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ozzi
3 years ago
6

A rifle fires a bullet at a velocity of 78 m/s, 40 degrees above the horizontal. Determine the height (h) above the starting pos

ition of the bullet 6 s after it has been fired.
Physics
1 answer:
qwelly [4]3 years ago
7 0

Answer:

H = Vy t - 1/2 g t^2  height of an object with an initial "vertical" velocity

                                 at t sec after firing

Vy = 78 m/s * sin 40 = .643 * 78 m/s = 50.1 m/s

H = 50.1 * 6 - 1/2 * 9.8 * 6^2 = 300 m - 176 m = 124 m

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Coherent light that contains two wavelengths, 660 nm and 470 nm , passes through two narrow slits with a separation of 0.280 mm
bekas [8.4K]

Answer:

λ1 = 0.0129m = 1.29cm

λ2 = 0.00923m = 0.92 cm

Explanation:

To find the distance between the first order bright fringe and the central peak, can be calculated by using the following formula:

y_m=\frac{m\lambda D}{d}    (1)

m: order of the bright fringe = 1

λ: wavelength of the light = 660 nm, 470 nm

D: distance from the screen = 5.50 m

d: distance between slits = 0.280mm = 0.280 *10^⁻3 m

ym: height of the m-th fringe

You replace the values of the variables in the equation (1) for each wavelength:

For λ = 660 nm = 660*10^-9 m

y_1=\frac{(1)(660*10^{-9}m)(5.50m)}{0.280*10^{-3}m}=0.0129m=1.29cm

For λ = 470 nm = 470*10^-9 m

y_1=\frac{(1)(470*10^{-9}m)(5.50m)}{0.280*10^{-3}m}=0.00923m=0.92cm

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3 years ago
A system absorbs 194 kj of heat and the surroundings do 120 kj of work on the system. internal eneergy change
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We can solve the problem by using the first law of thermodynamics, which states that:
\Delta U = Q-W
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In our problem, the heat absorbed by the system is Q=+194 kJ, while the work done is W=-120 kJ, where the negative sign means the work is done by the surroundings on the system. Therefore, the variation of internal energy is
\Delta U= Q-W=+194 kJ - (-120 kJ)=+314 kJ
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I think the answer is b
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2 years ago
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Answer:

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If you multiply the acceleration with time you get the average speed.

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2 years ago
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