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maksim [4K]
3 years ago
7

A particle moves along a line so that its position at any time is t>0 or t=0 is given by the function s(t)=(t^3)-(6t^2)+(8t)+

2 where s is measured in meters and t is measured in seconds. a) find instantaneous velocity at any time t. b) find acceleration of the particle at any time t. c) when is the particle at rest?. d) describe motion of the particle. at what values of t does the particle change directions
Physics
1 answer:
Butoxors [25]3 years ago
6 0
A. The instantaneous velocity at any time is  3t^2 -12t +8.
b. The acceleration of the particle at any time is 6t - 12.
c. The acceleration when particle is at rest is 3t^2 -12t+8 = 0.
d. The particle travels like a cubic graph.
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Refraction occurs when __________.
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Refraction occurs when a wave enters a new medium and changes its speed.

i hope this helps 
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3 years ago
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Which of the following would be a valid method to increase the buoyant force acting on an object?
shepuryov [24]
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2 years ago
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Coherent light with wavelength 598 nm passes through two very narrow slits, and the interference pattern is observed on a screen
Stella [2.4K]

Answer:

1.196 μm

Explanation:

D = Screen distance = 3 m

\lambda = Wavelength = 598 m

y = Distance of first-order bright fringe from the center of the central bright fringe = 4.84 mm

d = Slit distance

tan\theta=\frac{y}{D}\\\Rightarrow \theta=tan^{-1}{\frac{y}{D}}\\\Rightarrow \theta=tan^{-1}{\frac{4.84\times 10^{-3}}{3}}\\\Rightarrow \theta=0.09243\ ^{\circ}

sin\theta=\frac{\lambda}{d}\\\Rightarrow d=\frac{\lambda}{sin\theta}\\\Rightarrow d=\frac{598\times 10^{-9}}{sin0.09243}\\\Rightarrow d=0.00037066\ m

For first dark fringe

dsin\theta=\frac{\lambda'}{2}\\\Rightarrow \lambda'=2dsin\theta\\\Rightarrow \lambda'=2\times 0.00037066\times sin0.09243\\\Rightarrow \lambda'=1.196\times 10^{-6}\\\Rightarrow \lambda'=1.196\ \mu m

Wavelength of first-order dark fringe observed at this same point on the screen is 1.196 μm

3 0
2 years ago
Given a 45 45 90 prism with index of 1.5, immersed in air. The hypotenuse acts as the reflecting face by TIR. A ray of light ent
Genrish500 [490]

Answer:

83.6°

Explanation:

For the ray to be totally internally reflected, at the boundary, the angle of refraction is 90. Using the law of refraction where

n₁sinθ₁ = n₂sinθ₂ where n₁ = refractive index of prism = 1.5, θ₁ = critical angle in prism, n₂ = refractive index of air = 1 and θ₂ = refractive angle = 90°.

So, substituting these values into the equation,

n₁sinθ₁ = n₂sinθ₂

1.5 × sinθ₁ = 1 × sin90

1.5 × sinθ₁  = 1

sinθ₁ = 1/1.5

sinθ₁ = 0.6667

θ₁  = sin*(0.6667)

θ₁  = 41.8°

So, for total internal reflection, an incidence angle of 41.8° is required. So, a full convergence angle of 2 × 41.8° = 83.6° is required for the whole bundle of rays.

5 0
3 years ago
What is the magnitude of the sum of the two vectors A = 36 units at 53 degrees, and B =47 units at 157 degrees.
Thepotemich [5.8K]

Answer:

51.82

Explanation:

First of all, let's convert both vectors to cartesian coordinates:

Va = 36 < 53° = (36*cos(53), 36*sin(53))

Va = (21.67, 28.75)

Vb = 47 < 157° = (47*cos(157), 47*sin(157))

Vb = (-43.26, 18.36)

The sum of both vectors will be:

Va+Vb = (-21.59, 47.11)   Now we will calculate the module of this vector:

|Va+Vb| = \sqrt{(-21.59)^2+(47.11)^2}=51.82

4 0
3 years ago
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