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Softa [21]
2 years ago
6

This is for 6th grade Science -Geology - weather and climate.

Physics
1 answer:
Ierofanga [76]2 years ago
3 0

Answer:

Explanation:

nahh

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Light of wavelength 559 nm is used to illuminate normally two glass plates 22.1 cm in length that touch at one end and are separ
umka21 [38]

Answer:

M = 222 fringes

Explanation:

given

λ = 559 n m = 559 × 10⁻⁹ m

radius = 0.026 mm = 0.026 ×10⁻³ m

length of the glass plate = 22.1 ×10⁻² m

using relation

2t=(m+\dfrac{1}{2})\lambda\ \ (m=0,1,2,3...)\\where\ 0\leq t\leq 2r\\m = \dfrac{2t}{\lambda}-\dfrac{1}{2}

m_{max} = \dfrac{2\times 2r}{\lambda}-\dfrac{1}{2}\\m_{max} = \dfrac{2\times 2\times 0.026\times 10^{-3}}{559\times 10^{-9}}-\dfrac{1}{2}

 = 221.79  

 = 221 (approx.)

hence no of bright fringe

M = m + 1

   = 221 +1

M = 222 fringes

6 0
3 years ago
After digging deeper and deeper, you would eventually hit solid
Slav-nsk [51]

Answer:

the answer is bedrock

Explanation:

Hope this helps : )

8 0
3 years ago
Read 2 more answers
A snail moves 18 millimeters every 24 seconds what is the snails rate of speed
Nataly_w [17]

The speed of the snail is given by:

v=\frac{S}{t}

where S is the distance covered by the snail and t the time taken.


The snail in the problem moves by S=18 mm in t=24 s, therefore its speed is

v=\frac{S}{t}=\frac{18 mm}{24 s}=0.75 mm/s


3 0
3 years ago
Two long parallel wires 40 cm apart are carrying currents of 10 A and 20 A in the opposite direction. What is the magnitude of t
Alex_Xolod [135]

Answer:

The magnitude of the magnetic field halfway between the wires is 3.0 x 10⁻⁵ T.

Explanation:

Given;

distance half way between the parallel wires, r = ¹/₂ (40 cm) = 20 cm = 0.2 m

current carried in opposite direction, I₁ and I₂ = 10 A and 20 A respectively

The magnitude of the magnetic field halfway between the wires can be calculated as;

B = \frac{\mu _oI_1}{2 \pi r} + \frac{\mu_oI_2}{2\pi r}

where;

B is magnitude of the magnetic field halfway between the wires

I₁ is current in the first wire

I₂ is current the second wire

μ₀ is permeability of free space

r is distance half way between the wires

B = \frac{\mu_o I_1}{2\pi r} + \frac{\mu_o I_2}{2\pi r} \\\\B = \frac{\mu_o }{2\pi r} (I_1 +I_2)\\\\B = \frac{4\pi *10^{-7} }{2\pi *0.2} (10 +20) = 3.0 *10^{-5}\  T

Therefore, the magnitude of the magnetic field halfway between the wires is 3.0 x 10⁻⁵ T.

5 0
3 years ago
What is an example of 'Centre of Gravity'?
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Water bottle and a paper towel role
5 0
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