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WINSTONCH [101]

3 years ago

7

If a diffraction grating has 3700 lines per cm, what is the spacing d between lines

1 answer:

Sonja [21]3 years ago

7 0

So first we find the gap between the slits by the formula d=1/N

<span>N is number of lines per metre so 3700 line/cm = 370000 lines/m </span>
<span>So d=2.7*10^-6 </span>

<span>Now we use the formula dsin(angle)=n(wavelength) </span>

<span>d is the same </span>
<span>n is the order of the diffraction pattern </span>

<span>so wavelenth=dsin(angle)/n </span>
<span>=[(2.7*10^-6)*sin30]/3 </span>
<span>=4.5*10^-7 m</span>

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During spring tide, the sun, earth, and moon are in a straight line. This causes ...............

enyata [817]

Answer:

This causes higher average tidal ranges. The gravitational pull of the Sun and moon on Earth combined cause high tides that will be higher and low tides that will be lower than average.

4 0

3 years ago

Zain is standing on a bridge. He tosses a ball upwards, which rises, stops, and then falls all the way to the water (30 meters b

IgorC [24]

Answer:

A. 2.41 s.

B. 24.3 m/s.

Explanation:

vf = vi + 2a*t

where, vf = final velocity

= 0 m/s

vi = final velocity

= 1.5 m/s

a = 9.8 m/s^2

A.

ti = 1.5/(9.8 * 2)

= 0.08 s

Vf^2 = Vi^2 + 2a*s

1.5^2 = 2 * 9.8 * s

S = 0.115 m

Time taken to drop into the water,

30.115 = 1.5*to + 4.9*to^2

to = 2.33 s

Total time taken = ti + to

= 2.33 + 0.08

= 2.41 s.

B.

Vo = 0 m/s

S = 30.115 m

Vf = ?

Using,

Vf^2 = Vo^2 + 2*a*s

= sqrt (2*9.8*30.115)

= 24.3 m/s.

7 0

4 years ago

How to get infinite thanks could you show me please ☺️

murzikaleks [220]

Answer: divide by zero, or square root of a negative

Explanation: If your question is how to get infinity as an answer to a problem, that generally means that the answer is undefined or doesn't exist.

A couple of ways to get that...

You try to divide by zero. In other words, a problem that asks you to perform something like this: 5/0= or 23/(4-4)= Such a problem will give you an error on a calculator because the answer is infinity or doesn't exist.
Another way is to try to get the square root of a negative number. That answer doesn't exist as a real number, so $\sqrt{-4\\}$ will also give you an error on a calculator.

7 0

3 years ago

1. A perspex box has a 10 cm square base and contains water to a height of 10 cm. A piece of rock of mass 600g is lowered into t

AnnZ [28]

Answer:

(a) The volume of water is 100 cm³

(b) The volume of the rock is 20 cm³

(c) The density of the rock is 30 g/cm³

Explanation:

The given parameters of the perspex box are;

The area of the base of the box, A = 10 cm²

The initial level of water in the box, h₁ = 10 cm

The mass of the rock placed in the box, m = 600 g

The final level of water in the box, h₂ = 12 cm

(a) The volume of water in the box, 'V', is given as follows;

V = A × h₁

∴ The volume of water in the box, V = 10 cm² × 10 cm = 100 cm³

The volume of water in the box, V = 100 cm³

(b) When the rock is placed in the box the total volume, $V_T$ , is given by the sum of the rock, $V_r$ , and the water, V, is given as follows;

$V_T$ = $V_r$ + V

$V_T$ = A × h₂

∴ $V_T$ = 10 cm² × 12 cm = 120 cm³

The total volume, $V_T$ = 120 cm³

The volume of the rock, $V_r$ = $V_T$ - V

∴ $V_r$ = 120 cm³ - 100 cm³ = 20 cm³

The volume of the rock, $V_r$ = 20 cm³

(c) The density of the rock, ρ = (Mass of the rock, m)/(The volume of the rock)

∴ The density of the rock, ρ = 600 g/(20 cm³) = 30 g/cm³

8 0

3 years ago

I) A circular coil with radius 20 cm is placed with it's plane parallel and between two straight

Mamont248 [21]

Answer:

$B=15.433\ T$ inwards when viewing from the left side.

Explanation:

Given:

radius of the circular loop, $r=0.2\ m$

current in the coil, $I_c=0.5\ A$

Direction of current is clockwise when viewed from left.

Distance of wire P from the loop, $d_p=0.4\ m$

Distance of wire Q from the loop, $d_q=0.8\ m$

current in each wires P & Q, $I=0.2\ A$

Now the magnetic field in coil will be inwards when viewed from left by the Maxwell's right hand thumb rule.

<u>Magnitude is given by:</u>

$B_c=\frac{\mu_0.I_c}{2R}$

$B_c=\frac{4\pi\times 10^{-7}\times (0.5)}{2\times 0.2}$

$B_c=15.7\times 10^{-7}\ T$

<u>Now the effect of magnetic field due to wire P at the center of the loop:</u>

(We get the effective distance as 0.4+0.2=0.6 m)

$B_P=\frac{\mu_0.I}{2\pi.d_p}$

$B_P=\frac{4\pi\times 10^{-7}\times (0.2)}{2\pi\times 0.6}$

$B_P=6.67\times 10^{-8}\ T$ coming out of the loop when viewed from left.

<u>Now the effect of magnetic field due to wire Q at the center of the loop:</u>

(We get the effective distance as 0.8+0.2=1 m)

$B_P=\frac{\mu_0.I}{2\pi.d_q}$

$B_P=\frac{4\pi\times 10^{-7}\times (0.2)}{2\pi\times 1}$

$B_P=4\times 10^{-8}\ T$ going in to the loop when viewed from left.

<u>Now the net resultant effect all the magnetic fields:</u>

$B=B_c-B_P+B_Q$

$B=15.7-0.667+0.4$

$B=15.433\ T$ inwards when viewing from the left side.

7 0

3 years ago