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GarryVolchara [31]
3 years ago
6

Reflection off of a smooth surface like a mirror is an example of diffuse reflection.

Physics
2 answers:
Akimi4 [234]3 years ago
8 0

Answer:

true

Explanation:

Reflection is when light bounces off an object. If the surface is smooth and shiny, like glass, water or polished metal, the light will reflect at the same angle as it hit the surface. ... This is called diffuse reflection. Diffuse reflection is when light hits an object and reflects in lots of different directions.

Varvara68 [4.7K]3 years ago
7 0

Answer:

false

its regular reflection.

hope it helps.

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A soccer player pumps air into a soccer ball until no more air can be pushed inside. Describe the air inside the soccer ball com
inessss [21]

Answer:

the filling stops when the pressure of the pump equals the pressure of the interior air plus the pressure of the walls.

Explanation:

This exercise asks to describe the inflation situation of a spherical fultball.

Initially the balloon is deflated, therefore the internal pressure is equal to the pressure of the air outside, atmospheric pressure, when it begins to inflate the balloon with a pump this creates a pressure in the inlet valve and as it is greater than the pressure inside, the air enters it, this is repeated in each filling cycle, manual pump.

When the ball is full we have two forces, the one created by the external walls and the one aired by the pressure of the pump, these forces are directed towards the inside, but the air molecules exert a pressure towards the outside, which translates into a force. When these two forces are equal, the pump is no longer able to continue introducing air into the balloon.

Consequently the filling stops when the pressure of the pump equals the pressure of the interior air plus the pressure of the walls.

4 0
3 years ago
An electron is confined in a harmonic oscillator potential well. What is the longest wavelength of light that the electron can a
kipiarov [429]

Answer:

The longest wavelength of light is 209 nm.

Explanation:

Given that,

Spring constant = 74 N/m

Mass of electron m= 9.11\times10^{-31}\ kg

Speed of light c= 3\times10^{8}\ m/s

We need to calculate the frequency

Using formula of frequency

f =\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

Where, k= spring constant

m = mass of the particle

Put the value into the formula

f=\dfrac{1}{2\pi}\sqrt{\dfrac{74}{9.11\times10^{-31}}}

f=1.434\times10^{15}\ Hz

We need to calculate the longest wavelength that the electron  can absorb

\lambda=\dfrac{c}{f}

Where, c = speed of light

f = frequency

Put the value into the formula

\lambda =\dfrac{3\times10^{8}}{1.434\times10^{15}}

\lambda=2.092\times10^{-7}\ m

\lambda=209\ nm

Hence, The longest wavelength of light is 209 nm.

6 0
3 years ago
A car is stopped at a traffic light. When the light turns green at t=0, a truck with a constant speed passes the car with a 20m/
s344n2d4d5 [400]

Answer:

At t = (70 / 3) \; {\rm s} (approximately 23.3 \; {\rm s}.)

Explanation:

Note that the acceleration of the car between t = 0\; {\rm s} and t = 20\; {\rm s} (\Delta t = 20\; {\rm s}) is constant. Initial velocity of the car was v_{0} = 0\; {\rm m\cdot s^{-1}}, whereas v_{1} = 35\; {\rm m\cdot s^{-1}} at t = 20\; {\rm s}\!. Hence, at t = 20\; {\rm s}\!\!, this car would have travelled a distance of:

\begin{aligned}x &= \frac{(v_{1} - v_{0})\, \Delta t}{2} \\ &= \frac{(35\; {\rm m\cdot s^{-1}} - 0\; {\rm m\cdot s^{-1}}) \times (20\; {\rm s})}{2} \\ &= 350\; {\rm m}\end{aligned}.

At t = 20\; {\rm s}, the truck would have travelled a distance of x = v\, t = 20\; {\rm m\cdot s^{-1}} \times 20\; {\rm s} = 400\; {\rm m}.

In other words, at t = 20\; {\rm s}, the truck was 400\; {\rm m} - 350\; {\rm m} = 50\; {\rm m} ahead of the car. The velocity of the car is greater than that of the truck by 35\; {\rm m\cdot s^{-1}} - 20\; {\rm m\cdot s^{-1}} = 15 \; {\rm m\cdot s^{-1}}. It would take another (50\; {\rm m}) / (15\; {\rm m\cdot s^{-1}}) = (10/3)\; {\rm s} before the car catches up with the truck.

Hence, the car would catch up with the truck at t = (20 + (10/3))\; {\rm s} = (70 / 3)\; {\rm s}.

3 0
1 year ago
Hey everyone<br><br><br>what is optics??​
lana66690 [7]

Explanation:

Optics is a branch of physics that is the study of light and vision. ... The branch of physics dealing with the nature and properties of electromagnetic energy in the light spectrum and the phenomena of vision. In the broadest sense, optics deals with infrared light, visible light, and ultraviolet light.

7 0
3 years ago
A 36.0 kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130
konstantin123 [22]

Answer:

(a) W = 650J

(b) Wf = 529.2J

(c) W = 0J

(d) W = 0J

(e) ΔK = 120.8J

(f) v2 = 2.58 m/s

Explanation:

(a) In order to find the work done by the applied force you use the following formula:

W=Fd      (1)

F: applied force = 130N

d: distance = 5.0m

W=(130N)(5.0m)=650J

The work done by the applied force is 650J

(b) The increase in the internal energy of the box-floor system is given by the work done of the friction force, which is calculated as follow:

W_f=F_fd=\mu Mgd       (2)

μ: coefficient of friction = 0.300

M: mass of the box = 36.0kg

g: gravitational constant = 9.8 m/s^2

W_f=(0.300)(36.0kg)(9.8m/s^2)(5.0m)=529.2J

The increase in the internal energy is 529.2J

(c) The normal force does not make work on the box because the normal force is perpendicular to the motion of the box.

W = 0J

(d) The same for the work done by the normal force. The work done by the gravitational force is zero because the motion of the box is perpendicular o the direction of the gravitational force.

(e) The change in the kinetic energy is given by the net work on the box. You use the following formula:

\Delta K=W_T         (3)

You calculate the total work:

W_T=Fd-F_fd=(F-F_f)d     (4)

F: applied force = 130N

Ff: friction force

d: distance = 5.00m

The friction force is:

F_f=(0.300)(36.0kg)(9.8m/s^2)=105.84N

Next, you replace the values of all parameters in the equation (4):

W_T=(130N-105.84N)(5.00m)=120.80J

\Delta K=120.80J

The change in the kinetic energy of the box is 120.8J

(e) The final speed of the box is calculated by using the equation (3):

W_T=\frac{1}{2}M(v_2^2-v_1^2)       (5)

v2: final speed of the box

v1: initial speed of the box = 0 m/s

You solve the equation (5) for v2:

v_2 = \sqrt{\frac{2W_T}{M}}=\sqrt{\frac{2(120.8J)}{36.0kg}}=2.58\frac{m}{s}

The final speed of the box is 2.58m/s

5 0
3 years ago
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