g n diffraction, the formula for minima is given by a times s i n (theta )equals m lambda, where a is the width of the slit, theta is the angle of dispersion, m is the order, and lambda is the wavelength. For a wavelength of 630 nm, m equals 3, and a equals 5 times 10 to the power of negative 6 end exponent m. Calculate the angular spread in Degrees. Enter only the numerical value (i.e. omit the unit abbreviations when inputting your answer into Blackboard).

Explanation:

When there is a high frequency sound the speed of the vibrations is faster and makes a high pitch sound. When there is a low frequency sound the speed of the vibrations is slower and makes a lower pitch sound.

Answer:

Instantaneous velocity  [tex]= 20[/tex] meter per second

Instantaneous acceleration  [tex]= 6[/tex] meter per second square

Explanation:

Given equation of distance X = [tex]2t+3t^2[/tex]

Instantaneous velocity [tex]= \frac{dX}{dt}[/tex]  [tex]= 2 + 6 t[/tex]

Substituting the value of t = 3 seconds, we get -

[tex]\frac{dX}{dt} = 2 + 6*3 = 20[/tex] meter per second

Instantaneous acceleration  [tex]= \frac{d^2X}{dt^2}[/tex]  [tex]= 6[/tex] meter per second square

Answer:

-40 degrees

Explanation:

I just took the Temperature Quiz and got it right. :)

Answer:

A.

Explanation:

Some of these photons are absorbed by the atoms in the ground state and the photons are lost to the laser process. However, some photons cause stimulated emission in excited-state atoms, releasing another coherent photon. In effect, this results in optical amplification

Answer:

2000 N.

Explanation:

From the question given above, the following data were obtained:

Load (L) = 8000 N

Mechanical advantage (MA) = 4

Effort (E) =?

The mechanical advantage of a machine is simply defined as:

Mechanical advantage = Load / Effort

MA = L / E

With the above formula, we can obtain the effort used to raise the load of 8000 N as follow:

Load (L) = 8000 N

Mechanical advantage (MA) = 4

Effort (E) =?

MA = L / E

4 = 8000 / E

Cross multiply

4 × E = 8000

Divide both side by 4

E = 8000 / 4

E = 2000 N

Thus, the effort used to raised the load is 2000 N.

Answer:

1. Rₑq = 4 Ω

2. R₂ = 6 Ω

3. Vₜ = 12 V, V₁ = 12 V, V₂ = 12 V

4. Iₜ = 3 A, I₁ = 1 A, I₂ = 2 A

Explanation:

1. Determination of the equivalent resistance

Voltage (V) = 12 V

Current (I) = 3 A

Resistance (Rₑq) =?

V= IRₑq

12 = 3 × Rₑq

Divide both side by 3

Rₑq = 12 / 3

Rₑq = 4 Ω

Thus, the equivalent resistance (Rₑq) = 4 Ω

2. Determination of R₂.

Equivalent resistance (Rₑq) = 4 Ω

Resistance 1 (R₁) = 12 Ω

Resistance 2 (R₂)

Since the resistor are in parallel arrangement, the value of R₂ can be obtained as follow:

Rₑq = R₁ × R₂ / R₁ + R₂

4 = 12 × R₂ / 12 + R₂

Cross multiply

4(12 + R₂) = 12R₂

48 + 4R₂ = 12R₂

Collect like terms

48 = 12R₂ – 4R₂

48 = 8R₂

Divide both side by 8

R₂ = 48 / 8

R₂ = 6 Ω

3. Determination of the total voltage (Vₜ), V₁ and V₂.

From the question given above, the total voltage is 12 V

Since the resistors are arranged in parallel connection, the same voltage will go through them.

Thus,

Vₜ = V₁ = V₂ = 12 V

4. Determination of the total current (Iₜ), I₁ and I₂

From the question given above, the total current (Iₜ) is 3 A

Next, we shall determine I₁. Since the resistors are arranged in parallel connection, different current will pass through each resistor respective.

Vₜ = V₁ = 12 V

R₁ = 12 Ω

I₁ =?

