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zlopas [31]
4 years ago
14

WILL MARK BRAINLIEST!!!

Physics
1 answer:
Neporo4naja [7]4 years ago
4 0
The floating of ice helps keep the ocean warm because it keeps the ice at the top of the ocean in certain places. But it also helps aquatic life keep in there Necessary environmental conditions.
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Perfect Elastic Collisions
ira [324]
What is the quation?
7 0
3 years ago
The distance between the ruled lines on a diffraction grating is 1900 nm. The grating is illuminated at normal incidence with a
SashulF [63]

Answer:

3.28 degree

Explanation:

We are given that

Distance between the ruled lines on a diffraction grating, d=1900nm=1900\times 10^{-9}m

Where 1nm=10^{-9} m

\lambda_2=400nm=400\times10^{-9}m

\lambda_1=700nm=700\times 10^{-9}m

We have to find  the angular width of the gap between the first order spectrum and the second order spectrum.

We know that

\theta=sin^{-1}(\frac{m\lambda}{d})

Using the formula

m=1

\theta_1=sin^{-1}(\frac{1\times700\times 10^{-9}}{1900\times 10^{-9}})

\theta=21.62^{\circ}

Now, m=2

\theta_2=sin^{-1}(\frac{2\times400\times 10^{-9}}{1900\times 10^{-9}})

\theta_2=24.90^{\circ}

\Delta \theta=\theta_2-\theta_1

\Delta \theta=24.90-21.62

\Delta \theta=3.28^{\circ}

Hence, the angular width of the gap between the first order spectrum and the second order spectrum=3.28 degree

6 0
3 years ago
For a given velocity of projection in a projectile motion, the maximum horizontal distance is possible only at ө = 45°. Substant
myrzilka [38]

First let us imagine the projectile launched at initial velocity V and at angle θ relative to the horizontal. (ignore wind resistance)

Vertical component y:

The initial vertical velocity is given as Vsinθ
The moment the projectile reaches the maximum height of h, the vertical velocity will be 0, therefore the time t taken to attain this maximum height is:

h = Vsinθ - gt
0 = Vsinθ - gt
t = (Vsinθ)/g

where g is  acceleration due to gravity

Horizontal component x:
The initial horizontal velocity is given as Vcosθ. However unlike the vertical component, this horizontal velocity remains constant because this is unaffected by gravity. The time to travel the horizontal distance D is twice the value of t times the horizontal velocity.
D = Vcosθ*[(2Vsinθ)/g] 
D = (2V²sinθ cosθ)/g 
 D = (V²sin2θ)/g

In order for D (horizontal distance) to be maximum, dD/dθ = 0
That is,

2V^2 cos2θ / g = 0
And since 2V^2/g must not be equal to zero, therefore cos(2θ) = 0
This is true when 2θ = π/2  or  θ = π/4


Therefore it is now<span> shown that the maximum horizontal travelled is attained when the launch angle is π/4 radians, or 45°.</span>

6 0
3 years ago
The lightest car in the world was built in London and had a mass of less than 10 kg. it's maximum speed was 25.0 km/h. Suppose t
Softa [21]
25km/h = 6.94 m/s
suvat
s=16
u=6.94
v=0
a=a
v^2=u^2+2as
(v^2-u^2)/2s = a =1.5ms^-2
6 0
3 years ago
An electron is pushed into an electric field where it acquires a 1-v electrical potential. suppose instead that two electrons ar
Lapatulllka [165]
1V electrical potential 
4 0
4 years ago
Read 2 more answers
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