(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:
A mass of 10 kilograms lifted 10 meters in 5 seconds.
Explanation:
Power can be defined as the energy required to do work per unit time.
Mathematically, it is given by the formula;
But Energy = mgh
Substituting into the equation, we have

Given the following data;
Mass = 10kg
Height = 10m
Time = 5 seconds
We know that acceleration due to gravity is equal to 9.8 m/s²

Hence, a mass of 10 kilograms lifted 10 meters in 5 seconds would produce the most power.
The three physical forms are:
Solid, liquid, or gas.
Answer:
The voltage across the capacitor is 1.57 V.
Explanation:
Given that,
Number of turns = 10
Diameter = 1.0 cm
Resistance = 0.50 Ω
Capacitor = 1.0μ F
Magnetic field = 1.0 mT
We need to calculate the flux
Using formula of flux

Put the value into the formula


We need to calculate the induced emf
Using formula of induced emf

Put the value into the formula

Put the value of emf from ohm's law





We know that,


We need to calculate the voltage across the capacitor
Using formula of charge


Put the value into the formula


Hence, The voltage across the capacitor is 1.57 V.