V₁ = I₁R₁

12 = I₁ ×12

Divide both side by 12

I₁ = 12 / 12

I₁ = 1 A

Next, we shall determine I₂. This can be obtained as follow:

Iₜ = 3 A

I₁ = 1 A

I₂ =?

Iₜ = I₁ + I₂

3 = 1 + I₂

Collect like terms

I₂ = 3 – 1

I₂ = 2 A

cold, dry, and stable air masses .

Answer:

 v = 4.8 10⁻⁴ m / s

Explanation:

To solve this exercise we can use the concepts of energy. In this case the potential energy is transformed into kinetic energy

          U = K

          q V = ½ m v²

          v = [tex]\sqrt { \frac{2qV}{m} }[/tex]

in the exercise they indicate the value of the charge q₁ = 1 pC = 1 10⁻¹² C

let's calculate

          v = [tex]\sqrt{ \frac{2 \ 1 \ 10^{-12} 120 \ 10^3}{500 ^{2} }[/tex]

          v = 4.8 10⁻⁴ m / s

Answer:

100 milliliters

Explanation:

Answer:

correct answer is C

Explanation:

In this exercise, you are asked to complete the sentences so that the sentence makes sense.

1) in projectile launching, the only force that acts is gravity in the vertical direction, so the time of going up is EQUAL to the time of going down

correct answer C

2) when a body falls freely, the acceleration is the ratio of gravity, therefore if it starts from rest, its height is

            y = v₀ t - ½ gt²

v₀ = 0

             y = -1/2 g t²

so the position is not proportional to the square of the time

correct answer is C

Answer:

[tex]6.28\times 10^{-5}\ \text{T}[/tex]

[tex]1.92\times 10^{-6}\ \text{Nm}[/tex]

Explanation:

[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi 10^{-7}\ \text{H/m}[/tex]

[tex]I_l[/tex] = Current in circular loop = 13 A

[tex]r_l[/tex] = Radius of circular loop = 13 cm

[tex]N[/tex] = Number of turns = 58

[tex]r_c[/tex] = Radius of coil = 0.94 cm

[tex]I_c[/tex] = Current in coil = 1.9 A

[tex]\theta[/tex] = Angle between loop and coil = [tex]90^{\circ}[/tex]

Magnitude of magnetic field in circular loop

[tex]B_l=\dfrac{\mu_0I_l}{2r_l}\\\Rightarrow B_l=\dfrac{4\pi 10^{-7}\times 13}{2\times 13\times 10^{-2}}\\\Rightarrow B_l=6.28\times 10^{-5}\ \text{T}[/tex]

The magnetic field produced by the loop at its center is [tex]6.28\times 10^{-5}\ \text{T}[/tex].

Torque is given by

[tex]\tau=\pi NI_cr_c^2B_l\sin\theta\\\Rightarrow \tau=\pi 58\times 1.9\times (0.94\times 10^{-2})^2\times 6.28\times 10^{-5}\sin90^{\circ}\\\Rightarrow \tau=1.92\times 10^{-6}\ \text{Nm}[/tex]

The torque on the coil due to the loop [tex]1.92\times 10^{-6}\ \text{Nm}[/tex].

Answer:

C

Explanation:

if you measure your body's flexibility then you can keep track of how flexible you have gotten over time

[tex] \Large {\underline { \sf {Required \; Solution :}}}[/tex]

We have ―

We've been asked to calculate acceleration.

[tex]\qquad \implies\boxed{\red{\sf{ a = \dfrac{v-u}{t} }}}\\[/tex]

[tex] \twoheadrightarrow \quad \sf {a = \dfrac{30-18}{3} \; ms^{-2} } \\ [/tex]

[tex] \twoheadrightarrow \quad \sf {a =\cancel{ \dfrac{12}{3} \; ms^{-2} }} \\ [/tex]

[tex]\twoheadrightarrow \quad \boxed{\red{\sf{ a = 4 \; ms^{-2} }}}\\[/tex]

Therefore, acceleration of the car is 4 m/s².