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Roman55 [17]
3 years ago
13

How do lenses and mirrors compare in their interactions with light?

Physics
2 answers:
slamgirl [31]3 years ago
7 0
D. lenses focus light , mirrors do not
Alisiya [41]3 years ago
4 0

Answer:

C. lenses refract light; mirrors do not

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You will now examine the relationship between the number of field lines through a surface and the tangle betwcen A and E) angle
nikitadnepr [17]

Answer:

a. cosθ b. E.A

Explanation:

a.The electric flux, Φ passing through a given area is directly proportional to the number of electric field , E, the area it passes through A and the cosine of the angle between E and A. So, if we have a surface, S of surface area A and an area vector dA normal to the surface S and electric field lines of field strength E passing through it, the component of the electric field in the direction of the area vector produces the electric flux through the area. If θ the angle between the electric field E and the area vector dA is zero ,that is θ = 0, the flux through the area is maximum. If  θ = 90 (perpendicular) the flux is zero. If θ = 180 the flux is negative. Also, as A or E increase or decrease, the electric flux increases or decreases respectively. From our trigonometric functions, we know that 0 ≤ cos θ ≤ 1 for  90 ≤ θ ≤ 0 and -1 ≤ cos θ ≤ 0 for 180 ≤ θ ≤ 90. Since these satisfy the limiting conditions for the values of our electric flux, then cos θ is the required trigonometric function. In the attachment, there is a graph which shows the relationship between electric flux and the angle between the electric field lines and the area. It is a cosine function  

b. From above, we have established that our electric flux, Ф = EAcosθ. Since this is the expression for the dot product of two vectors E and A where E is the number of electric field lines passing through the surface and A is the area of the surface and θ the angle between them, we write the electric flux as Ф = E.A  

4 0
3 years ago
An organ system is best described by which of the following?
earnstyle [38]
A group of cells together
3 0
3 years ago
Read 2 more answers
A 25 kg child is riding on a swing. If the child travels 8.9 m/s at the bottom of their swing, how high into the air is the chil
Setler [38]

Answer:

h = 4.04 m

Explanation:

Given that,

Mass of a child, m = 25 kg

The speed of the child at the bottom of the swing is 8.9 m/s

We need to find the height in the air is the child is able to swing. Let the height is h. Using the conservation of energy such that,

mgh=\dfrac{1}{2}mv^2\\\\h=\dfrac{v^2}{2g}

Put all the values,

h=\dfrac{(8.9)^2}{2\times 9.8}\\\\h=4.04\ m

So, the child is able to go at a height of 4.04 m.

7 0
3 years ago
Four solid plastic cylinders all have radius 2.41 cm and length 5.94 cm. Find the charge of each cylinder given the following ad
Paladinen [302]

Answer:

Check explanation

Explanation:

QUICK NOTE: THE QUESTION IS NOT COMPLETE. Although it is not, we can make assumptions, since we only need values for the UNIFORM CHARGE DENSITY.

SO, LET US BEGIN;

To solve this question we are to use the equation (1) below;

Charge,Q = uniform charge density,p × Total area of the cylinder,A ------------------------------------------------------------------------(1).

From the question, we are given radius, R to be 2.41 cm and length, L to be 5.94 cm.

Step one: calculate for the total area of the cylinder, A.

Total area of the cylinder, A= area of the top surface + area of the buttom + area of the curved surface of the cylinder.

Hence, total area of the cylinder,A is;

==> πR^2 + πR^2 + 2πRL. -------------------------------------------------------------------------(2).

Then, total area of the cylinder,A is;

==> (L + R)2πR.

Step two: find the charge of each cylinder.

===> For the first cylinder; we have the uniform charge density to be 35 nC/m^2.

Therefore, the combination of equation (1) and (3) gives;

Charge Q= p × (L + R)2πR...----------------------------(4)

Hence, Q= 35 × [(5.94 + 2.41) 2× 3.143 × 5.94].= 10912.615 coulumb.

====> For the second cylinder, we have a uniform charge density of 50 nC/m^2.

Using equation (4), charge,Q= 15,589.45 Coulumb

=====> For THE third cylinder, the uniform charge density is 600, we make use of equation (4);

Charge,Q= 600×311.789.

Charge,Q= 187,073.4 coulumb.

====> For THE fourth cylinder, the uniform charge density is 750 nC/m^2.., we make use of equation (4);

Charge,Q= 233,841.75 coulumb.

7 0
3 years ago
Compare and contrast the health benefits and the amount of physical or mental energy needed between active leisure activities an
dolphi86 [110]
This is the answer "Some active leisure activities, such as yoga and running, require a great deal of physical energy, while other activities, like playing chess, require a substantial amount of mental energy. Passive leisure activities, such as watching TV or going to the movies, require little to no physical or mental energy. Because of the exertion of physical or mental energy, active leisure activities have more health benefits than passive leisure activities." hope it help you?:)
4 0
3 years ago
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