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2 years ago

In your own words, describe the three steps in ray tracing used to identify where an image forms.

Physics

2 answers:

Dima020 [189]2 years ago

6 0

Answer:

A ray coming from the object parallel to the principle axis after reflection from the mirror would pass through focus or refracting through the lens would converge or appear to diverge from focus.
An incident ray passing through the focus would reflect/refract parallel to the principle axis.
A ray passing through the pole of the lens would transmit through it.

A ray which is incident at an angle to the mirror or lens would reflect or refract respectively and the angle of reflection or refraction would follow the laws of reflection or Snell's law of refraction respectively.

Nostrana [21]2 years ago

6 0

The important point is in image tracing:

1) The rays travelling from very far, meet at focus after interaction with the lens/mirror.
2) The rays travelling through focus are parallel to the principal axis.
3) The object must be kept facing the front part of the mirror.

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A 25.0 kg bag of peat moss sits in the back of a flatbed truck, driving up a hill. The bag experiences a 225N normal force. The

malfutka [58]

Answer:

$\theta = 23.32^o$

$\mu_s = 0.27$

$s = 0.948 \ m$

Explanation:

From the question we are told that

The mass of the bag is $m_b = 25.0 \ kg$

The normal force experienced is $F_n = 225 \ N$

The maximum acceleration of the bag is $a = 2.40 \ m/s^2$

Generally this normal force experience by the bag is mathematically represented as

$F_n = mg cos \theta$

=> $225 = (25 * 9.8) cos \theta$

=> $0.9183 = cos \theta$

=> $\theta = cos^{-1}[0.9183]$

=> $\theta = 23.32^o$

Generally for the bag not to slip , it means that the frictional force is equal to the sliding force

$F_f = F_s$

Hence $F_f$ is mathematically represented as

$F_f = \mu_s * F_n$

While $F_s$ is mathematically represented as

$F_s = m * a$

$\mu_s * F_n = m * a$

=> $\mu_s * 225 = 25 * 2.40$

=> $\mu_s = 0.27$

Generally from the workdone equation we have that

$KE_f - KE_i = W_f$

Here $W_f$ is the work done by friction which is mathematically represented as

$W_f = m * g * \mu_k * s$

Here s is the distance covered by the bag

$KE_f$ is zero given that velocity at rest is zero

and

$KE_i = \frac{1}{2} * m* v_i^2$

$\frac{1}{2} * m* v_i^2 = m * g * \mu_k * s$

=> $\frac{1}{2} * v_i^2 = g * \mu_k * s$

substituting 2.55 m/s for v_i and 0.350 for \mu_k we have that

$\frac{1}{2} * 2.55^2 = 9.8 * 0.350 * s$

=> $s = 0.948 \ m$

4 0

2 years ago

a crowbar of 2 meter is used to lift an object of 800N if the effort arm is 160cm , calculste the effort applied

Vitek1552 [10]

Answer:

200 N

Explanation:

The crowbar is 2 meter, or 200 cm. The effort arm is 160 cm, so the moment arm of the object is 40 cm.

(800 N) (40 cm) = F (160 cm)

F = 200 N

5 0

2 years ago

Is the statement "An object always moves in the direction of the net force acting on it" true or false

Trava [24]

Answer:

False

Explanation:

This is because according to newtons second law which says the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object. So take for example a net a net force in opposite direction will cause an object to slow down.

velocity vector here is not the same as acceleration vector

7 0

2 years ago

Intensity: Radiation of a single frequency reaches the upper atmosphere of the earth with an intensity of 1350 W/m2. What is the

sesenic [268]

Answer:

The right answer is "1010 V/m".

Explanation:

The given values are:

Intensity,

$I=1350 \ W/m^2$

$c = 3.00\times 108 \ m/s$

$\mu_0= 4\pi\times 10^{-7} \ T.m/A$

Now,

The electric field's maximum value will be:

= $\sqrt{2\times u\times c\times I}$

On substituting the values in the above formula, we get

= $\sqrt{2\times 4\times \pi\times 10^{-7}\times 3\times 10^8\times 1350}$

= $\sqrt{32400\times 3.14\times 10^{-7}\times 10^8}$

= $1010 \ V/m$

3 0

1 year ago

For the meter stick shown in figure 10-4, the force F1 10.0 N acts at 10.0 cm. What is the magnitude of torque due to F1 about a

Phantasy [73]

Torque is equal position vector times (r) times force vector (F). Since F= 10 N and r = 0.1 m, so the torque is equal to (10 N) x ( 0.1 m) = 1Nm. The direction of the torque would be into the screen, clockwise rotation.

8 0

2 years ago