Coherent monochromatic light falls perpendicularly on two slits (each of width 0.10 mm) separated by 0.50 mm. In the resulting interference pattern on a screen 2.80 m away, adjacent bright fringes are separated by 2.80 mm. (a)What is the wavelength of the light that falls on the slits

Answer:

B. They are inversely proportional to each other

[tex] \frac{momentum}{time} = force \\ \\ \frac{mass \times velocity}{time} = force \\ \\ \frac{mass \times velocity}{time} = mass \times acceleration[/tex]

Answer:

a) Ink X is likely to be pure because it only contain 1 spot.

b) The chromatography tell us about ink Y that it is a mixture as it contain more than 1 spot.

c) The three different spots are separated out from ink Y at different heights beacaus different substance have different solubility.

The different spots from Y are found at various heights because they represent different compounds.

The term chromatography has to do with a method of separating the component of a substance. The term chromatography originally means color writing.

We can see that the pure ink is the ink marked X. We can see from the chromatogram that Y is a mixture of colors. The different spots from Y are found at various heights because they represent different compounds.

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Answer:

F = 10.78 N

Hence q₂ will move away from the charge q₁ towards right side.

Explanation:

The force between two charged particles can be found by using Colomb's Law:

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

where,

F = Force = ?

k = Colomb Constant = 8.99 x 10⁹ N.m²/C²

q₁ = charge on first particle = 6 μC = 6 x 10⁻⁶ C

q₂ = charge on second particle = 2 μC = 2 x 10⁻⁶ C

r = distance between particles = 0.1 m

Therefore,

[tex]F = \frac{(8.99\ x\ 10^9\ N.m^2/C^2)(6\ x\ 10^{-6}\ C)(2\ x\ 10^{-6}\ C)}{(0.1\ m)^2}[/tex]

F = 10.78 N

Since both particles have a positive charge. Therefore this force will be the force of repulsion.

Hence q₂ will move away from the charge q₁ towards right side.

Answer:

Explanation:

E2020

Answer:

52.5 J

Explanation:

Applying,

Hook's law,

F = ke............... Equation 1

Where F = Force, k = spring constant, e = extension.

make k the subject of the equation

k = F/e............ Equation 2

From the question,

Given: F = 350 Newtons, e = 30 cm = 0.3 m

Substitute these values into equation 2

k = 350/0.3 N/m

Also,

W = 1/2(ke²).................. Equation 3

Where W = work done in stretching the spring.

Also given: e = (50-20) cm = 30 cm = 0.3 m, k = 350/0.3 N/m

Substitute these values into equation 3

W = 1/2(350/0.3)(0.3²)

W = 350×0.3/2

W = 52.5 J

Answer:

[tex]E=6.86\times 10^6\ N/C[/tex]

Explanation:

Given that,

Mass of the sphere, m = 2.1 g = 0.0021 kg

Charge, q = 3 nC

We need to find the magnitude of the electric field that balanced the weight of sphere. Let it is E. So,

qE = mg

[tex]E=\dfrac{mg}{q}[/tex]

Put all the values,

[tex]E=\dfrac{0.0021\times 9.8}{3\times 10^{-9}}\\\\E=6.86\times 10^6\ N/C[/tex]

So, the magnitude of the elecric field is [tex]6.86\times 10^6\ N/C[/tex].

Answer:

 F_net = 9.87 10⁻⁴ N

Explanation:

Let's use that force is a vector magnitude

         ∑ F = F₁₃ + F₂₃

De bold arfe vectros. The force is the electric force, we use that charges of the same sign repel and when the charges are of a different sign they attract

the charges q1 and q2 are negative and the charge q3 is positive with the positions y1 = 38 m, y2 = -7m, y3 = 16 m

       ∑ F = F₁₃ - F₂₃

       F_net = [tex]k \frac{q_1q_3}{r_{13}^2 } - k \frac{q_2q_3}{r_{23}^2 }[/tex]

in this case q₁ = q₂ = q

       F_net = k q q₃  (  )

      let's look for the distance

      r₂₃ = y₂ - y₃

      r₂₃ = -7 -16

      r₂₃ = - 23 m

      r₁₃ = 38 - 16

      r₁₃ = 22 m

let's calculate

      F_net = 9 10⁹ 24 26 10⁻¹² ( )

      F_net = 5.616 ( 1.758 10⁻⁴ )

      F_net = 9.87 10⁻⁴ N

The resulting wave has the largest possible amplitude when Wave-1 and Wave-2 are exactly in step ... their peaks both happen at the same time and their troughs both happen at the same time.

