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2 years ago

14

The figure in Figure 1 shows two single-slit diffraction patterns. The distance between the slit and the viewing screen is the s

ame in both cases. Which of the following could be true

1 answer:

V125BC [204]2 years ago

8 0

Answer:

"The wavelengths are the same for both. The width of slit 1 is larger than the width of slit 2."

Explanation:

The full question has not been provided, so I just copied this into the web and found this answer and explanation on quizlet:

"The wavelengths are the same for both. The width of slit 1 is larger than the width of slit 2.

D sin θ = m λ

if the wavelengths are the same, then if the angle is smaller, the slit width must be larger. The top photo shows a pattern that is more closely spaced. That means the angle is smaller. The slit width must be larger."

This answer/explanation should be correct, as we are looking at bright fringes and the formula being used corresponds to the parameters of the question.

Hope this helps!

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What is the speed of a proton whose kinetic energy is 3.4 kev ?

Andreas93 [3]

The kinetic energy of the proton is 3.4 kev
1 kev = 1.602 × 10^-16 joules
therefore 3.4 kev is equivalent to;
3.4 × (1.602 ×10^-16)= 5.4468 × 10^-16 J
Kinetic energy is calculated by the formula 1/2mv² where m is the mass and v is the velocity.
Therefore V = √((2 × ( 5.4468×10^-16))/ (1.67 ×10^-27))
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3 years ago

Which would be most reliable source for information about the toxity of an industrial chemical

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Read 2 more answers

What is the purpose of the rorschach test

krok68 [10]

Answer:

to examine a person's personality characteristics and emotional functioning

Explanation:

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3 years ago

2. A Plate 0.02 mm distance from a fixed Plate moves at a velocity Of 0.6mls and requires a force of 1.962 N Per unit area to ma

Marizza181 [45]

Answer:

6.54 × 10⁻⁵ Pa-s

Explanation:

Since the shear force, F = μAu/y where μ = viscosity of fluid between plates, A = area of plates, u = velocity of fluid = 0.6 m/s and y = separation of plates = 0.02 mm = 2 × 10⁻⁵ m

Since F = μAu/y

F/A = μu/y where F/A = force per unit area

Since we are given force per unit area, F/A = 1.962 N per unit area = 1.962 N/m²

So, μ = F/A ÷ u/y

substituting the values of the variables into the equation, we have

μ = F/A ÷ u/y

μ = 1.962 N/m² ÷ 0.6 m/s/2 × 10⁻⁵ m

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μ = 6.54 × 10⁻⁵ Ns/m²

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5 0

3 years ago

A two-liter bottle of your favorite beverage has just been removed from the trunk of your car. The temperature of the beverage i

Ksivusya [100]

Answer:

a) 209.3 kilojoules must be removed from two liter of beverage, b) A rate of heat removal of 1.163 kilowatts is required to cool down 10 2-liter bottles, c) Cooling 10 2-L bottles during 30 minutes costs 4.9 cents.

Explanation:

a) <em>How much heat energy must be removed from your two liters of beverage?</em>

At first we suppose that the beverage has the mass and specific heat of water and that there are no energy interactions between the bottle and its surroundings.

From the First Law of Thermodynamics and definition of sensible heat, we get that amount of removed heat ( $Q$ ), measured in kilojoules, is represented by the following formula:

$Q = \rho \cdot V\cdot c\cdot (T_{o}-T_{f})$ (Eq. 1)

Where:

$\rho$ - Density of the beverage, measured in kilograms per cubic meter.

$V$ - Volume of the bottle, measured in cubic meters.

$c$ - Specific heat of water, measured in kilojoules per kilogram-Celsius.

$T_{o}$ , $T_{f}$ - Initial and final temperatures, measured in Celsius.

If we know that $\rho = 1000\,\frac{kg}{m^{3}}$ , $V = 2\times 10^{-3}\,m^{3}$ , $c = 4.186\,\frac{kJ}{kg\cdot ^{\circ}C}$ , $T_{o} = 35\,^{\circ}C$ and $T_{f} = 10\,^{\circ}C$ , then:

$Q = \left(1000\,\frac{kg}{m^{3}}\right)\cdot (2\times 10^{-3}\,m^{3})\cdot \left(4.186\,\frac{kJ}{kg\cdot ^{\circ}C} \right) \cdot (35\,^{\circ}C-10\,^{\circ}C)$

$Q = 209.3\,kJ$

209.3 kilojoules must be removed from two liter of beverage.

b) <em>You are having a party and need to cool 10 of these two-liter bottles in one-half hour. What rate of heat removal, in kW, is required?</em>

The total amount of heat that must be removed from 10 2-L bottles is:

$Q_{T} = 10\cdot (209.3\,kJ)$

$Q_{T} = 2093\,kJ$

If we suppose that bottles are cooled at constant rate, then, rate of heat removal is determined by this formula:

$\dot Q = \frac{Q_{T}}{\Delta t}$ (Eq. 2)

Where:

$Q_{T}$ - Total heat, measured in kilojoules.

$\Delta t$ - Time, measured in seconds.

$\dot Q$ - Rate of heat removal, measured in kilowatts.

If we know that $Q_{T} = 2093\,kJ$ and $\Delta t = 1800\,s$ , we find that rate of heat removal is:

$\dot Q = \frac{2093\,kJ}{1800\,s}$

$\dot Q = 1.163\,kW$

A rate of heat removal of 1.163 kilowatts is required to cool down 10 2-liter bottles.

c) <em>Assuming that your refrigerator can accomplish this and that electricity costs 8.5 cents per kW-hr, how much will it cost to cool these 10 bottles (in $)?</em>

A kilowatt-hour equals 3600 kilojoules. The electricity cost is equal to the removal heat of 10 bottles ( $Q_{T}$ ), measured in kilojoules, and unit electricity cost ( $c$ ), measured in US dollars per kilowatt-hour. That is:

$C = c\cdot Q_{T}$

If we know that $c = 0.085\,\frac{USD}{kWh}$ and $Q_{T} = 2093\,kJ$ , the total cost of cooling 10 bottles is:

$C = \left(0.085\,\frac{USD}{kWh}\right)\cdot \left(2093\,kJ\right)\cdot \left(\frac{1}{3600}\,\frac{kWh}{kJ} \right)$

$C = 0.049\,USD$

Cooling 10 2-L bottles during 30 minutes costs 4.9 cents.

3 0

4 years ago