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erastovalidia [21]
4 years ago
15

The distance between the 1st and 2nd maxima is 0.3 cm when the screen is located 2 m away. If d = 0.4 mm, a) what is the wavelen

gth?
b) what is the size of central peak?
Physics
1 answer:
Marrrta [24]4 years ago
7 0

Answer:

a) The wavelength of light is 6×10^-7 m.

b) The size of the central peak is 6×10^-3 m.

Explanation:

let:

α be the distance between the 1st and 2nd maxima

d = 0.4 mm be the grating slit space

X be the distance to the screen

a) α = λ×X/d

   λ = α×d/X

      = (0.3×10^-2)×(0.4×10^-3)/(2)

      = 6×10^-7 m

Therefore, the wavelength of light is 6×10^-7 m.

b) the size of the central peak is given by:

 D = 2×λ×X/d

     = 2×(6×10^-7)×(2)/(0.4×10^-3)

     = 6×10^-3 m

Therefore, the size of the central peak is 6×10^-3 m.

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Elina [12.6K]

Answer: F = 1235 N

Explanation: Newton's Second Law of Motion describes the effect of mass and net force upon acceleration: F_{net}=m.a

Acceleration is the change of velocity in a period of time: a=\frac{\Delta v}{\Delta t}

Velocity of the car is in km/h. Transforming it in m/s:

v=\frac{47.10^{3}}{36.10^{2}}

v = 13 m/s

At the moment the car decelerates, acceleration is

a=\frac{13}{0.2}

a = 65 m/s²

Then, force will be

F_{net}=19(65)

F_{net} = 1235 N

The horizontal net force the straps of the restraint chair exerted on the child to hold her is 1235 newtons.

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3 years ago
How many laws does newton have?
liq [111]
Newton has 3 Laws specifically The Three Laws of Motion
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4 years ago
WILL MARK BRAINLIST. HELP ASAP
Pachacha [2.7K]

Answer:

Less than the distance

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3 years ago
A block with m = 300 grams oscillates at the end of a linear spring with k = 6.5 N/m. (assume this is a top-down view and the bl
inna [77]

Answer:

a) T=1.35s

b) amplitude = 0.0923m

Explanation:

m=300 gr

k=6.5 N/m

first we need to get the angular frequency of the motion

so we have that

ω = √(k/m)

in this case motion is a simple harmonic so the period is defined by:

T= 2π / ω

T= 2π / √(k/m)

replacing the variables...

T= 2π / √(6.5/0.3)

T=1.35s   (period of the block's motion)

and...

α max = | ω²r max |

2 = (2π/1.35)² * r max

r max= 0.0923m

6 0
3 years ago
The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of g
Morgarella [4.7K]

Here is the full question:

The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given by:  

k=\sqrt{\frac{I}{M} }

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.

Answer:

a) 0.85 m

b) 0.98 m

c) 0.76 m

Explanation:

Given that: the radius of gyration  k=\sqrt{\frac{I}{M} }

So, moment of rotational inertia (I) of a cylinder about it axis = \frac{MR^2}{2}

k=\sqrt{\frac{\frac{MR^2}{2}}{M} }

k=\sqrt{{\frac{MR^2}{2}}* \frac{1}{M} }

k=\sqrt{{\frac{R^2}{2}}

k={\frac{R}{\sqrt{2}}

k={\frac{1.20m}{\sqrt{2}}

k = 0.8455 m

k ≅ 0.85 m

For the spherical shell of radius

(I) = \frac{2}{3}MR^2

k = \sqrt{\frac{\frac{2}{3}MR^2}{M}  }

k = \sqrt{\frac{2}{3} R^2}

k = \sqrt{\frac{2}{3} }*R

k = \sqrt{\frac{2}{3}}  *1.20

k = 0.9797 m

k ≅ 0.98 m

For the solid sphere of  radius

(I) = \frac{2}{5}MR^2

k = \sqrt{\frac{\frac{2}{5}MR^2}{M}  }

k = \sqrt{\frac{2}{5} R^2}

k = \sqrt{\frac{2}{5} }*R

k = \sqrt{\frac{2}{5}}  *1.20

k = 0.7560

k ≅ 0.76 m

6 0
4 years ago
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