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Troyanec [42]
3 years ago
12

Which of the following is accurate in describing converging and diverging lenses?

Physics
2 answers:
kenny6666 [7]3 years ago
3 0

B. Both of these types of lenses have the ability to produce real images.

koban [17]3 years ago
3 0

Answer:

A. Both of these types of lenses have the ability to produce upright images.

Explanation:

Let's analyze each statement proposed:

A. Both of these types of lenses have the ability to produce upright images.  --> TRUE. In fact, depending on the position of the object relative to the lens, both converging and diverging lenses can produce an image which is upright.

B. Both of these types of lenses have the ability to produce real images.  --> FALSE. In fact, diverging lenses can produce virtual images only, because the rays of light are not focused into a single point.

C. Parallel rays converge into a single point at the focal point of a diverging lens, whereas parallel rays diverge at the focal point of a converging lens and spread out.  --> FALSE. It is actually the opposite: in a diverging lens, parallel rays diverge and spread out, while in a converging lens, parallel rays converge into a single point (the focal point)

D. Diverging lenses will produce images behind the lens, and converging lenses will produce images in front of the lens.  --> FALSE. It's actually the opposite: the image produced by a diverging lens is in front of the lens, while the image produced by a converging lens is behind the lens.


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A little boy is standing at the edge of a cliff 1000 m high. He throws a ball straight downward at an initial speed of 20 m/s, a
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Answer:

The ball will be at 700 m above the ground.

Explanation:

We can use the following kinematic equation

y(t) = \ y_0 \ + \ v_0 \ t \ + \frac{1}{2} \ a \ t^2.

where y(t) represent the height from the ground. For our problem, the initial height will be:

y_0 \ = \ 1000 m.

The initial velocity:

v_0 = - 20 \frac{m}{s},

take into consideration the minus sign, that appears cause the ball its thrown down.  The same minus appears for the acceleration:

a=-10\frac{m}{s}

So, the equation for our problem its:

y(t) = \ 1000 m \ - \ 20 \ \frac{m}{s} \ t \ - \frac{1}{2} \ 10 \frac{m}{s^2} \ t^2.

Taking t=6 s:

y(6 \ s) = \ 1000 m \ - \ 20 \ \frac{m}{s} \ * \ 6 \ s \ - \frac{1}{2} \ 10 \frac{m}{s^2} \ * \ (6 s)^2.

y(6 \ s) = \ 1000 m \ - 120 m - \frac{1}{2} \ 10 \frac{m}{s^2} \ * \ 36 s^2.

y(6 \ s) = \ 1000 m \ - 120 m - 180 m.

y(6 \ s) = \ 1000 m \ - 300 m.

y(6 \ s) = \ 700 m.

So this its the height of the ball 6 seconds after being thrown.

6 0
3 years ago
Why does water frozen in the cracks of a rock help to break down the rock? a) Water expands when frozen and chemically forces th
Gre4nikov [31]

Answer: B. water frozen in the cracks of a rock help to break down the rock because water expands when frozen, and physically forces the rock apart

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3 years ago
How much gravitational potential energy does a 1.0 kg hammer have on a shelf 1.5 m above the ground?
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Answer:

14.7 J

Explanation:

PE=MGH

PE= 1.0 x 9.8 x 1.5 = 14.7 J

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The magnetic field in a cyclotron is 1.25 T, and the maximum orbital radius of the circulating protons is 0.40 m. (a) What is th
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Answer:

1.92 x 10⁻¹²J

Explanation:

The magnetic force from the magnetic field gives the circulating protons gives the particle the necessary centripetal acceleration to keep it orbiting round the circular path. And from Newton's second law of motion, the force(F) is equal to the product of the mass(m) of the proton and the centripetal acceleration(a). i.e

F = ma

Where;

a = \frac{v^2}{r}             [v = linear velocity, r = radius of circular path]

=> F = m\frac{v^2}{r}           ------------(i)

We also know that the magnitude of this magnetic force experienced by the moving charge (proton) in a magnetic field is given by;

F = q v B sin θ       ----------(ii)

Where;

q = charge of the particle

v = velocity of the particle

B = magnetic field

θ = the angle between the velocity and the magnetic field.

Combining equations (i) and (ii) gives

m\frac{v^2}{r} = q v B sin θ           [θ = 90° since the proton is orbiting at the maximum orbital radius]

=> m\frac{v^2}{r} = q v B sin 90°

=> m\frac{v^2}{r} = q v B

Divide both side by v;

=> m\frac{v}{r} = qB

Make v subject of the formula

v = \frac{qBr}{m}

From the question;

B = 1.25T

m = mass of proton = 1.67 x 10⁻²⁷kg

r = 0.40m

q = charge of a proton = 1.6 x 10⁻¹⁹C

Substitute these values into equation(iii) as follows;

v = \frac{(1.6*10^{-19})(1.25)(0.4)}{(1.67*10^{-27})}

v = 4.79 x 10⁷m/s

Now, the kinetic energy, K, is given by;

K = \frac{1}{2}mv²

m = mass of proton

v = velocity of the proton as calculated above

K = \frac{1}{2}(1.67*10^{-27} * (4.79 * 10^7)^2 )

K = 1.92 x 10⁻¹²J

The kinetic energy is 1.92 x 10⁻¹²J

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