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3 years ago

How are the three lines of defense the same

Physics

2 answers:

DedPeter [7]3 years ago

7 0

Answer:They all involve fighting pathogens.

Explanation:

xz_007 [3.2K]3 years ago

5 0

Here is an example of how the three lines of defense are the same:

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a pennyfarthing is a style of a bicycle with a very large front wheel and a small real wheel, the cyclist who sit high above and

s344n2d4d5 [400]

Answer:

In one rotation, the large wheel turns 4m.

Explanation:

The given values are:

Input distance,

= 0.64 m

Mechanical advantage,

= 0.16

As we know,

⇒ $Out. \ Distance = \frac{Inp. \ distance}{Mechanical \ advantage}$

On putting the values, we get

⇒ $=\frac{0.64}{0.16}$

⇒ $=4 \ m$

4 0

3 years ago

As your hand moves back and forth to generate longitudinal pulses in a spiral spring, your hand completes 3.19 back-and-forth cy

lesantik [10]

Explanation:

Since, it is given that one hand completes 3.19 vibrations in 8.46 sec. Therefore, in one second the number of vibrations will be as follows.

$\frac{3.19}{8.46}$

= 0.377 vibrations

Hence, frequency (f) = 0.38 Hz

Now, formula to calculate the speed is as follows.

v = $f \times \lambda$

or, $\lambda = \frac{v}{f}$

= $\frac{0.6 cm/s}{0.38}$

= 1.57 cm

Thus, we can conclude that the wavelength is 1.57 cm.

3 0

3 years ago

A 4.80 Kg watermelon is dropped from rest from the roof of an 18.0 m building. Calculate the work done by gravity on the waterme

Nonamiya [84]

Answer:

Work, W = 846.72 Joules

Explanation:

Given that,

Mass of the watermelon, m = 4.8 kg

It is dropped from rest from the roof of 18 m building. We need to find the work done by the gravity on the watermelon from the roof to the ground. It is same as gravitational potential energy i.e.

W = mgh

$W=4.8\ kg\times 9.8\ m/s^2\times 18\ m$

W = 846.72 Joules

So, the work done by the gravity on the watermelon is 846.72 Joules. Hence, this is the required solution.

7 0

3 years ago

What is common between the quantities of area, density and speed?

Rainbow [258]

Answer:

They are all s

calar quantities

5 0

3 years ago

Read 2 more answers

xConsider the following reduction potentials: Cu2+ + 2e– Cu E° = 0.339 V Pb2+ + 2e– Pb E° = –0.130 V For a galvanic cell employi

slega [8]

Answer:

Approximately $\rm 90\; kJ$ .

Explanation:

Cathode is where reduction takes place and anode is where oxidation takes place. The potential of a electrochemical reaction ( $E^{\circ}(\text{cell})$ ) is equal to

$E^{\circ}(\text{cell}) = E^{\circ}(\text{cathode}) - E^{\circ}(\text{anode})$ .

There are two half-reactions in this question. $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ and $\rm Pb^{2+} + 2\,e^{-} \rightleftharpoons Pb$ . Either could be the cathode (while the other acts as the anode.) However, for the reaction to be spontaneous, the value of $E^{\circ}(\text{cell})$ should be positive.

In this case, $E^{\circ}(\text{cell})$ is positive only if $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ is the reaction takes place at the cathode. The net reaction would be

$\rm Cu^{2+} + Pb \to Cu + Pb^{2+}$ .

Its cell potential would be equal to $0.339 - (-0.130) = \rm 0.469\; V$ .

The maximum amount of electrical energy possible (under standard conditions) is equal to the free energy of this reaction:

$\Delta G^{\circ} = n \cdot F \cdot E^{\circ} (\text{cell})$ ,

where

$n$ is the number moles of electrons transferred for each mole of the reaction. In this case the value of $n$ is $2$ as in the half-reactions.
$F$ is Faraday's Constant (approximately $96485.33212\; \rm C \cdot mol^{-1}$ .)

$\begin{aligned}\Delta G^{\circ} &= n \cdot F \cdot E^{\circ} (\text{cell})\cr &= 2\times 96485.33212 \times (0.339 - (-0.130)) \cr &\approx 9.0 \times 10^{4} \; \rm J \cr &= 90\; \rm kJ\end{aligned}$ .

5 0

3 years ago