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2 years ago

8

A ball is thrown horizontally from the roof of a building 7.8 m tall and lands 9.5 m from the base. what was the ball's initial

speed

1 answer:

spin [16.1K]2 years ago

5 0

173 was the initial speed

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. A PNP transistor is connected in a circuit so that the collector-base junction remains reverse biased and the emitter-base jun

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A PNP transistor is connected in a circuit so that the collector-base junction remains reverse biased and the emitter-base junction is forward biased.

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In an inkjet printer, letters and images are created by squirting drops of ink horizontally at a sheet of paper from a rapidly m

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Answer:

$q = 6.48 \times 10^{-14} C$

Explanation:

Deflection in the drop is due to electric field force

so we will have

$F = qE$

acceleration of the drop is given as

$a = \frac{qE}{m}$

$a = \frac{q(7.75 \times 10^4)}{1.00 \times 10^{-11}}$

$a = 7.75 \times 10^{15} q$

now we know that time to cross the plates is given as

$t = \frac{D}{v}$

$t = \frac{0.02}{18}$

$t = 1.11 \times 10^{-3} s$

now the deflection is given as

$d = \frac{1}{2}at^2$

$0.310 \times 10^{-3} = \frac{1}{2}(7.75 \times 10^{15} q)(1.11 \times 10^{-3})^2$

$0.310 \times 10^{-3} = 4.78 \times 10^9 q$

$q = 6.48 \times 10^{-14} C$

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g An electron enters a region of space containing a uniform 1.63 × 10 − 5 T magnetic field. Its speed is 121 m/s and it enters p

kolbaska11 [484]

Answer:

i. The radius 'r' of the electron's path is 4.23 × $10^{-5}$ m.

ii. The frequency 'f' of the motion is 455.44 KHz.

Explanation:

The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.

r = $\frac{mv}{qB}$

Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.

From the question, B = 1.63 × $10^{-5}$ T, v = 121 m/s, Θ = $90^{0}$ (since it enters perpendicularly to the field), q = e = 1.6 × $10^{-19}$ C and m = 9.11 × $10^{-31}$ Kg.

Thus,

r = $\frac{mv}{qB}$ ÷ sinΘ

But, sinΘ = sin $90^{0}$ = 1.

So that;

r = $\frac{mv}{qB}$

= (9.11 × $10^{-31}$ × 121) ÷ (1.6 × $10^{-19}$ × 1.63 × $10^{-5}$ )

= 1.10231 × $10^{-28}$ ÷ 2.608 × $10^{-24}$

= 4.2266 × $10^{-5}$

= 4.23 × $10^{-5}$ m

The radius 'r' of the electron's path is 4.23 × $10^{-5}$ m.

B. The frequency 'f' of the motion is called cyclotron frequency;

f = $\frac{qB}{2\pi m}$

= (1.6 × $10^{-19}$ × 1.63 × $10^{-5}$ ) ÷ (2 × $\frac{22}{7}$ × 9.11 × $10^{-31}$ )

= 2.608 × $10^{-24}$ ÷ 5.7263 × $10^{-30}$

= 455442.4323

f = 455.44 KHz

The frequency 'f' of the motion is 455.44 KHz.

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