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3 years ago

An incident ray parallel to the axis and one through the focal point reflect and converge at the image.

Physics

2 answers:

Grace [21]3 years ago

3 0

Answer:

The answer is true.

Explanation:

A ray coming from the object hits the surface of a mirror and then passes through focus point after reflection. Since the image is formed by the actual meeting of the reflected rays therefore it is called as real image. the direction of arrow will be upward.

Same is true if the object is placed at center of curvature the coming ray along the parallel axis passes through the focus and bends downward. The point where the two reflecting rays will meet is the point of image. And the direction of image arrow will be downward. In this case image is formed between F and C.

melomori [17]3 years ago

3 0

Answer:

TRUE

Explanation:

In mirrors, all rays of light incidented on a plane surface (mirror) and parallel to the principal axis converges at the focus of the mirror after reflection.

When forming the image of an object placed in front of a mirror (for example curved mirrors), there are two steps:

1. , all rays of light incidented on a plane surface (mirror) and parallel to the principal axis converges at the focus of the mirror after reflection.

2. Light rays from the object passes through the centre of curvature undeflected and reflect back on the same line of incidence

The two reflected rays from formed will intersect at a point and this point is where the image will be formed.

Based on the conclusion above the statement in question is TRUE i.e

an incident ray parallel to the axis and one through the focal point reflect and converge at the image.

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The half-life of Co-55 is 175 hours. How much of a 4000 g of Cobalt-55 sample would be left after 525 hours?

Harrizon [31]

We know that whatever amount we start with, half of it decays and forms atoms of other elements in 175 hours. So in order to figure out how much is left after 525 hours, we'll need to know how many half-lifes pass in that amount of time.

Well, (525 divided by 175) is exactly 3 half-lifes. So this will be easy.

-- After 1 half-life . . .

. . . . . 50% decays, 50% is still there.

-- After the 2nd half-life . . .

. . . . . (half of the leftover 50%) = another 25% decays, 25% is left.

-- After the 3rd half-life . . .

. . . . . (half of the leftover 25%) = another 12.5% decays, 12.5% is left.

12.5% of 4,000g = (0.125 x 4,000g) = 500 g .

============================================

Another way:

After 1 half-life, 1/2 is left.

After 2 half-lifes, 1/4 is left.

After 3 half-lifes, 1/8 is left.

1/8 of 4,000g = (4,000g/8) = 500 g .

5 0

2 years ago

un tanque de gasolina de 40 litros fue llenado por la noche, cuando la temperatura era de 68 grados farenheit. Al dia siguiente

Sedaia [141]

Answer:

Volume of gasoline that expands and spills out is 1.33 ltr

Explanation:

As we know that when temperature of the liquid is increased then its volume will expand and it is given as

$\Delta V = V_o\gamma \Delta T$

here we know that

$V_o = 40 Ltr$

volume expansion coefficient of the gasoline is given as

$\gamma = 950 × 10^{–6}$

change in temperature is given as

$\Delta T = (131 - 68) \times \frac{5}{9}$

$\Delta T = 35 ^oC$

Now we have

$\Delta V = 40(950 \times 10^{-6})(35)$

$\Delta V = 1.33 Ltr$

3 0

3 years ago

You are at a train station, standing next to the train at the front of the first car. The train starts moving with constant acce

Aleksandr [31]

The time needed for the 7th car to pass is 13.2 s

Explanation:

The motion of the train is a uniformly accelerated motion, therefore we can use suvat equations.

We start by analzying the motion of the first car, by using the equation:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered by the first car in a time t, which corresponds to the length of one car

u = 0 is the initial velocity

a is the acceleration

t = 5.0 s is the time

The equation can be rewritten as

$a=\frac{2s}{t^2}=\frac{2L}{(5.0)^2}=0.08L[m/s^2]$

where L is the length of one car.

The same equation can be written considering the first 7 cars:

$7L = ut+\frac{1}{2}at'^2$

where

7L is the distance covered by the 7 cars

t' is the time needed

We still have

u = 0

And the acceleration is constant so it is

$a=0.08L$

Substituting into the equation, we can find t':

$7L = \frac{1}{2}(0.08L)t'^2\\7=0.04t'^2\\t'=\sqrt{\frac{7}{0.04}}=13.2 s$

In attachment the graph of the distance covered versus the time taken: since the motion is uniformly accelerated, the relationship between the two variables is quadratical.

Learn more about accelerated motion:

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