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3 years ago

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A supercapacitor is an electrical energy storage device. A supercapacitor, initially charged to 2.1 thousand millivolts, supplie

s power to an emergency beacon. The power draw from the load causes the capacitor to lose 9.9% of its charge every minute. When the charge drops below 8 hundred millivolts the beacon will cease functioning. How long will t

1 answer:

svet-max [94.6K]3 years ago

6 0

Answer:

the time taken t is 9.25 minutes

Explanation:

Given the data in the question;

The initial charge on the supercapacitor = 2.1 × 10³ mV = 2.1 V

now, every minute, the charge lost is 9.9 %

so we need to find the time for which the charge drops below 800 mV or 0.8 V

to get the time, we can use the formula for compound interest in basic mathematics;

A = P × ( (1 - r/100 )ⁿ

where A IS 0.8, P is 2.1, r is 9.9

so we substitute

0.8 = 2.1 × ( 1 - 0.099 )ⁿ

0.8/2.1 = 0.901ⁿ

0.901ⁿ = 0.381

n = 9.25 minutes

Therefore, the time taken t is 9.25 minutes

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. Consider the equation =0+0+02/2+03/6+04/24+5/120, where s is a length and t is a time. What are the dimensions and SI units of

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Answer:

See Explanation

Explanation:

Given

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Solving (a): Units and dimension of $s_0$

From the question, we understand that:

$s \to L$ --- length

$t \to T$ --- time

Remove the other terms of the equation, we have:

$s=s_0$

Rewrite as:

$s_0=s$

This implies that $s_0$ has the same unit and dimension as $s$

Hence:

$s_0 \to L$ --- dimension

$s_o \to$ Length (meters, kilometers, etc.)

Solving (b): Units and dimension of $v_0$

Remove the other terms of the equation, we have:

$s=v_0t$

Rewrite as:

$v_0t = s$

Make $v_0$ the subject

$v_0 = \frac{s}{t}$

Replace s and t with their units

$v_0 = \frac{L}{T}$

$v_0 = LT^{-1}$

Hence:

$v_0 \to LT^{-1}$ --- dimension

$v_0 \to$ $m/s$ --- unit

Solving (c): Units and dimension of $a_0$

Remove the other terms of the equation, we have:

$s=\frac{a_0t^2}{2}$

Rewrite as:

$\frac{a_0t^2}{2} = s_0$

Make $a_0$ the subject

$a_0 = \frac{2s_0}{t^2}$

Replace s and t with their units [ignore all constants]

$a_0 = \frac{L}{T^2}\\$

$a_0 = LT^{-2$

Hence:

$a_0 = LT^{-2$ --- dimension

$a_0 \to$ $m/s^2$ --- acceleration

Solving (d): Units and dimension of $j_0$

Remove the other terms of the equation, we have:

$s=\frac{j_0t^3}{6}$

Rewrite as:

$\frac{j_0t^3}{6} = s$

Make $j_0$ the subject

$j_0 = \frac{6s}{t^3}$

Replace s and t with their units [Ignore all constants]

$j_0 = \frac{L}{T^3}$

$j_0 = LT^{-3}$

Hence:

$j_0 = LT^{-3}$ --- dimension

$j_0 \to$ $m/s^3$ --- unit

Solving (e): Units and dimension of $s_0$

Remove the other terms of the equation, we have:

$s=\frac{S_0t^4}{24}$

Rewrite as:

$\frac{S_0t^4}{24} = s$

Make $S_0$ the subject

$S_0 = \frac{24s}{t^4}$

Replace s and t with their units [ignore all constants]

$S_0 = \frac{L}{T^4}$

$S_0 = LT^{-4$

Hence:

$S_0 = LT^{-4$ --- dimension

$S_0 \to$ $m/s^4$ --- unit

Solving (e): Units and dimension of $c$

Ignore other terms of the equation, we have:

$s=\frac{ct^5}{120}$

Rewrite as:

$\frac{ct^5}{120} = s$

Make $c$ the subject

$c = \frac{120s}{t^5}$

Replace s and t with their units [Ignore all constants]

$c = \frac{L}{T^5}$

$c = LT^{-5}$

Hence:

$c \to LT^{-5}$ --- dimension

$c \to m/s^5$ --- units

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The electric field at the surface of the cylinder is 51428V/m

Given data:

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The expression to calculate the electric field is given as,

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Substitute the values in the above equation,

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sergejj [24]

Answer:

$\alpha =54.7$ º

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Since $v_{y}=0$ and $y_{o}=0$

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Knowing that

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