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3 years ago

10

A change in the direction of a wave when the wave finds an obstacle or an edge, such as an opening bending of waves around a bar

rier is called __________________.

1 answer:

WARRIOR [948]3 years ago

5 0

A diffracation is the correct answer. :D

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Find h , the maximum height attained by the projectile. express the maximum height in terms of v 0 , Î¸ , and g .

lutik1710 [3]

solution:

the utmost height would be comprehensive while y'(t)=0. (on the suitable of the trajectory, the y speed is 0.) $y(t)=(88sin20)t-16t^2 y'(t)=88sin20-32t So y'(t)=0 while t=(88sin20)/32$

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3 years ago

If the concentration of Ag+ is 0.0115 M, the concentration of H+ is 0.355 M, and the pressure of H2 is 1.00 atm, calculate the c

scoundrel [369]

<u>Answer:</u> The cell potential is 0.712 V

<u>Explanation:</u>

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, silver will undergo reduction reaction will get reduced. Hydrogen will undergo oxidation reaction and will get oxidized

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> $H_2(1.00atm)\rightarrow 2H^{+}(0.355M)+2e^-;E^o_{H^{+}/H_2}=0.00V$

<u>Reduction half reaction:</u> $Ag^{+}(0.0115M)+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V$ ( × 2 )

<u>Net reaction:</u> $H_2(1.00atm)+2Ag^{+}(0.0115M)\rightarrow 2H^{+}(0.355M)+2Ag(s)$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=0.80-(0.00)=0.80V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^2}{[Ag^{+}]^2}$

where,

$E_{cell}$ = electrode potential of the cell = ?V

$E^o_{cell}$ = standard electrode potential of the cell = +0.80 V

n = number of electrons exchanged = 2

$[Ag^{+}]=0.0115M$

$[H^{+}]=0.355M$

Putting values in above equation, we get:

$E_{cell}=0.80-\frac{0.059}{2}\times \log(\frac{(0.355)^2}{(0.0115)^2})\\\\E_{cell}=0.712V$

Hence, the cell potential is 0.712 V

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3 years ago

3. Why is the term cold blooded a misconception? Explain

jok3333 [9.3K]

Answer:

The term “cold-blooded” implies that these animals are in a never-ending struggle to stay warm. That really isn't correct. A cold-blooded animal can warm up their blood by being in the sun for hours.

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Vera_Pavlovna [14]

They got back in the Lunar Explorer Module

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True or False : Waves lose most of their energy when traveling across the ocean.

klemol [59]

Answer:

true

Explanation:

that's the answear

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2 years ago

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