This means that Wave-1 and Wave-2 have the same frequency, and the phase shift from one wave to the other is zero.

When all of that happens, the amplitude of the resulting wave is the sum of the amplitudes of Wave-1 and Wave-2.  If Wave-1 and Wave-2 have the same amplitude, then the resulting wave will have double that amplitude.

(a) 4.20 J

(b) 16.74 J

For a parallel plate vacuum capacitor with area A and whose plates are separated by by a distance of d, its capacitance C is given by;

C = A∈₀ / d              --------------------(i)

Where;

∈₀ = constant called permittivity of vacuum.

The energy U stored in such capacitor is given by;

U = [tex]\frac{1}{2}[/tex]CV²             ----------------------(ii)

or

U =  [tex]\frac{1}{2}[/tex](Q²/C)        -------------------(**)

Where;

V = potential difference or voltage across the plates.

Q = charge on the plates.

(a) If the charge is held constant

Combine equations (i) and (**) to give;

U =  [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / d)     -----------------------(iii)

From the question;

The parallel plate capacitor has 8.40J energy stored and distance between plates is 2.30mm i.e

U = 8.40J

d = 2.30mm = 0.023m

Substitute these values into equation (iii)

8.40 =  [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / 0.023)

8.40 =  [tex]\frac{1}{2}[/tex]Q² x (0.023 / A∈₀)

Multiply through by 2

2 x 8.40 = Q² x (0.023 / A∈₀)

16.80 = Q² x (0.023 / A∈₀)

Divide through by 0.023

16.80 / 0.023 = Q² x (0.023 / A∈₀) / 0.023

730.4 = Q² / (A∈₀)

Make Q² subject of the formula

Q² = 730.4(A∈₀)

Now, if the separation distance is decreased to 1.15mm and the voltage is held constant i.e

d = 1.15mm = 0.0115m

Q = constant [this means that Q² still remains 730.4(A∈₀) ]

The energy stored is found by substituting these values of d and Q² into equation (iii) as follows;

U =   [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / d)

U = [tex]\frac{1}{2}[/tex](730.4(A∈₀)) / (A∈₀ / 0.0115)

U = [tex]\frac{1}{2}[/tex](730.4(A∈₀))(0.0115 / A∈₀)

U = [tex]\frac{1}{2}[/tex](730.4)(0.0115)

U = 4.20J

Therefore, the energy stored if the charge Q on the plates is held constant is 4.20 J

(b) If the voltage is held constant

Combine equations (i) and (ii) to give;

U =  [tex]\frac{1}{2}[/tex](A∈₀ / d)V²     -----------------------(iv)

From the question;

The parallel plate capacitor has 8.40J energy stored and distance between plates is 2.30mm i.e

U = 8.40J

d = 2.30mm = 0.023m

Substitute these values into equation (iv)

8.40 =  [tex]\frac{1}{2}[/tex](A∈₀ / 0.023)V²

Multiply through by 2 x 0.023

2 x 0.023 x 8.40 = (A∈₀)V²

2 x 0.023 x 8.40 = (A∈₀)V²

0.385 = (A∈₀)V²

Make V² subject of the formula

V² = 0.385/(A∈₀)

Now, if the separation distance is decreased to 1.15mm and the voltage is held constant i.e

d = 1.15mm = 0.0115m

V = constant [this means that V² still remains 0.385/(A∈₀) ]

The energy stored is found by substituting these values of d and V² into equation (iv) as follows;

U =  [tex]\frac{1}{2}[/tex](A∈₀ / 0.0115)[0.385/(A∈₀)]  

U = [tex]\frac{1}{2}[/tex](0.385/0.0115)

U = 16.74

Therefore, the energy stored if the voltage V across the plates is held constant is 16.74 J

Answer:

b. is the correct answer ....

Answer:

Explanation:

The one above is incorrect, and I know this is late. Even if it doesn't help you I hope it helps people in the future! YES I AM TALKING ABOUT YOU FUTURE PEOPLE!! I know this is the answer because I have taken 5.11 Quiz: Uncrewed Spacecraft in K12. There will only be the questions and correct answers below.

1. Which planetary body was Spirit designed to explore?

Mars.

2. What is the name of the most distant manmade object in space? (Credit: shathaadnan64/lak521)

Voyager 1.

3. Which group was designed to study Saturn? (Credit: Brainly User/snowballandtigoya1xa

Voyager 1, Huygens, and Cassini.

4. Why are scientists interested in exploring Mars?

Possible evidence of life.

5. What do interplanetary space missions study?

Planets in the solar system.

Have an amazing day!!

