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3 years ago

PLEASE HELP! TRUE OR FALSE!

Physics

2 answers:

SpyIntel [72]3 years ago

8 0

Answer:

<h3>About the firs question.</h3>

The first statement is completely true. An image is attached. You can observe than a concave mirror like it's shown, the focal points is on the optic axis, and all rays converge in that point.

<h3>About the second question.</h3>

<em>A dispersive medium refers to mediums where waves of different frequencies travel at different velocities. </em>In addition, the relation between frequency and degree of refraction of the medium is that waves with lower frequency will bend less when passing through the dispersive medium. For example, if we pass a red light through a medium, and then we pass a violet light, we'll see the difference in the angle of refraction, the violet light will bend more than the red one.

Therefore, the answer to the second question is "<em>dispersive media waves of high frequency refract more than waves of low frequency do".</em>

<em />

<h3>About the third question.</h3>

The reason why Roemer's measurement of the speed of light was inaccurate is:<em> "He did not know the accurate diameter of the Earth's orbit".</em>

At that time, during 1668-1674, the diameter of Earth's orbit wasn't accurate, which added error of delay to his Roemer's measurement. However, he was the first scientist in deduct a finite speed of light, before him, it was known that the speed of light was infinite.

<h3>About the fourth question.</h3>

For convex lenses we have the relation between distances and highnesses:

$\frac{d_{o} }{d_{i} }=\frac{h_{o}}{h_{i}}$

We have:

$f=10.0 \ cm$

$d=5.0 \ cm$

$h_{o}=3.0 \ cm$

So, we can deduct that the object is $d_{o}=15.0 \ cm$ away from the lens. But, we need to calculate the distance of the image $d_{i}$ with this relation:

$\frac{1}{f}=\frac{1}{d_{o} } +\frac{1}{d_{i} }$

Replacing all values, we have:

$\frac{1}{10}=\frac{1}{15} +\frac{1}{d_{i}}\\\frac{1}{10}-\frac{1}{15} =\frac{1}{d_{i}}\\\frac{1}{d_{i}}=\frac{15-10}{150}=\frac{5}{150}\\d_{i}=\frac{150}{5}=30$

Now, we can calculate the highness of the image:

$\frac{d_{o} }{d_{i} }=\frac{h_{o}}{h_{i}}\\\frac{15}{30}=\frac{3}{h_{i}}\\h_{i}=\frac{3(30)}{15}=6$

Therefore, the image height is 6 centimetres.

Fed [463]3 years ago

4 0

For the first true or false question, it is true.

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To answer this problem, we will use the equations of motions.

Part (a):
For the ball to start falling back to the ground, it has to reach its highest position where its final velocity will be zero.
The equation that we will use here is:
v = u + at where
v is the final velocity = 0 m/sec
u is the initial velocity = 160 m/sec
a is acceleration due to gravity = -9.8 m/sec^2 (the negative sign is because the ball is moving upwards, thus, its moving against gravity)
t is the time that we want to find.
Substitute in the equation to get the time as follows:
v = u + at
0 = 160 - 9.8t
9.8t = 160
t = 160/9.8 = 16.3265 sec
Therefore, the ball would take 16.3265 seconds before it starts falling back to the ground

Part (b):
First, we will get the total distance traveled by the ball as follows:
s = 0.5 (u+v)*t
s = 0.5(160+0)*16.3265
s = 1306.12 meters
The equation that we will use to solve this part is:
v^2 = u^2 + 2as where
v is the final velocity we want to calculate
u is the initial velocity of falling = 0 m/sec (ball starting falling when it reached the highest position, So, the final velocity in part a became the initial velocity here)
a is acceleration due to gravity = 9.8 m/sec^2 (positive as ball is moving downwards)
s is the distance covered = 1306.12 meters
Substitute in the above equation to get the final velocity as follows:
v^2 = u^2 + 2as
v^2 = (0)^2 + 2(9.8)(1306.12)
v^2 = 25599.952 m^2/sec^2
v = 159.99985 m/sec
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Answer:

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Explanation:

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Wave travelling along right with speed of 0. 5 m/s time 4 second point p

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The wavelength of the wave which is travelling along right with speed of 0. 5 m/s time 4 second point is 2.

<h3>What is the speed of wave?</h3>

Speed of wave is the rate by which the wave travels. It can be described as the distance cover by the crest or thrust of it per unit of time. It can be calculated with the following formula,

$v=f\times\lambda$

Here, (f) is the frequency and (λ) is the wavelength.

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