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Otrada [13]
3 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]3 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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Two resistances, R1 and R2, are connected in series across a 9-V battery. The current increases by 0.450 A when R2 is removed, l
Rina8888 [55]

Answer:

a. R1 = 0.162 Ω

b. R2 = 0.340 Ω

Explanation:

Since the resistors R1 and R2 are connected in series, the current flowing through them when the 9 V battery is applied is 9/R1 + R2.

When the current increases by 0.450 A wen only R1 is in the circuit, the current is

9/R1 + R2 + 0.450 A = 9/R1       (1)

When the current increases by 0.225 A when only R2 is in the circuit, the current is

9/R1 + R2 + 0.225 A = 9/R2       (2)

equation (1) - (2) equals

9(1/R1 - 1/R2) = 0.450 A - 0.225

9(1/R1 - 1/R2) = 0.125

(1/R1 - 1/R2) = 0.125 A/9 = 0.0138

1/R1 = 0.0138 + 1/R2

R1 = R2/(1 + 0.0138R2)     (3)

From (1)

9/R1 - 9/R1 + R2 = 0.450 A

9R2/[R1(R1 + R2)] = 0.450 A

R2/[R1(R1 + R2)] = 0.450 A/9 = 0.5

R2/[R1(R1 + R2)] = 0.5    (4)

From (3) R2/R1 = (1 + 0.0138R2) and from (4) R2/R1 = 0.5(R1 + R2). So,

(1 + 0.0138R2) = 0.5(R1 + R2)

0.5R1 + 0.5R2 = 1 + 0.0138R2

0.5R1 = 1 + 0.0138R2 - 0.5R2

0.5R1 = 1 - 0.4862R2        (5)

Substituting (3) into (5) we have

0.5R2/(1 + 0.0138R2) = 1 - 0.4862R2

R2 = (1 + 0.0138R2)(1 - 0.4862R2)

R2 = 1 - 0.4724R2 - 0.0067R2²

Collecting like terms, we have

0.0067R2² + 0.4724R2 + R2 - 1 = 0

0.0067R2² + 1.4724R2 - 1 = 0

Using the quadratic formula,

R_{2} = \frac{-1.4724 +/-\sqrt{(1.4724)^{2} - 4 X 0.0067 X -1} }{2 X 0.0067}  \\= \frac{-1.4724 +/-\sqrt{2.1680 + 0.0268} }{0.0268}\\= \frac{-1.4724 +/-\sqrt{2.1948} }{0.0268}\\= \frac{-1.4724 +/- 1.4815 }{0.0268}\\= \frac{-1.4724 + 1.4815 }{0.0268} or \frac{-1.4724 - 1.4815 }{0.0268}\\= \frac{0.0091 }{0.0268} or \frac{-2.9539}{0.0268}\\= 0.340 or -110.22

We choose the positive answer.

So R2 = 0.340 Ω

From (5)

R1 = 0.5 - 0.9931R2

   = 0.5 - 0.9931 × 0.340

   = 0.5 - 0.338

   = 0.162 Ω

a. R1 = 0.162 Ω

b. R2 = 0.340 Ω

5 0
3 years ago
6. A girl pushes her little brother on his sled with a force of 300 N for 750 m.
Oduvanchick [21]

w=225000 J  

E=225000 J  

P=9000 W  

P=22500 W

Ew=Fd=E  

P=E/t

5 0
3 years ago
How many type of expansion physic
Yuki888 [10]

Answer:

Three types of thermal expansion are linear expansion,s superficial expansion,cubical expansion

4 0
3 years ago
When a bus is moving (especially with a high speed) on the road suddenly stops or suddenly changes its direction, the luggage on
Masja [62]

Answer:

First law of motion

Explanation:

I say this because this example shows how an object is staying persistent unless it's compled to change

(not sure)

7 0
3 years ago
A tetrahedron has an equilateral triangle base with 25.0-cm-long edges and three equilateral triangle sides. The base is paralle
Jet001 [13]

Answer:

-7.03645640575 Nm²/C

2.34548546858 Nm²/C

Explanation:

S = Surface area

Electric flux through the base is given by

\phi_b=-\int Eds\\\Rightarrow \phi_b=-ES\\\Rightarrow \phi_b=-260\dfrac{\sqrt{3}}{4}0.25^2\\\Rightarrow \phi_b=-7.03645640575\ Nm^2/C

The electric flux through the base is -7.03645640575 Nm²/C

Inside the tetrahedron there is no charge so total flux will be zero

\phi_b+3\phi_s=0\\\Rightarrow \phi_s=-\dfrac{\phi_b}{3}\\\Rightarrow \phi_s=-\dfrac{-7.03645640575}{3}\\\Rightarrow \phi_s=2.34548546858\ Nm^2/C

The electric flux through the sides is 2.34548546858 Nm²/C

6 0
3 years ago
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