(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
A :-) for this question , we should apply
F = ma
( i ) Given - m = 2 kg
a = 15 m/s^2
Solution :
F = ma
F = 2 x 15
F = 30 N
( ii ) Given - m = 2 kg
a = 10 m/s^2
Solution :
F = ma
F = 2 x 10
F = 20 N
.:. The net force of object ( i ) has greater force compared to object ( ii ) by
( 30 - 20 ) 10 N
(a) 0.96 m/s
The period of the wave corresponds to the time taken for one complete oscillation of the boat, from the highest point to the highest point again. Since the time between the highest point and the lowest point is 2.5 s, the period is twice this time:

The frequency of the waves is the reciprocal of the period:

The wavelength instead is just the distance between two consecutive crests, so

And the wave speed is given by:

(b) 0.265 m
The total distance between the highest point of the wave and its lowest point is
d = 0.53 m
The amplitude is just the maximum displacement of the wave from the equilibrium position, so it is equal to half of this distance. So, the amplitude is

(c) Amplitude: 0.15 m, wave speed: same as before
In this case, the amplitude of the wave would be lower. In fact,
d = 0.30 m
So the amplitude would be

Instead, the wave speed would not change, since neither the frequency nor the wavelength of the wave have changed.