To calculate the amount of water mass from an individual's body that would need to evaporate to reduce their body temperature, we can use the principle of thermodynamic heat transfer. The basic equation is Q = mLv, where Q is the heat absorbed or lost, m is the mass, and Lv is the latent heat of vaporization.

In this case, knowing that at body temperature of 37.0°C, the latent heat of water vaporization (Lv) is approximately 2430 kJ/kg, we substitute these numbers. Given our desire to reduce body temperature by 0.750°C in a 60 kg human, we first calculate the amount of heat to dissipate (Q) using Q = mcΔT, where c is the specific heat capacity of the human body (roughly equivalent to that of water, 4.184 kJ/kg°C), m is the mass, and ΔT is the change in temperature.

The calculation is as follows:

Q = (60 kg)(4.184 kJ/kg°C)(0.750°C) = ~189 kJ

Next, we substitute Q into the Q = mLv equation to determine the mass m:

m = Q / Lv = 189 kJ / 2430 kJ/kg = 0.078 kg, or 78 grams

Hence, around 78 grams of water would need to evaporate from a 60.0-kg person to lower their body temperature by 0.750ºC.

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