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WINSTONCH
1 day ago
8

The distance between two slits is 1.50 *10-5 m. A beam of coherent light of wavelength 600 nm illuminates these slits, and the d

istance between the slit and the screen is 2.00 m. What is the distance on the screen between the central bright fringe and the fourth-order bright fringe?
Physics
1 answer:
Yuliya22 [2.4K]1 day ago
8 0
The result is y = 2.4×10^-6 m or y = 2.4 μm. The equation for the distance from the central bright fringe to any other fringe in the pattern is described by the formula y = R×mλ/d, where y represents the distance between the nth fringe and the central bright fringe. Here, m indicates the fringe position, which is 4; λ denotes the wavelength of light, which is 600 nm (or 600×10^-9 m); d is the distance separating the slits, set at 1.50×10^-5 m; and R represents the distance from the slit to the screen, which is 2 m. Therefore, y = 2 × 4 × 600×10^-9/2 = 2400 × 10^-9 = 2.4×10^-6 m or y = 2.4 μm.
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hows a map of Olivia's trip to a coffee shop. She gets on her bike at Loomis and then rides south 0.9mi to Broadway. She turns e
Softa [2029]

The inquiry is unfinished. Below is the complete question.

The attached map depicts Olivia's journey to the coffee shop. She rides her bike south for 0.9mi to Broadway after starting at Loomis. Next, she turns east onto Broadway, traveling 0.8mi until the road curves, and then she proceeds another 1.4mi to reach the shop.

What is the total displacement magnitude of her journey?

What direction does the total displacement of her journey take?

Response: Magnitude = 2.6mi

              Direction: 54.65° east

Clarification: Displacement refers to the change in position of a moving entity.

There are various methods to calculate total displacement. In this scenario, the method of Perpendicular Components of a Vector will be applied.

This method calculates total displacement as follows:

\Delta d_{t}=\sqrt{(\Delta d_{x})^{2}+(\Delta d_{y})^{2}}

\theta=tan^{-1}(\frac{\Delta d_{y}}{\Delta d_{x}})

\Delta d_{x} represents the x-component of the total displacement, which is the total of all individual x-components;

\Delta d_{y} stands for the y-component of the total displacement, calculated by summing all individual y-components;

θ signifies the angle of the resultant displacement;

In Olivia's case, there is no x-component for the first leg, and for the last segment, the biking path acts as the hypotenuse of a right triangle. Therefore, the x-component of that triangle is:

sin30=\frac{x}{1.4}

\frac{1}{2} =\frac{x}{1.4}

x = 0.7

Consequently,

\Delta d_{x} = 0 + 0.8 + 0.7

\Delta d_{x} = 1.5

Regarding the y-component, there’s no y-component in the second portion of Olivia's ride, and in the last segment:

cos30=\frac{y}{1.4}

\frac{\sqrt{3} }{2} =\frac{y}{1.4}

y = 1.21

Thus,

\Delta d_{y} = 0.9 + 0 + 1.21

\Delta d_{y} = 2.11

The total displacement is

\Delta d_{t}=\sqrt{(1.5)^{2}+(2.11)^{2}}

\Delta d_{t}=\sqrt{6.7021}

\Delta d_{t}= 2.6

The magnitude of Olivia's total displacement is 2.6mi

Connecting the starting and ending points on the map forms a vector directed eastward, at an angle of:

\theta=tan^{-1}(\frac{2.11}{1.5})

\theta=tan^{-1}(1.41)

θ = 54.65°

The direction of the total displacement is 54.65° East.

6 0
1 month ago
Argelia has a stack of schoolbooks sitting in the backseat of her car. When Argelia makes a sharp right turn, the books slide to
Yuliya22 [2420]

Answer:

The books are displaced to the left due to inertia and ultimately halt when impacted by the car door.

Explanation:

The movement of the books can be understood through Newton's first two laws:

- The first law (Law of Inertia): an object will remain at rest or continue moving in a straight line unless an unbalanced force acts upon it.

- The second law: if unbalanced forces act on an object, it experiences an acceleration that can be described by the formula

F=ma

where F is the object's net force, m its mass, and a its acceleration.

Now let's relate this to the scenario:

- When Argelia makes a sharp right turn, the books, which are not secured in the car, maintain their straight-line motion due to inertia so they appear to move left as the car shifts right.

