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Juli2301 [7.4K]
3 years ago
15

In a car, 75 percent of the chemical energy of gasoline is lost as thermal

Physics
1 answer:
LenaWriter [7]3 years ago
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The spreading of waves behind an aperture is more for long wavelengths and less for short wavelengths.Less for long wavelengths
Allisa [31]

Answer:

Increase in wavelength of incident wave also increases the spread angle or spread of the interference pattern.

Explanation:

Solution:-

- The diffraction occurs when light bends in the same medium. The bending is the result of light waves "squeezing" through small openings or "curving" around sharp edges.

- Moreover, waves diffract best when the size of the diffraction opening (or grting or groove) corresponds to the size of the wavelength. Hence, light diffracts more through small openings than through larger openings.

- The formula for diffraction shows a direct relationship between the angle of diffraction (theta) and wavelength:

                                         d sin (θ) = m λ

Where,

     λ : Wavelength , θ : The spread angle , d : Slit opening or grating

- We can see that the wavelength λ and spread angle θ are related proportionally. So if we increase the wavelength of incident wave we also increase the spread angle or spread of the interference pattern.

5 0
3 years ago
Help ASAP
EastWind [94]

Answer:

Elastic potential energy, E=2.35\times 10^{-8}\ J

Explanation:

Charge, q=9.4\times 10^{-10}\ C

Potential, V = 50 V

It is required to find the electric potential energy in a capacitor stored in it. The formula of the electric potential energy in a capacitor is given by :

E=\dfrac{1}{2}qV\\\\E=\dfrac{1}{2}\times 9.4\times 10^{-10}\times 50\\\\E=2.35\times 10^{-8}\ J

So, the electric potential energy stored in the capacitor is 2.35\times 10^{-8}\ J

8 0
3 years ago
If two waves pass a point every second what is the frequency of the waves
marishachu [46]
[two waves] pass a point [every second]... The answer is in the question (B)
7 0
3 years ago
An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of th
jeyben [28]

Answer:

the shooting angle ia 18.4º

Explanation:

For resolution of this exercise we use projectile launch expressions, let's see the scope

      R = Vo² sin (2θ) / g

      sin 2θ = g R / Vo²

      sin 2θ = 9.8 75/35²

      2θ = sin⁻¹ (0.6)

      θ = 18.4º

To know how for the arrow the tree branch we calculate the height of the arrow at this point

       X2 = 75/2 = 37.5 m

We calculate the time to reach this point since the speed is constant on the X axis

       X = Vox t

       t2 = X2 / Vox = X2 / (Vo cosθ)

        t2 = 37.5 / (35 cos 18.4)

        t2 = 1.13 s

With this time we calculate the height at this point

        Y = Voy t - ½ g t²

        Y = 35 sin 18.4   1.13 - ½ 9.8 1,13²

        Y = 6.23 m

With the height of the branch is 3.5 m and the arrow passes to 6.23, it passes over the branch

8 0
3 years ago
Read 2 more answers
B. Jerome plays middle linebacker for South's varsity football team. In a game against
weqwewe [10]

Answer:

Option D

670 Kg.m/s

Explanation:

Initial momentum is given by mv=82*5.6=459.2 Kg.m/s (taking eastward as positive)

Final momentum is also mv but v being westward direction, we take it negative

Final momentum=82*-2.5= -205 Kg.m/s

Change in momentum=Final momentum-Initial momentum=-205-459.2=-664.2 Kg.m/s

Impulse=change in momentum=664.2 Kg.m/s rounded off as 670 Kg.m/s

4 0
3 years ago
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