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3 years ago

Why are images reflected from a rough surface not as clear as those reflected from a smooth

Physics

1 answer:

ioda3 years ago

7 0

Answer:

Just as images are reflected from the surface of a mirror, light reflected from a smooth water surface also produced a clear image. ... Consequently, the outgoing rays are reflected at many different angles and the image is disrupted. Reflection from such a rough surface is called diffuse reflection and appears matte.

Explanation:

hi po I hope it's help you

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How does temperature,barometric pressure,humidity,wind speed and direction, and precipitation determine the weather in a particu

Deffense [45]

Temperature is how hot or cold something is,barometric is air pressure(when the pressures high,the weather is dry)humidity is how moist the air is and how many water particles are there, wind speed and direction is how the hot and cold air is moving and how fast,which makes wind (high to low, hot to cold) precipitation is how much it is raining at that point.

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3 years ago

It took a crew 3 h 20 min to row 5 km upstream and back again. if the rate of flow of the stream was 2 km/h, what was the rowing

Basile [38]

Suppose rowing speed is V km/s.

So, upstream speed = V-2

Downstream speed = V+ 2

Total time is taken 3 h 20 min to cover up and down journey of 5 km.

So, total time = upstream time + downstream time

$3. 3 \ hours = \frac{5 km}{ V- 2} + \frac{5 km}{V+2} \\\\\ 3.3 (V^2-2^2) = 10V\\\\V^2- 4-3V= 0$

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Thus, the the rowing speed of the crew in still water is 1 km/s.

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3 years ago

How many normal modes of oscillation or natural frequencies does each if the following have: (

Vadim26 [7]

<span>Each of these systems has exactly one degree of freedom and hence only one natural frequency obtained by solving the differential equation describing the respective motions. For the case of the simple pendulum of length L the governing differential equation is d^2x/dt^2 = - gx/L with the natural frequency f = 1/(2π) √(g/L). For the mass-spring system the governing differential equation is m d^2x/dt^2 = - kx (k is the spring constant) with the natural frequency ω = √(k/m). Note that the normal modes are also called resonant modes; the Wikipedia article below solves the problem for a system of two masses and two springs to obtain two normal modes of oscillation.</span>

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4 years ago

Answer the questions to help you understand your parachute and forces experiment. Use the data table below to record your data.

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Ohh so hexagon is. Tool use the data

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3 years ago

A block of mass m = 2.20 kg slides down a 30.0° incline which is 3.60 m high. At the bottom, it strikes a block of mass M = 6.80

Bogdan [553]

Answer:

(a) The speed of the lighter block is $v_{2x} = 3.7~m/s$ .

The speed of the heavier block is $v_{3x} = 3.55~m/s$ .

(b) The smaller block goes up to 0.69 m.

Explanation:

We will divide this question into three parts: Part A is for the smaller mass from the top of the incline to the collision. Part B is the collision. And Part C is for the smaller mass from the bottom to the highest point it can achieve.

In order to solve this question, I will assume that the smaller mass is initially at rest.

Part A:

We will use conservation of energy.

$K_1 + U_1 = K_2 + U_2\\0 + mgh = \frac{1}{2}mv_1^2 + 0\\(2.2)(9.8)(3.6) = \frac{1}{2}(2.2)v_1^2\\v_1 = 8.4 ~m/s$

This is the speed of the smaller mass just before the collision. The velocity of the mass is directed 30° above horizontal, since the mass is sliding down the incline.

Part B:

Momentum is a vector identity, so the x- and y-components of momentum are to be investigated separately. Since the collision occurs at the horizontal surface, only the x-component of momentum is conserved.

$P_1 = P_2\\mv_{1x} = mv_{2x} + Mv_{3x}\\(2.2)(8.4\cos(30^\circ)) = (2.2)v_{2x} + (6.8)v_{3x}\\16 = 2.2v_{2x} + 6.8v_{3x}$

During the collision kinetic energy is also conserved. Since kinetic energy is a scalar quantity, we don't have to separate its components.

$K_{initial} = K_{final}\\\frac{1}{2}mv_1^2 = \frac{1}{2}mv_{2}^2 + \frac{1}{2}Mv_{3x}^2\\(2.2)(8.4)^2 = (2.2)v_{2}^2 + (6.8)v_{3x}^2\\155.23 = 2.2v_{2}^2 + 6.8v_{3x}^2$