(1) The image of an object placed further from the lens than the focal point will be upside down and smaller than the object.
(2) When light rays reflect, they bounce back.
(3) Images formed by a concave lens will look magnified.
(4) When light rays enter a different medium, they bend.
<h3>
1.0 Object placed further from the lens than the focal point</h3>
The image of an object placed further from the lens than the focal point will be diminished and inverted.
Thus, the correct answer will be "upside down and smaller than the object".
<h3>2.0 What is reflection of light?</h3>
The ability of light to bounce back when it strike a hard surface is known as refection.
<h3>3.0 Image formed by concave lens</h3>
A concave lens is diverging lens is usually virtual, erect and magnified.
<h3>4.0 Refraction of light</h3>
The change in speed of light when it travels from medium to another medium is known as refraction. Refraction is also, the ability of light to bend around obstacles.
Learn more about reflection and refraction of light here: brainly.com/question/1191238
Answer:
The answer is not able to be solved, because we dont know what objects are in it, and how heavy they are. More information please!
Explanation:
Answer:
Fc = 89.67N
Explanation:
Since the rope is unstretchable, the total length will always be 34m.
From the attached diagram, you can see that we can calculate the new separation distance from the tree and the stucked car H as follows:
L1+L2=34m
Replacing this value in the previous equation:
Solving for H:
We can now, calculate the angle between L1 and the 2m segment:
If we make a sum of forces in the midpoint of the rope we get:
where T is the tension on the rope and F is the exerted force of 87N.
Solving for T, we get the tension on the rope which is equal to the force exerted on the car:
Answer:
.
Explanation:
The frequency of a wave is equal to the number of wave cycles that go through a point on its path in unit time (where "unit time" is typically equal to one second.)
The wave in this question travels at a speed of . In other words, the wave would have traveled in each second. Consider a point on the path of this wave. If a peak was initially at that point, in one second that peak would be
How many wave cycles can fit into that ? The wavelength of this wave gives the length of one wave cycle. Therefore:
.
That is: there are wave cycles in of this wave.
On the other hand, Because that of this wave goes through that point in each second, that wave cycles will go through that point in the same amount of time. Hence, the frequency of this wave would be
Because one wave cycle per second is equivalent to one Hertz, the frequency of this wave can be written as:
.
The calculations above can be expressed with the formula:
,
where
- represents the speed of this wave, and
- represents the wavelength of this wave.
The correct answer would be compound