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4 years ago

15

A diffraction grating with 210 lines per mm is used in an experiment to study the visible spectrum of a gas discharge tube. At w

hat angle from the beam axis will the first order peak occur if the tube emits light with wavelength of 506.0 nm

1 answer:

11Alexandr11 [23.1K]4 years ago

8 0

Answer:

Explanation:

Width of slit a = 1 x 10⁻³ / 210

a = 4.76 x 10⁻⁶ m .

angle at which first order peak is formed

= λ / a (where λ is wavelength and a is width of slit)

given λ = 506 x 10⁻⁹ m

a = 4.76 x 10⁻⁶

θ = 506 x 10⁻⁹ / 4.76 x 10⁻⁶

= 106.3 x 10⁻³ radian .

first order peak is formed at an angle of 106.3 x 10⁻³ radian .

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3 years ago

A uniform sphere (I = 2/5 MR 2 ) rolls down an incline. (a) What must be the incline angle if the linear acceleration of the cen

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Answer:

Part a)

$\theta = 8.05 degree$

Part b)

$a = 1.37 m/s^2$

Explanation:

As the uniform sphere is rolling down the inclined plane then the net force on the sphere is given as

$mg sin\theta - F_f = ma$

also we have torque equation on it

$F_f R = I\alpha$

for pure rolling

$a = R \alpha$

$F_f = \frac{Ia}{R^2}$

now we have

$mg sin\theta = ma + \frac{Ia}{R^2}$

now we have

$mg sin\theta = (m + \frac{2}{5}m)a$

$a = \frac{5}{7}g sin\theta$

now given that

$a = 0.10 g$

so we have

$0.10 g = \frac{5}{7} g sin\theta$

$sin\theta = 0.14$

$\theta = 8.05 degree$

Part b)

If the inclined plane is frictionless then the acceleration is given as

$a = g sin\theta$

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$a = 1.37 m/s^2$

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4 years ago

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Answer:

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3 years ago

What distance will a car traveling at a speed of 15 m/s cover in 300 s?

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Distance formula: d = st

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3 years ago

2. A 2000 kg car with speed 12.0 m/s hits a tree. The tree does not move or

krek1111 [17]

a) The work done by the tree is $-1.44\cdot 10^5 J$

b) The amount of force applied is 2880 N

Explanation:

a)

According to the work-energy theorem, the work done on the car is equal to the change in kinetic energy of the car. Therefore, we can write:

$W=K_f - K_i = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

where

W is the work done on the car

m is the mass of the car

u is its initial speed

v is its final speed

For the car in this problem, we have:

m = 2000 kg

u = 12.0 m/s

v = 0 (since the car comes to a stop, after the crash)

Therefore, the work done by the tree on the car is:

$W=0-\frac{1}{2}(2000)(12.0)^2=-1.44\cdot 10^5 J$

The work is negative because it is done in the direction opposite to the direction of motion of the car.

b)

The work done by the tree on the car can also be rewritten as

$W=Fd$

where

F is the force applied on the car

d is the displacement of the car during the collision

In this situation, we have:

$W=-1.44\cdot 10^5 J$ is the work done

$d=50.0 cm = 0.50 m$ is the displacement of the car during the collision

Solving the equation for F, we find the force exerted by the tree on the car:

$F=\frac{W}{d}=\frac{-1.44\cdot 10^5 J}{0.50}=-2880 N$

Where the negative sign means the force is applied opposite to the direction of motion of the car. Therefore, the magnitude of the force applied is 2880 N.

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

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3 years ago