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Kruka [31]
2 years ago
7

Older televisions display a picture using a device called a cathode ray tube, where electrons are emitted at high speed and coll

ide with a phosphorescent surface, causing light to be emitted. The paths of the electrons are altered by magnetic fields. Consider one such electron that is emitted with an initial velocity of 2.00 x10^7 m/s in the horizontal direction when magnetic forces deflect the electron with a vertically upward acceleration of 5.25 x 10^15 m/s^2. The phosphorescent screen is a horizontal distance of 6 cm away from the point where the electron is emitted.
Required:
a. How much time does the electron take to travel from the emission point to the screen?
b. How far does the electron travel vertically before it hits the screen?
Physics
1 answer:
Vaselesa [24]2 years ago
5 0

Answer:

a) 3*10^{-9}\ s

b) 2.36 cm

Explanation:

a) The horizontal distance = x = 6 cm

1 cm = 0.01 m

6 cm = 6 cm * 0.01 m/cm = 0.6 m

Therefore the time taken (t) by the electron  to travel from the emission point to the screen can be gotten from:

x = t * V_i

V_i=initial\ velocity=2*10^7\ m/s

x=tV_i\\\\t=\frac{x}{V_i}\\ \\t=\frac{0.06}{2*10^7}\\ \\t=3*10^{-9}\ m/s

b) The vertical distance (y) traveled by the electon before it hits the screen is given by:

y=\frac{1}{2}at^2\\ \\a=upward\ acceleration=5.25*10^{15}\ m/s^2\\\\Substituting:\\\\y=\frac{1}{2}*5.25*10^{15}*  (3*10^{-9})\\\\y=0.0236\ m\\y=2.36\ cm

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The eyes of many older people have lost the ability to accommodate, and so an older person’s near point may be more than 25 cm f
tensa zangetsu [6.8K]

Answer:

Smaller refractive power

Explanation:

The refractive power of an eye is the extent to which it can converge or diverge the light rays.

Near point is the the closest point for an eye such that when an object is placed at that point the image it forms is sharp and clearly visible to the eye.

A the person ages, the ciliary muscles of the eyes weakens and as a result the lens contracts and the formation of the image takes place behind the retina instead of forming at the retina.

Thus the near point also increases and the refractive power becomes smaller.

4 0
2 years ago
In an attempt to impress its friends, an acrobatic beetle runs and jumps off the bottom step of a flight of stairs. The step is
Aleks04 [339]

Answer:

0.3677181864 m

Explanation:

u = Velocity = 1.5 m/s

\theta = Angle = 20°

y = -20 cm

Velocity components

u_x=ucos\theta\\\Rightarrow u_x=1.5cos20\\\Rightarrow u_x=1.40953\ m/s

u_y=usin\theta\\\Rightarrow u_y=1.5sin20\\\Rightarrow u_y=0.51303\ m/s

Acceleration components

a_x=0

a_y=-9.81\ m/s^2

y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow -0.2=0.51303\times t+\dfrac{1}{2}\times -9.81t^2\\\Rightarrow 4.905t^2-0.51303t-0.2=0

t=\frac{-\left(-0.51303\right)+\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}, \frac{-\left(-0.51303\right)-\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}\\\Rightarrow t=0.26088, -0.15629

Time taken is 0.26088 seconds

x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow x=1.40953\times 0.26088\\\Rightarrow x=0.3677181864\ m

The distance the beetle travels on the ground is 0.3677181864 m

6 0
2 years ago
In the ENGR 10 lab (E391), there are 50 long light bulbs (P=100 W) and 30 regular bulbs (P=60 W). How much energy is consumed li
Alenkinab [10]

Answer:

Total energy saving will be 0.8 KWH

Explanation:

We have given there are 50 long light bulbs of power 100 W so total power of 50 bulb = 100×50 = 5000 W = 5 KW

30 bulbs are of power 60 W

So total power of 30 bulbs = 30×60 = 1800 W = 1.8 KW

Total power of 80 bulbs = 1.8+5 = 6.8 KW

Total time = 3 hour

We know that energy E=power\times time=6.8\times 3=20.4KWH

Now power of each CFL bulb = 25 W

So power of 80 bulbs = 80×25 = 2000 W = 2 KW

Energy of 80 bulbs = 2×3 = 6 KWH

So total energy saving = 6.8-6 = 0.8 KWH

6 0
2 years ago
A uniform metre rule of weight 0.9 N is suspended horizontally by two vertical loops of thread A and B placed at 20cm and 30cm f
podryga [215]

Answer:

(a) 29 cm

(b) 43.5 cm

Explanation:

(a) when loop A is slack, there are three forces acting on the metre rule.

-0.9 N at 50 cm mark

T at 70 cm mark

-2 N at x

Taking the sum of the torques about B:

∑τ = Iα

(-0.9 N) (50 cm − 70 cm) + (-2 N) (x − 70 cm) = 0

18 Ncm − 2 N (x − 70 cm) = 0

2 N (x − 70 cm) = 18 Ncm

x − 70 cm = 9 cm

x = 79 cm

The distance from the center is |50 cm − 79 cm| = 29 cm.

(b) when loop B is slack, there are three forces acting on the metre rule.

-0.9 N at 50 cm mark

T at 20 cm mark

-2 N at x

Taking the sum of the torques about A:

∑τ = Iα

(-0.9 N) (50 cm − 20 cm) + (-2 N) (x − 20 cm) = 0

-27 Ncm − 2 N (x − 20 cm) = 0

2 N (x − 20 cm) = -27 Ncm

x − 20 cm = -13.5 cm

x = 6.5 cm

The distance from the center is |50 cm − 6.5 cm| = 43.5 cm

3 0
2 years ago
Compare the benefits of wildfires to grasslands, northern forests, and deciduous forests.
Dimas [21]
Wildfires benefit grasslands, northern forests, and deciduous forests. Grasslands are benefited by improved soil quality and control of tree cover. Invertebrate species diversity is maintained through wildfire as well. Northern forests, like grasslands, experience increased production and nutritional quality of food as a result of wildfires. Deciduous forests experience an increase in the nutritional quality of food as well, but the effects are more temporary. The amount of shrubs in deciduous forests is reduced as a result of wildfires, allowing more herbaceous plants such as mosses and lichens to grow.
6 0
2 years ago
Read 2 more answers
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