All categories

3 years ago

The image formed by a lens may be real or virtual. The image formed by a lens is always virtual.

2 answers:

8 0

There are types of lens; concave and convex lens.
The concave lens is a lens which has an inward curve in the middle, that is, the edges of the curve are thicker than the center of the lens, because of this, any light that enter the lens will spread out [diverge]. An image will look smaller and upright when viewed by a concave lens. Image formed by concave lens are usually VIRTUAL.
A concave lens will produce a real image ONLY if the object is located beyond the focal point of the lens.
A convex lens is a converging lens, this is because, the center of the lens is thicker than its edges. Any ray of light that passes through the lens will converge at the middle of the lens at point called principal focus. A convex lens produce a VIRTUAL image when the object is placed infront of the focal point. The virtual image formed is always magnified and upright.

daser333 [38]3 years ago

5 0

The image formed by a convex lens may be real or virtual. The image formed by a concave lens is always virtual.

You might be interested in

Which of these is a characteristic of science? (5 points) Question 1 options: 1) It cannot be reproduced by any scientist. 2) It

Alex17521 [72]

Answer:

Explanation:

It is based on empirical evidence

4 0

3 years ago

10.0 grams of water are heated during the preparation of a cup of coffee 1.0x 103 j of the heat are added to the water. which is

katovenus [111]

Answer: The final temperature of the coffee is 43.9°C

Explanation:

To calculate the final temperature, we use the equation:

$q=mC(T_2-T_1)$

where,

q = heat released = $1.0\times 10^3J=1000J$

m = mass of water = 10.0 grams

C = specific heat capacity of water = 4.184 J/g°C

$T_2$ = final temperature = ?

$T_1$ = initial temperature = 20°C

Putting values in above equation, we get:

$1000J=10.0g\times 4.184J/g^oC\times (T_2-20)\\\\T_2=43.9^oC$

Hence, the final temperature of the coffee is 43.9°C

6 0

3 years ago

Individual particles move with the wave.

MArishka [77]

Answer:

No.

Explanation:

No, individual particles do not move with the wave, it only oscillates back and forth its mean position. The particles in the medium transfer its energy to their neighboring particles and in that way the energy moves in the form of wave. The particles only vibrates on its means position instead of moving from one place to another. So we can conclude that Individual particles do not move with the wave.

3 0

3 years ago

A motorcycle travel at 6.0 m/s. After 3.0 second, the motorcycle travel at 15.0 m/s. Which of the following was the average acce

Lerok [7]

Hello!

$\large\boxed{a = 3m/s^{2}}$

Use the following equation to solve for the average acceleration of the motorcycle:

$a = \frac{v_{f}-v_{i}}{t}$

Plug in the given final, initial velocities, and the time:

$a = \frac{15-6}{3}\\\\a = \frac{9}{3}\\\\a = 3m/s^{2}$

7 0

3 years ago

The rate of disappearance of HBr in the gas phase reaction2HBr(g) → H2(g) + Br2(g)is 0.301 M s-1 at 150°C. The rate of appearanc

jok3333 [9.3K]

Answer: 0.151

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$2HBr(g)\rightarrow H_2(g)+Br_2(g$

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

$Rate=-\frac{1d[HBr]}{2dt}=+\frac{1d[H_2]}{2dt}=+\frac{1d[Br_2]}{dt}$

Given: $-\frac{d[HBr]}{dt}]=0.301$

Putting in the values we get:

$\frac{0.301}{2}=+\frac{1d[Br_2]}{dt}$

$+\frac{1d[Br_2]}{dt}=0.151$

Thus the rate of appearance of $Br_2$ is 0.151

8 0

3 years ago