- Upon contacting the car door, the books cease moving due to the second law: the door exerts an unbalanced force, causing the books to decelerate and ultimately come to rest.

4 0
1 month ago
Read 2 more answers
Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T
kicyunya [2264]

Answer:

v_{2} will be less than v_{1} and P_{2} will be greater than P_{1}.

Explanation:

The law of conservation of mass states that the rate of fluid mass (m_{1}) entering a system equals the rate at which the fluid mass (m_{2}) exits the system.

The mass flow rate can be expressed as follows:

m = \rho A v

where \rho denotes the fluid density, A signifies the cross-sectional area through which fluid flows, and v represents the fluid's velocity.

Based on the problem conditions, as the fluid's density remains constant, we can write:

&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}

where A_{1} and A_{2} are the cross-sectional areas for the fluid flow, while v_{1} and v_{2} are the corresponding velocities across those areas.

Given the conditions in the problem, A_{2} > A_{1}, leading from the formula to v_{2} < v_{1}.

Furthermore, fluid pressure arises from the fluid's movement through any specific area. When the fluid accelerates, part of its energy increases its speed in the direction of flow, resulting in lower pressure.

Thus, in this instance, v_{2} < v_{1} the pressure in the larger cross-sectional area P_{2} will exceed the pressure P_{1} in the smaller cross-sectional area, implying

P_{2} > P_{1}.

6 0
25 days ago
A 100 cm3 block of lead weighs 11N is carefully submerged in water. One cm3 of water weighs 0.0098 N.
Keith_Richards [2256]

#1

The volume of lead measures 100 cm^3

with a density of lead at 11.34 g/cm^3

. Thus, the mass of the lead block equals density multiplied by volume

m = 100 * 11.34 = 1134 g

m = 1.134 kg

Therefore, its weight in air is noted as

W = mg = 1.134* 9.8 = 11.11 N

Next, the buoyant force acting on the lead is defined as

F_B = W - F_{net}

F_B = 11.11 - 11 = 0.11 N

We know that

F_B = \rho V g

0.11 = 1000* V * 9.8

After solving, we find

V = 11.22 cm^3

(ii) This corresponding volume of water exerts the same weight as the buoyant force, resulting in 0.11 N

(iii) The buoyant force measures 0.11 N

(iv) The lead block sinks in water due to its density being greater than that of water.


#2

The buoyant force acting on the lead block counterbalances its weight

F_B = W

\rho V g = W

13* 10^3 * V * 9.8 = 11.11

V = 87.2 cm^3

(ii) This volume of mercury corresponds to the buoyant force weight, confirming that the block floats within mercury, resulting in 11.11 N as its weight.

(iii) The buoyant force is recorded as 11.11 N

(iv) Given that lead's density is less than mercury's, the lead will float in the mercury medium.


#3

Indeed, an object that has lesser density than a liquid will float; otherwise, it will sink in the liquid.

3 0
4 days ago
Calculate the heat required when 2.50 mol of a reacts with excess b and a2b according to the reaction: 2a + b + a2b → 2ab + a2 g
Softa [2029]

Answer:

Q = 12.5 kJ

Explanation:

The formula used to compute heat is:

Q = H° * n

Where:

Q: heat (J or kJ)

H°: enthalpy of reaction (kJ/mol)

n: moles

Now, as noted in the comments, the question lacks completeness, here is the part that is missing:

Given:

2A + B  A2B (1)

ΔH° = – 25.0 kJ/mol

2A2B  2AB + A2 (2)

ΔH° = 35.0 kJ/mol

Using these two reactions, we can determine the heat change.

Using the above two reactions, we need to establish the overall reaction (the one presented in the question), so let’s combine (1) and (2):

2A + B -------> A2B   H°1 = -25 kJ/mol

2A2B --------> 2AB + A2   H°2 = 35 kJ/mol

When we add these equations, one A2B cancels out with one A2B from reaction 2, thus, we have:

2A + B + 2A2B -------> 2AB + A2

So, for the enthalpy, the values are summed:

H°3 = -25 + 35 = 10 kJ/mol

Now we can calculate the heat:

Q = 10 * 2.5 = 25 kJ

However, as we have 2A in the reaction, it does not maintain a 1:1 mole ratio, instead, it is 1:2, which requires us to adjust; thus:

Q = 25 / 2 = 12.5 kJ

3 0
8 days ago
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