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3 years ago

An electron travels 1.49 m in 7.4 µs (microsecWhat is its speed if 1 inch = 0.0254 m? Answer in units of in/min.

Physics

1 answer:

mezya [45]3 years ago

7 0

Explanation:

Write what you know

Speed = Distance / Time

micro- = 10^-6

write your conversions as fractions

1 in / 0.0254 m

1 min / 60 sec

First convert time to regular seconds

7.4 x 10^-6 seconds

Use Velocity

1.49m / (7.4 x 10^-6) s

We've written our conversions in fractions because units cancel out just like numbers

$\frac{1.49m}{7.4 \times {10}^{ - 6} } \times \frac{1in}{0.0254m} \times \frac{60sec}{1min}$

Multiply all the fractions accross and youll have your answer

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A 594 Ω resistor, an uncharged 1.3 μF capacitor, and a 6.53 V emf are connected in series. What is the current in milliamps afte

ivanzaharov [21]

Answer:

6.88 mA

Explanation:

Given:

Resistance, R = 594 Ω

Capacitance = 1.3 μF

emf, V = 6.53 V

Time, t = 1 time constant

Now,

The initial current, I₀ = $\frac{\textup{V}}{\textup{R}}$

I₀ = $\frac{\textup{6.53}}{\textup{594}}$

I₀ = 0.0109 A

also,

I = $I_0[1-e^{-\frac{t}{\tau}}]$

here,

τ = time constant

e = 2.717

on substituting the respective values, we get

I = $0.0109[1-e^{-\frac{\tau}{\tau}}]$

I = $0.0109[1-2.717^{-1}]$

I = 0.00688 A

I = 6.88 mA

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2 years ago

A normal human heart, beating about once per second, creates a maximum 4.00 mV potential across 0.400 m of a person's chest, pro

kvasek [131]

Answer:

The maximum electric field strength = 0.01 V/m

Explanation:

Given

ΔV(max) = 4.00 mV = 0.004 V

d = 0.400 m

f = 1.00 Hz

Maximum electric field = (maximum potential)/(length)

Maximum electric field = E(max)

Maximum potential = 4.00 mV = 0.004 V

Length = 0.400 m

E(max) = (0.004/0.4) = 0.01 V/m

Hope this Helps!!!

3 0

2 years ago

Read 2 more answers

A horizontal uniform bar of mass 2.7 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to t

Jlenok [28]

Answer:

14.36 N

Explanation:

$T_{1}$ = Tension in string 1

$T_{2}$ = Tension in string 2

$m_b$ = mass of the bar = 2.7 kg

$W_b$ = weight of the bar

weight of the bar is given as

$W_b = m_{b} g = (2.7) (9.8) = 26.46$ N

$m_m$ = mass of the bar = 1.35 kg

$W_m$ = weight of the monkey

weight of the monkey is given as

$W_m = m_{m} g = (1.35) (9.8) = 13.23$ N

Using equilibrium of torque about left end

$W_{m} (AB) + W_{b} (AB) = T_{2} (AC)\\W_{m} (AB) + W_{b} (AB) = T_{2} (AD - CD)\\(13.23) (1.5) + (26.46)(1.5) = T_{2} (3 - 0.65)\\\\T_{2} = 25.33$ N

Using equilibrium of force in vertical direction

$T_{1} + T_{2} = W_{b} + W_{m}\\T_{1} + 25.33 = 26.46 + 13.23\\T_{1} = 14.36$ N

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