So what would the answer for D be?

Robert Hook discovered cells when viewing a _____ under a microscope.

sp2606 [1]

<span>What you need to do while answering this questions, is ask yourself what has cells - only if a thing has cells can you see those cells under a microscope. Objects of animal and plant origin have cells, so blood, plant and cork (made of tree bark) can have cells, and a box too, if it's made of wood. So we can''t exclude any answers based on this. We must then know the story of Robert Hook - and it was in fact a cork. He did this discovery around 1655. At the time his main interest was the microscope rather than the cork, and he used to cork to demonstrate the function of the microscope. The correct answer is CORK.</span>

3 0

3 years ago

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The logarithm of x, written log(x), tells you the power to which you would raise 10 to get x. So, if y=log(x), then x=10^y. It i

fomenos

To solve this problem it is necessary to apply the rules and concepts related to logarithmic operations.

From the definition of logarithm we know that,

$Log_{10}(10) = 1$

In this way for the given example we have that a logarithm with base 10 expressed in the problem can be represented as,

$log_{10}(1,000,000)$

We can express this also as,

$log_{10}(10^6)$

By properties of the logarithms we know that the logarithm of a power of a number is equal to the product between the exponent of the power and the logarithm of the number.

So this can be expressed as

$6*log_{10}(10)$

Since the definition of the base logarithm 10 of 10 is equal to 1 then

$6*1=6$

The value of the given logarithm is equal to 6

8 0

3 years ago

What is the primary determinant of the voltage developed by a battery?

Fittoniya [83]

For the answer to the question above asking what is the primary determinant of the voltage developed by a battery?the answer is that the <span>the nature of the materials in the reaction that is the primary determinant of the voltage from a battery.</span>

5 0

3 years ago

4.77 Augment the rectifier circuit of Problem 4.70 with a capacitor chosen to provide a peak-to-peak ripple voltage of (i) 10% o

goblinko [34]

The question incomplete! The complete question along with answer and explanation is provided below.

Question:

Augment the rectifier circuit of Problem 4.68 with a capacitor chosen to provide a peak-to-peak ripple voltage of (i) 10% of the peak output and (ii) 1% of the peak output. In each case:

(a) What average output voltage results?

(b) What fraction of the cycle does the diode conduct?

(c) What is the average diode current?

(d) What is the peak diode current?

Problem 4.68:

A half-wave rectifier circuit with a 1-kΩ load operates from a 120-V (rms) 60-Hz household supply through a 10-to-1 step-down transformer. It uses a silicon diode that can be modeled to have a 0.7-V drop for any current.

Given Information:

Input voltage = 120 Vrms

10 to 1 step-down transformer

Voltage drop at diode = 0.7 V

Load resistance = R = 1 kΩ

Required Information:

(i) 10% of the peak output and (ii) 1% of the peak output. In each case:

(a) What average output voltage results?

(b) What fraction of the cycle does the diode conduct?

(c) What is the average diode current?

(d) What is the peak diode current?

Answer:

Case (i)

Vavg = 15.45 V

Conduction of diode = 7.11 %

Iavg = 0.232 A

Ip = 0.449 A

Case (ii)

Vavg = 16.18 V

Conduction of diode = 2.25 %

Iavg = 0.735 A

Ip = 1.453 A

Explanation:

Voltage at the secondary side of the transformer is

Vrms = Vpri/turn ratio

Vrms = 120/10 = 12 V

The relation between rms voltage and peak voltage is

Vp = Vrms/√2

Vp = 12√2 = 16.97 V

Vd = 0.7 V

First we will calculate all the required parameters for the 10% ripple voltage and then for 1% ripple voltage.

case (i) 10% of the peak output:

(a) What average output voltage results?

Average output voltage = Vavg = Vp - Vd - 0.5Vr

Where Vp is the peak output voltage Vd is the voltage drop of diode and Vr is the ripple voltage which is given as a percentage of Vp

Vavg = Vp - Vd - 0.5Vr

Vavg = 16.97 - 0.7 - 0.5[0.1(16.97 - 0.7)]

Vavg = 15.45 V

(b) What fraction of the cycle does the diode conduct?

ω = √2Vr/Vp - Vd

ω = √2*0.1(Vp-Vd)/Vp - Vd

ω = √2*0.1(16.97-0.7)/16.97 - 0.7

ω = 0.447 rad

Conduction of diode = (ω/2π)*100

Conduction of diode = (0.447/2π)*100

Conduction of diode = 7.11 %

(c) What is the average diode current?

Average current = Iavg = Vavg/R[ 1 + π( √2(Vp - Vd)/0.1(Vp-Vd))]

Average current = Iavg = 15.45/1000[ 1 + π( √2(16.97 - 0.7)/0.1(16.97-0.7))]

Average current = Iavg = 0.232 A

(d) What is the peak diode current?

Peak current = Ip = Vavg/R[ 1 + 2π( √2(Vp - Vd)/0.1(Vp-Vd))]

Peak current = Ip = 15.45/1000[ 1 + 2π( √2(16.97 - 0.7)/0.1(16.97-0.7))]

Peak current = Ip = 0.449 A

case (ii) 1% of the peak output:

(a) What average output voltage results?

Vavg = 16.97 - 0.7 - 0.5[0.01(16.97 - 0.7)]

Vavg = 16.18 V

(b) What fraction of the cycle does the diode conduct?

ω = √2*0.01(Vp-Vd)/Vp - Vd

ω = √2*0.01(16.97-0.7)/16.97 - 0.7

ω = 0.1417 rad

Conduction of diode = (0.1417/2π)*100

Conduction of diode = 2.25 %

(c) What is the average diode current?

Average current = Iavg = 16.18/1000[ 1 + π( √2(16.97 - 0.7)/0.01(16.97-0.7))]

Average current = Iavg = 0.735 A

(d) What is the peak diode current?

Peak current = Ip = 16.18/1000[ 1 + 2π( √2(16.97 - 0.7)/0.01(16.97-0.7))]

Peak current = Ip = 1.453 A

3 0

3 years ago

If Frequency F, velocity v, and density D are considered fundamental units, the dimensional formula for momentum will be :

gizmo_the_mogwai [7]

Let's see

Momentum be P

$\\ \rm\Rrightarrow P=[Frequency]^a[velocity]^b[Density]^c$

$\\ \rm\Rrightarrow [P]=[F]^a[v]^b[D]^c$

$\\ \rm\Rrightarrow [M^1L^1T^{-1}]=[T^{-1}]^a[L^1T^{-1}]^b[M^1L^{-3}]^c$

$\\ \rm\Rrightarrow MLT^{-1}=T^{-a}L^bT^{-b}M^cL^{-3c}$

$\\ \rm\Rrightarrow MLT^{-1}=T^{-a-b}L^{b-3c}M^c$

On comaparing

c=1

So

b-3c=1
b-3=1
b=1+3
b=4

and

-a-b=-1
-a-4=-1
-a=-1+4=3
a=-3

So the unit is

DV⁴/F³

5 0

2 years ago

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