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Misha Larkins [42]

3 years ago

13

Violet light (410 nm) and red light (685 nm) pass through a diffraction grating with d = 3.33 x 10-6 m. What is the angular sepa

ration between them for m = 2?

1 answer:

jeyben [28]3 years ago

7 0

Answer:10.03

Explanation:

right on acellus

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A frog jumps vertically upward from a 20m tall building with an initial velocity of 8.1m/s. How high above the ground will the f

topjm [15]

Answer:

Explanation:

Consider the initial position of the frog (20 m above ground) as the reference position. All measurements are positive measured upward.

Therefore,

u = 10 m/s, initial upward velocity.

H = - 20 m, position of the ground.

g = 9.8 m/s², acceleration due to gravity.

Part (a)

When the frog reaches a maximum height of h from the reference position, its velocity is zero. Therefore

u² - 2gh = 0

h = u²/(2g) = 10²/(2*9.8) = 5.102 m

At maximum height, the frog will be 20 + 5.102 = 25.102 m above ground.

Answer: 25.1 m above ground

Part (b)

Let v = the velocity when the frog hits the ground. Then

v² = u² - 2gH

v² = 10² - 2*9.8*(-20) = 492

v = 22.18 m/s

Answer: The frog hits the ground with a velocity of 22.2 m/s

8 0

4 years ago

A mountain slope and a skier make up a closed system with 21,800 J J of energy. When the skier starts her descent down the mount

belka [17]

Answer: 1,270 J

Explanation: APEX

4 0

3 years ago

The earth has a vertical electric field at the surface, pointing down, that averages 100 N/C. This field is maintained by variou

zlopas [31]

Answer:

q = 3.6 10⁵ C

Explanation:

To solve this exercise, let's use one of the consequences of Gauss's law, that all the charge on a body can be considered at its center, therefore we calculate the electric field on the surface of a sphere with the radius of the Earth

r = 6 , 37 106 m

E = k q / r²

q = E r² / k

q = $\frac{100 \ (6.37 \ 10^6)^2}{9 \ 10^9}$

q = 4.5 10⁵ C

Now let's calculate the charge on the planet with E = 222 N / c and radius

r = 0.6 r_ Earth

r = 0.6 6.37 10⁶ = 3.822 10⁶ m

E = k q / r²

q = E r² / k

q = $\frac{222 (3.822 \ 10^6)^2}{ 9 \ 10^9}$

q = 3.6 10⁵ C

4 0

3 years ago

In the figure, a weightlifter's barbell consists of two identical small but dense spherical weights, each of mass 50 kg. These w

kondaur [170]

The moment of inertia is $24.8 kg m^2$

Explanation:

The total moment of inertia of the system is the sum of the moment of inertia of the rod + the moment of inertia of the two balls.

The moment of inertia of the rod about its centre is given by

$I_r = \frac{1}{12}ML^2$

where

M = 24 kg is the mass of the rod

L = 0.96 m is the length of the rod

Substituting,

$I_r = \frac{1}{12}(24)(0.96)^2=1.84 kg m^2$

The moment of inertia of one ball is given by

$I_b = mr^2$

where

m = 50 kg is the mass of the ball

$r=\frac{L}{2}=\frac{0.96}{2}=0.48 m$ is the distance of each ball from the axis of rotation

So we have

$I_b = (50)(0.48)^2=11.5 kg m^2$

Therefore, the total moment of inertia of the system is

$I=I_r + 2I_b = 1.84+ 2(11.5)=24.8 kg m^2$

Learn more about inertia:

brainly.com/question/2286502

brainly.com/question/691705

#LearnwithBrainly

6 0

3 years ago

A pendulum of length L = 1 m is released from an initial angle of 15° . After 1000 s, its amplitude is reduced to 2.5º. What is

jonny [76]

Answer:

The value of the time constant is 558.11 sec.

Explanation:

Given that,

Pendulum length = 1 m

Initial angle = 15°

Time = 1000 s

Reduced amplitude = 2.5°

We need to calculate the value of the time constant

Using formula of damping oscillation

$\theta=\theta_{0}e^{\dfrac{t}{\tau}}$

Where, $\theta$ =amplitude

$\theta_{0}$ =amplitude at t = 0

Put the value into the formula

$2.5=15e^{\dfrac{-1000}{\tau}}$

$\dfrac{1}{6}=e^{\dfrac{-1000}{\tau}}$

$ln\dfrac{1}{6}=\dfrac{-1000}{\tau}$

$\tau=\dfrac{1000}{-ln\dfrac{1}{6}}$

$\tau=558.11\ sec$

Hence, The value of the time constant is 558.11 sec.

6 0

3 years ago