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Otrada [13]
4 years ago
12

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 2.10 107 m/s in the horizontal direction when magnetic forces deflect the electron with a vertically upward acceleration of 5.10 1015 m/s2. The phosphorescent screen is a horizontal distance of 5.5 cm away from the point where the electron is emitted.
(a) How much time does the electron take to travel from the emission point to the screen? s

(b) How far does the electron travel vertically before it hits the screen?
Physics
1 answer:
Len [333]4 years ago
4 0

Answer:

Explanation:

initial velocity v = 2.1  x 10⁷ m/s

acceleration a = 5.1 x 10¹⁵ m /s²

horizontal distance covered = 5.5 x 10⁻² m

time taken to cover horizontal distance =  5.5 x 10⁻² / 2.1  x 10⁷

= 2.62 x 10⁻⁹ s .

b )

vertical distance travelled due to vertical acceleration

= 1/2 a t²

= .5 x 5.1 x 10¹⁵ x (2.62 x 10⁻⁹)²

= 17.5 x 10⁻³ m

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D. Humidity refers to the amount of water vapor in the air

7 0
3 years ago
The 6 kg block is then released and accelerates to the right, toward the 4 kg block. The surface is rough and the coefficient of
KatRina [158]

Answer:

v =3.41 m/s

Explanation:

given,

mass of block 1 = 6 Kg

mass of another block 2 = 4 Kg

coefficient of friction = 0.3

Assuming 6 Kg block is attached to the spring of spring constant 350 N/m

and distance between the two block is equal to 0.5 m

using formula

U = \dfrac{1}{2}kx^2

U = \dfrac{1}{2}\times 350 \times 0.5^2

   U = 43.75 J

using conservation of energy

 KE = U - f.d

where f is the frictional force acting

\dfrac{1}{2}mv^2 = 43.75- \mu m g d

\dfrac{1}{2}\times 6 \times v^2 = 43.75- 0.3\times 6 \times 9.8 \times 0.5

v= \sqrt{11.643}

       v =3.41 m/s

4 0
4 years ago
How long does it take to get from a new moon to a first quarter moon
Andru [333]

'First Quarter' is the phase that appears one quarter of the
period of time from one new moon to the next one.

The total period of the moon's phases is 29.53 days (rounded).

One quarter of it is (29.53 / 4) =  <em>7 days  9.2 hours</em>   (rounded)


5 0
4 years ago
If you were to apply a force of 15N to a medicine ball with a mass of 5kg, what would its acceleration be
yaroslaw [1]

Given that

Force (F) = 15 N ,

mass (m) = 5 Kg ,

acceleartion = ?

              We know that, From Newtons II law

                      F = m. a

                    15 = 5 × a

                     a = 15÷ 5

                       a = 3 m/s²

acceleration of the ball is 3 m/s²

4 0
4 years ago
Using carson's rule, what is the required bw to transmit an fm signal if the maximum deviation is 60 khz and intelligence freque
VashaNatasha [74]

Carson's Rule says:

FM occupied bandwidth =

(2) · (Peak deviation + Highest modulating frequency)

so

FM bw = (2) · (60 kHz + 15 kHz)

FM bw = (2) · (75 kHz)

FM bw = 150 kHz

(I used to eat this stuff for lunch, but it's been almost 40 years.  Thanks for taking me back.  Those were the good old days.)

5 0
3 years ago
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