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alisha [4.7K]
3 years ago
8

Structures on a bird feather act like a diffraction grating having 7000 7000 lines per centimeter. What is the angle of the firs

t-order maximum for 602 nm 602 nm light shone through a feather?
Physics
1 answer:
Anestetic [448]3 years ago
8 0

Answer:

θ = 28.9°

Explanation:

We are given;

Wavelength; λ = 602nm = 602 x 10^(-9) m

Lines per centimetre = 7000 /cm = 700000 /m

Thus, the distance between 2 adjacent lines is;

d = 1/700000 = 1.43 x 10^(-6) m

The angle at which diffracted light is formed is given by the formula

mλ = d sinθ

Where;

m is the mth order of the diffraction

λ is the wavelength of the incident light

d is the distance separating the centres of 2 adjacent slits

θ is the angle at which diffraction occurs

From the question, m is 1 because it says first order.

Thus, plugging in the relevant values into mλ = d sinθ, we have;

1 x 602 x 10^(-9) = 1.43 x 10^(-6) sinθ

sinθ = 602 x 10^(-9)/(1.43 x 10^(-6))

sinθ = 0.42098

θ = sin^(-1) 0.42098

θ = 28.9°

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Answer:

1500 Joules

Explanation:

Work = Force x Distance

When multiplying by 10 you simply shift all the digits to the

left and append a 0 to the end.

so 150 x 10 = 1500 Joules

3 0
3 years ago
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Determine the current in the 7-ohm resistor for the circuit shown in the figure. Assume that the batteries are ideal and that al
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Answer:

I₁ = 1.6 A (through 7 Ohm Resistor)

I₂ = 1.3 A (through 8 Ohm Resistor)

I₃ = I₁ - I₂ = 1.6 A - 1.3 A = 0.3 A (through 4 Ohm Resistor)

Explanation:

Here we consider two loops doe applying Kirchhoff's Voltage Law (KVL). The 1st loop is the left side one with a voltage source of 12 V and the 2nd Loop is the right side one with a voltage source of 9 V. We name the sources and resistor's as follows:

R₁ = 7 Ω

R₂ = 4 Ω

R₃ = 8 Ω

V₁ = 12 V

V₂ = 9 V

Now, we apply KVL to 1st Loop:

V₁ = I₁R₁ + (I₁ - I₂)R₂

12 = 7I₁ + (I₁ - I₂)(4)

12 = 7I₁ + 4I₁ - 4I₂

I₁ = (12 + 4 I₂)/11   ------------ equation (1)

Now, we apply KVL to 2nd Loop:

V₂ = (I₂ - I₁)R₂ + I₂R₃

9 = (I₂ - I₁)(4) + 8I₂

9 = 4I₂ - 4I₁ + 8I₂

9 = 12I₂ - 4I₁   -------------- equation (2)

using equation (1)

9 = 12I₂ - 4[(12 + 4 I₂)/11]

99 = 132 I₂ -  48 - 16 I₂

147 = 116 I₂

I₂ = 147/116

I₂ = 1.3 A

use this value in equation 2:

9 = 12(1.3 A) - 4I₁

4I₁ = 15.6 - 9

I₁ = 6.6 A/4

I₁ = 1.6 A

Hence, the currents through all resistors are:

<u>I₁ = 1.6 A (through 7 Ohm Resistor)</u>

<u>I₂ = 1.3 A (through 8 Ohm Resistor)</u>

<u>I₃ = I₁ - I₂ = 1.6 A - 1.3 A = 0.3 A (through 4 Ohm Resistor)</u>

4 0
3 years ago
Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic colli
serious [3.7K]

Answer:

pf = 198.8 kg*m/s

θ = 46.8º N of E.

Explanation:

  • Since total momentum is conserved, and momentum is a vector, the components of the momentum along two axes perpendicular each other must be conserved too.
  • If we call the positive x- axis to the W-E direction, and the positive y-axis to the S-N direction, we can write the following equation for the initial momentum along the x-axis:

       p_{ox} = p_{oAx} + p_{oBx}  (1)

  • We can do exactly the same for the initial momentum along the y-axis:

       p_{oy} = p_{oAy} + p_{oBy}  (2)

  • The final momentum along the x-axis, since the collision is inelastic and both objects stick together after the collision, can be written as follows:

       p_{fx} =  (m_{A} + m_{B} ) * v_{fx}  (3)

  • We can repeat the process for the y-axis, as follows:

       p_{fy} =  (m_{A} + m_{B} ) * v_{fy}  (4)

  • Since (1) is equal to (3), replacing for the givens, and since p₀Bₓ = 0, we can solve for vfₓ as follows:

       v_{fx} = \frac{p_{oAx}}{(m_{A}+ m_{B)}} = \frac{m_{A}*v_{oAx} }{(m_{A}+ m_{B)}} =\frac{17.0kg*8.00m/s}{46.0kg} =  2.96 m/s (5)

  • In the same way, we can find the component of the final momentum along the y-axis, as follows:

       v_{fy} = \frac{p_{oBy}}{(m_{A}+ m_{B)}} = \frac{m_{B}*v_{oBy} }{(m_{A}+ m_{B)}} =\frac{29.0kg*5.00m/s}{46.0kg} =  3.15 m/s (6)

  • With the values of vfx and vfy, we can find the magnitude of the final speed of the two-object system, applying the Pythagorean Theorem, as follows:

      v_{f} = \sqrt{v_{fx} ^{2} + v_{fy} ^{2}} = \sqrt{(2.96m/s)^{2} + (3.15m/s)^{2}} = 4.32 m/s (7)

  • The magnitude of the final total momentum is just the product of the combined mass of both objects times the magnitude of the final speed:

       p_{f} = (m_{A} + m_{B})* v_{f}  = 46 kg * 4.32 m/s = 198.8 kg*m/s (8)

  • Finally, the angle that the final momentum vector makes with the positive x-axis, is the same that the final velocity vector makes with it.
  • We can find this angle applying the definition of tangent of an angle, as follows:

       tg \theta = \frac{v_{fy}}{v_{fx}} = \frac{3.15 m/s}{2.96m/s} = 1.06 (9)

       ⇒ θ = tg⁻¹ (1.06) = 46.8º N of E

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Answer:

Because of the location, humidity and temperatures.

Explanation:

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they form the lower floor of the upper Jungle, in the Central Andes, mostly in Peru and Bolivia. The  yungas are in contact with the rainforests of the lowlands in Amazonia, where it has been  started to expand coca cultivation recently (Dourojeanni, 1988). The optimum altitude is 1000 a  2000 meters (where cocaine content is higher), with optimal annual average precipitation, is 2000 meters  mm, but it is grown between 700 and 2000 msnm and with an average annual rainfall of 1000 to 4200 mm.

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