Please help me i need it it’s the only way i’ll pass

Categorize each ray tracing statement as relating to ray 1, ray 2, or ray 3.

Sliva [168]

Answer:

statement 1 with answer C

statement 2 with answer F

statement 3 with answer B

Statement 1 with E

Statement 2 with A

Statement 3 with D

Explanation:

In this exercise you are asked to relate each with the answers

In general, in the optics diagram,

* Ray 1 is a horizontal ray that after stopping by the optical system goes to the focal point

* Ray 2 is a ray that passes through the intercept point between the optical axis and the system and does not deviate

* Ray 3 is a ray that passes through the focal length and after passing the optical system, it comes out horizontally.

With these statements, let's review the answers

statement 1 with answer C

statement 2 with answer F

statement 3 with answer B

Statement 1 with E

Statement 2 with A

Statement 3 with D

7 0

3 years ago

An unknown charged particle passes without deflection throughcrossed electric and magnetic fields of strengths 187,500 V/m and0.

UNO [17]

Explanation:

The given data is as follows.

Electric field strength (E) = 187,500 V/m

Magnetic field strength (B) = 0.125 T

Diameter (d) = 25.05 cm = 0.2505 m (as 1 m = 100 cm)

Radius (r) = $\frac{d}{2}$

= $\frac{0.2505}{2}$

= 0.12525 m

Formula to calculate the magnetic force ( $F_{M}$ ) is as follows.

$F_{M} = Bqv$ ............ (1)

Electrical force is calculated as follows.

$F_{E} = qE$ ............ (2)

On both electric and magnetic fields the velocity is perpendicular.

$F_{M} - F_{E}$ = 0

or, $F_{M} = F_{E}$

Hence, from equations (1) and (2)

Bqv = qE

or, v = $\frac{E}{B}$ ............. (3)

= $\frac{187500 V/m}{0.125 T}$

= 1,500,000 m/s

As the particle is moving in a semi-circular trajectory and motion of charged particle is given by the electric field as follows.

$F_{c} = \frac{mv^{2}}{r}$ ........... (4)

where, $F_{c}$ = centripetal force

$F_{M} = F_{c}$

Using equation (1) and (4) as follows.

$F_{M} = F_{c}$

Bqv = $\frac{mv^{2}}{r}$

$\frac{q}{m} = \frac{v}{Br}$

= $\frac{15 \times 10^{5}}{0.125 \times 0.12525}$

= $958.08 \times 10^{5} C/kg$

Thus, we can conclude that charge-to-mass ratio of the given particle is $958.08 \times 10^{5} C/kg$ .

8 0

4 years ago

Calculate the force required to accelerate a 600 g ball from rest to 14 m/s in 0.1 s.

Evgen [1.6K]

Statement:

A force is required to accelerate a 600 g ball from rest to 14 m/s in 0.1 s.

To find out:

The force required to accelerate the ball.

Solution:

Mass of the ball (m) = 600 g = 0.6 Kg
Initial velocity (u) = 0 m/s [it was at rest]
Final velocity (v) = 14 m/s
Time (t) = 0.1 s

Let the acceleration be a.
We know the equation of motion,
v = u + at

Therefore, putting the values in the above formula, we get
14 m/s = 0 m/s + a × 0.1 s
or, 14 m/s ÷ 0.1 s = a
or, a = 140 m/s²

Let the force be F.
We know, the formula : F = ma

Putting the values in the above formula, we get
F = 0.6 Kg × 140 m/s²
or, F = 84 N

Answer:

The force required to accelerate the ball is 84 N and this force acts along the direction of motion.

Hope you could understand.

If you have any query, feel free to ask.

5 0

3 years ago

Now the elevator is moving downward with a velocity of v = -2.8 m/s but accelerating upward with an acceleration of a = 5.5 m/s2

borishaifa [10]

Answer:

160.75 N

Explanation:

The downward velocity has no effect on the force situation, it is only changes in velocity (plus, of course, gravity, which is always there) that require a force. At constant velocity, the bottom spring s_3 is supporting its mass m_3 to balance gravity.

As the elevator slows, though, it also ends up slowing down the spring arrangement, too. However, because the stretching takes time, it means that some damped harmonic motion will be set up in the spring chain.

When the motion has finally damped out, the net force the bottom spring s3 exerts on m3 has two components--that of gravity and of the deceleration of the elevator:

F_3net = m3 * (g + a) = 10.5×(9.81+5.5)= 10.5×15.31= 160.75 N

5 0

3 years ago

When the north pole of a bar magnet is moved into a solenoid, the needle on a galvanometer attached to the solenoid moves to the

djyliett [7]

answer BBBBBBBBBBBB on EDGE

4 0

3 years ago

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