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3 years ago

Describe the shape of a convex lens and explain what it does to light.

1 answer:

4 0

Convex lenses are thicker at the centers than the edges, they are known as the converging lenses. Rays of light that pass through the lens are brought closer together (they converge). When rays of light that are parallel pass through a convex lens they are refracted, the refracted rays converge at one point called the principal focus.

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Transverse waves are sent along a 4.50 m long string with a speed of 85.00 m/s. The string is under a tension of 20.00 N. What i

frutty [35]

Answer:

m = 0.0125 kg

Explanation:

Let us apply the formula for the speed of a wave on a string that is under tension:

$v = \sqrt{\frac{F}{\mu} }$

where F = tension force

μ = mass per unit length

Mass per unit length is given as:

μ = m / l

where m = mass of the string

l = length of the string

This implies that:

$v = \sqrt{\frac{F}{m/l} }\\ \\v = \sqrt{\frac{F * l}{m} }$

Let us make mass, m, the subject of the formula:

$v^2 = \frac{F * l}{m}\\\\m = \frac{F * l}{v^2}$

From the question:

F = 20 N

l = 4.50 m

v = 85 m/s

Therefore:

$m = \frac{20 * 4.5}{85^2}\\\\m = \frac{90}{7225}\\ \\m = 0.0125 kg$

5 0

3 years ago

Does the size of a paper airplane affect how far it flies

astra-53 [7]

No the only thing that affects it is how it is built

4 0

3 years ago

11. Which of the following phenomena is taking place when sound waves are reflected from a surface along parallel lines? A. Refr

allsm [11]

The Answer is C. Focusing

4 0

3 years ago

In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co

labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

$m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m$

The final velocity of cart 1 after collision is given as:

$v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\ Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s$

The final velocity of cart 2 after collision is given as:

$v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\ Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s$

b) Using the law of conservation of energy:

$\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm$

7 0

3 years ago

A steel ball and a wooden ball of the same diameter are released together from the top of a tower. In the absence of air resista

ella [17]

Answer:

False

Explanation:

The steel ball and the wooden ball do not have the same force acting on them because their masses are different. But, they have the same acceleration which is the acceleration due to gravity g = 9.8 m/s².

Using the equation of motion under freefall, s = ut +1/2gt². Since u = 0,

s = 1/2gt² ⇒ t = √(2s/g)

Since. s = height is the same for both objects, they land at the same time neglecting air resistance.

8 0

3 years ago