Answer:

b 4.5

Explanation:

time=distance/speed

area= length × length

area = 4m × 9m

ans 36

Answer:

D

Explanation:

Potential energy involves the change of an object's position, which in this case a rocket is increasing its vertical displacement from the ground.

When a rocket is increasing its vertical displacement from the ground, it exhibits both potential and kinetic energy. Therefore option D is correct.

At the initial stage, when the rocket is on the ground and not moving, it possesses potential energy. This potential energy is in the form of stored energy due to its elevated position above the ground.

As the rocket launches and gains altitude, it continues to accumulate potential energy because it is moving higher against the force of gravity.

Simultaneously, as the rocket moves upward, it also gains kinetic energy. Kinetic energy is the energy associated with the rocket's motion.

The faster the rocket moves, the greater its kinetic energy becomes. As the rocket ascends, its speed increases, resulting in an increase in kinetic energy.

Therefore, in the context of a rocket increasing its vertical displacement from the ground, both potential energy (due to its height) and kinetic energy (due to its motion) are present.

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Answer:

68.6 J

Explanation:

Applying,

P.E = mgh............... Equation 1

Where P.E = Potential Energy of the basket, m = mass of the basket, g = acceleration due to gravity of the basket, h = height of the basket

From the question,

Given: m = 3.5 kg, h = 2.00 m

Constant: g = 9.8 m/s²

Substitute these values into equation 1

P.E = 3.5×2×9.8

P.E = 68.6 J

Hence the potential energy of the basket is 68.6 J

This question is incomplete, the complete question is;

In a softball windmill pitch, the pitcher rotates her arm through just over half a circle, bringing the ball from a point above her shoulder and slightly forward to a release point below her shoulder and slightly forward. (Figure 1) shows smoothed data for the angular velocity of the upper arm of a college softball pitcher doing a windmill pitch; at time t = 0 her arm is vertical and already in motion. For the first 0.15 s there is a steady increase in speed, leading to a final push with a greater acceleration during the final 0.05 s before the release. In each part of the problem, determine the corresponding quantity during the first 0.15 s of the pitch.

Angular Velocity at time 0s = 12 rad/s

Angular Velocity at time 0.15s = 24 rad/s

a) What is the angular acceleration?

b) If the ball is 0.60 m from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g

Answer:

a) the angular acceleration is 80 rad/s²

b) the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

Explanation:

Given the data in the question;

from the graph below;

Angular Velocity at time 0s [tex]w_o[/tex] = 12 rad/s

Angular Velocity at time 0.15s [tex]w_f[/tex] = 24 rad/s

a) What is the angular acceleration;

Angular acceleration ∝ = ( [tex]w_f[/tex] - [tex]w_o[/tex] ) / dt

we substitute

Angular acceleration ∝ = ( 24 - 12 ) / 0.15

Angular acceleration ∝ = 12 / 0.15

Angular acceleration ∝ = 80 rad/s²

Therefore, the angular acceleration is 80 rad/s²

b)

If the ball is 0.60 m from her shoulder, i.e s = 0.6 m

the tangential acceleration of the ball will be;

a = ∝ × s

we substitute

a = 80 × 0.6

a = 48 m/s²

a = ( 48 / 9.8 )g

a = 4.9 g

Therefore, the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

Answer:

sieving is when you separate particles of different sizes.

Explanation:

separating sand mixtures

separating chaffs from local garri

Answer:

1.07 × 10⁸ m/s

Explanation:

Using the relativistic Doppler shift formula which can be expressed as:

[tex]\lambda_o = \lambda_s \sqrt{\dfrac{c+v}{c-v}}[/tex]

here;

[tex]\lambda _o[/tex] = wavelength measured in relative motion with regard to the source at velocity v

[tex]\lambda_s =[/tex] observed wavelength from the source's frame.

Given that:

[tex]\lambda _o[/tex] = 656.3 nm

[tex]\lambda_s =[/tex] 953.3 nm

We will realize that [tex]\lambda _o[/tex] > [tex]\lambda_s[/tex]; thus, v < 0 for this to be true.

