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Artyom0805 [142]
3 years ago
7

A beaker of vegetable oil contains a beam of light that is aimed at a surface at an angle of 34 degrees as shown. If the index o

ff refraction of the oil is 1.47, what is the angle of reflection?
Physics
1 answer:
Over [174]3 years ago
4 0

Answer:

Angle of reflection of light is 34 degree

Explanation:

As per law of reflection of light we know that

angle of incidence of light = angle of reflection of light

So here we know that

angle of incidence on the surface of oil is given as

\theta_i = 34 degree

so we know that

\theta_i = \theta_r

so here we can say that reflection angle of light will be same as angle of incidence

\theta_r = 34 degree

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Two blocks are placed at the ends of a horizontal massless board, as in the drawing. The board is kept from rotating and rests o
Andrew [12]

Answer:

The magnitude of the angular acceleration ∝ = \frac{rxF}{2.8[tex]r^{2}}[/tex]

Explanation:

The angular acceleration ∝ is equal to the torque (radius multiplied by force) divided by the mass times the square of the radius. The magnitude of angular acceleration ∝ will have the equation above but we have to replace the mass in the equation by 2.8kg as stated.

7 0
3 years ago
Pls I need help for the problem solving part.
Bogdan [553]

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3 0
3 years ago
A block of mass 0.510 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x. T
maria [59]

Answer:

x=0.46m, speed=7.9m/s

Explanation:

Using the concept of conservation of energy:

1. kinetic energy of mass m and velocity v: E_k=\frac{1}{2}mv^2

2. gravitational potential energy of mass m, grav. acc. g and height h: E_g=mgh

3. potential energy in a spring with spring constant k and displacement from equilibrium x: E_s=\frac{1}{2}kx^2

Calculating x:

\frac{1}{2}mv_a^2=\frac{1}{2}kx^2

x=\sqrt{\frac{m}{k}}v_a

Calculating the speed:

\frac{1}{2}mv_a^2 +mgh_a=\frac{1}{2}mv_b^2+mgh_b + W_{friction}

h_a=0, h_b=2R,W_{friction}=F_{friction}\times distance=7\pi R

\frac{1}{2}mv_a^2=\frac{1}{2}mv_b^2+2mgR+7\pi R

Solving for v_b:

v_b=\sqrt{v_a^2-4gR-14\pi\frac{R}{m}}

7 0
3 years ago
As more lamps are put into a series circuit, the overall current in the circuit a. Increasesb. Decreasesc. Remains the same
erik [133]

Answer:

b. Decreases

Explanation:

The total resistance of a series circuit is equal to the sum of the individual resistances:

R_T=R_1+R_2+...+R_n (1)

Therefore, as we add more lamps, the total resistance increases (because we add more positive tems in the sum in eq.(1).

The current in a circuit is given by Ohm's law:

I=\frac{V}{R_T}

where V is the voltage provided by the power source and R_T is the total resistance. We notice that the current, I, is inversely proportional to the total resistance: therefore, when more lamps are added to the series circuit, the total resistance increases, and therefore the current in the circuit decreases.

8 0
3 years ago
ANSWER One end of a string is attached to an object of mass M, and the other end of the string is secured so that the object is
qaws [65]

Answer:

Explanation:

It is a case of oscillation by simple pendulum . Expression for simple pendulum is given as follows

T = 2\pi\sqrt{\frac{l}{g} }

where T is time period , l is length of pendulum and g is acceleration due to gravity .

\frac{1}{f} =2\pi\sqrt{\frac{l}{g} }  , f is frequency of oscillation

For the given case

\frac{1}{f_o} =2\pi\sqrt{\frac{l}{g} }

subsequently length becomes half so

\frac{1}{f} =2\pi\sqrt{\frac{l}{2g} }

dividing

\frac{f}{f_o} = \sqrt{\frac{2}{1} }

f = \sqrt{2} f_o

frequency of oscillation becomes √2 times.

4 0
3 years ago
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