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Andreyy89
3 years ago
9

A wave on the ocean has a wavelength of 200 m, the wave is a deep-water wave __________. only if the water is deeper than 200 m

if the water is deeper than 100 m if the water is less than 100 m deep if the water is shallower than 50 m
Physics
1 answer:
klasskru [66]3 years ago
3 0
The answer is to this question is 250 because I'm very smarr
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The corona
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If the distance between slits on a diffraction grating is 0.50 mm and one of the angles of diffraction is 0.25°, how large is th
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Answer:

-  path differnce = 2.18*10^-6

-  1538 lines

Explanation:

- The path difference for the waves that produce the pattern of diffraction, is given by the following formula:

path\ difference\ = dsin\theta           (1)

d: separation between slits = 0.50mm = 0.50*10^-3 m

θ: angle of a diffraction = 0.25°

Then, the path difference is:

path\ difference\ =(0.50*10^{-3}m)sin(0.25\°)=2.18*10^{-6}m

- The maximum number of bright lines are calculated by using the following formula:

m\lambda = dsin\theta           (2)

m: order of the bright

λ: wavelength = 650nm

The maximum bright is calculated for an angle of 90°:

m=\frac{(0.50*10^{-3}m)sin90\°}{650*10^{-9}m} \approx 769

The maxium number of bright lines are twice the previous result, that is, 1538 lines

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The efficiency of conversion of glucose to ATP during glycolysis is approximately:
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2 years ago
A projectile is launched from ground level with an initial speed of 47 m/s at an angle of 0.6 radians above the horizontal. It s
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Answer:

30.67m

Explanation:

Using one of the equations of motion as follows, we can describe the path of the projectile in its horizontal or vertical displacement;

s = ut ± \frac{1}{2} at^2               ------------(i)

Where;

s = horizontal/vertical displacement

u = initial horizontal/vertical component of the velocity

a = acceleration of the projectile

t = time taken for the projectile to reach a certain horizontal or vertical position.

Since the question requires that we find the vertical distance from where the projectile was launched to where it hit the target, equation (i) can be made more specific as follows;

h = vt ± \frac{1}{2} at^2               ------------(ii)

Where;

h = vertical displacement

v = initial vertical component of the velocity = usinθ

a = acceleration due to gravity (since vertical motion is considered)

t = time taken for the projectile to hit the target

<em>From the question;</em>

u = 47m/s, θ = 0.6rads

=> usinθ = 47 sin 0.6

=> usinθ = 47 x 0.5646 = 26.54m/s

t = 1.7s

Take a = -g = -10.0m/s   (since motion is upwards against gravity)

Substitute these values into equation (ii) as follows;

h = vt - \frac{1}{2} at^2

h = 26.54(1.7) - \frac{1}{2} (10)(1.7)^2

h = 45.118 - 14.45

h = 30.67m

Therefore, the vertical distance is 30.67m        

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