From the above equation, let's make (v/c) the subject of the formula: we have:

[tex]\dfrac{\lambda_o}{\lambda_s}=\sqrt{\dfrac{c+v}{c-v}}[/tex]

[tex]\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2=\dfrac{c+v}{c-v}[/tex]

[tex]\dfrac{v}{c} =\dfrac{\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2-1}{\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2+1}[/tex]

[tex]\dfrac{v}{c} =\dfrac{\Big(\dfrac{656.3}{953.3} \Big)^2-1}{\Big(\dfrac{656.3}{953.3} \Big)^2+1}[/tex]

[tex]\dfrac{v}{c} =0.357[/tex]

v = 0.357 c

To m/s:

1c = 299792458 m/s

∴

0.357c = (299 792 458 × 0.357) m/s

= 107025907.5 m/s

= 1.07 × 10⁸ m/s

Answer:

Explanation:

A parachute works by forcing air into the front of it and creating a structured 'wing' under which the canopy pilot can fly. Parachutes are controlled by pulling down on steering lines that change the shape of the wing, cause it to turn. The main forces acting on a parachute are gravity and drag. When you first release the parachute, the force of gravity pulls it downward, and the parachute speeds toward the ground. The faster the parachute falls, though, the more drag it creates.

Answer:

Magnetism is a physical phenomenon that manifests itself in a force acting between magnets or other magnetized or magnetisable objects, and a force acting on moving electric charges, such as in current-carrying cables. The force action takes place by means of a magnetic field, which is generated by the objects themselves or otherwise. There are natural and artificial magnets. All magnets have two poles called the north pole and the south pole. The north pole of one magnet repels the north pole of another magnet and attracts the south pole of another magnet; the same with south poles.  

SOLVED DOWN BELOW

Explanation:

In series the same current goes thru both resistors, equiv resistance is 200 ohms, then using ohms law

I = 25/200

I= .125 amps or 125 ma

__________

R= r1 * r2 / r1 +r2

R= 100 * 100 / 100 + 100

R= 10000 / 200

R= 50 ohms

Answer:

Less

Explanation:

Weight is a force measurement. The object's mass is 5kg not its weight. To find its weight you have to take the mass of an object and multiply it by the acceleration of gravity. The acceleration of gravity is greater on earth than on the moon so therefore the object will weigh less on the moon.

Answer:

36.11 degrees

Explanation:

index of refraction n = sin i/sinr

i is the angle of incidence

r is the angle of refraction

Substitute into the expression

1.39 = sin55/sin(r)

1.39 = 0.8191/sin(r)

sin(r) = 0.8191/1.39

sin(r) = 0.5893

r = arcsin(0.5893)

r = 36.11

hence the angle of refraction of light is 36.11 degrees

Answer:

F = 312 N

Explanation:

Given that,

The mass of a crate, m = 40 kg

Acceleration of the crate, a = 2 m/s²

As the carte is falling downward, the net force exerted by the rope on the carte is given by :

F = m(g-a)

Put all the values,

F = 40(9.8-2)

F = 312 N

Hence, the required force exerted by the rope on the crate is equal to 312 N.

Answer:

the correct answer is B

Explanation:

The quantum numbers are the constants obtained when solving the Schrodinger equation, the first quantum number or principal quantum number (n), can take values ​​from zero to infinity.

This quantum number is placed as a coefficient in the quantum distribution.

In this case for phosphorus, the number is n = 3

the correct answer is B

Answer:

E = 16581.6 J

Explanation:

Given that,

Current, I = 4.9 A

Time for which the current is set up, I = 4.7 min = 282 s

The voltage of the battery, V = 12 V

We need to find how much chemical energy of the battery reduced. Let It is E. We know that,

E = P t

Where

P is power of battery, P = VI

So,

[tex]E=VIt[/tex]

Put all the values,

[tex]E=12\times 4.9\times 282\\E=16581.6\ J[/tex]

So, 16581.6 J of chemical energy of the battery is reduced.

Answer:

Z-number

Explanation:

The Z number is the number of protons in an atom, and this does not change when an isotope is created. I got it right on the test.

Answer:

The rotation rate is 15.3 rad/s.

Explanation:

maximum voltage, V = 120 V

Magnetic field, B = 1.5 T

length, L = 10.2 cm

width, W = 12.8 cm

Number of loops, N = 400

Let the rate of rotation is w.

The maximum voltage is given by

V = N B A w

120 = 400 x 1.5 x 0.102 x 0.128 x w

w = 15.3 rad/s

Answer:

(a) I = 1650000 A

(b) 4.125 T

Explanation:

Magnetic field, B = 5.5 T

distance, r = 0.06 m

(a) Let the current is I.

The magnetic field due to a long wire is given by

[tex]B =\frac{\mu o}{4\pi }\frac{2 I}{r}\\5.5= 10^{-7}\times \frac{2\times I}{0.06}\\I =1650000 A[/tex]

(b) Let the magnetic field is B' at distance r = 0.08 m.

[tex]B =\frac{\mu o}{4\pi }\frac{2 I}{r}\\B = 10^{-7}\times \frac{2\times 1650000}{0.08}\\B'= 4.125 T[/tex]