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3 years ago

What is a benefit to the immune system from vitamin A

Physics

1 answer:

SashulF [63]3 years ago

5 0

Vitamin A may reduce the risk for certain types of cancers. It also increases the activity of immune system cells.

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You heat up a 3.0 kg aluminum pot with 1.5 kg of water from 5 to 90o Celsius. The specific heat capacity of Aluminum is 900 J/kg

Evgen [1.6K]

Answer:

765,000Joules or 765kJ

Explanation:

The Quantity of heat required is expressed as;

Q = (mcΔt)al + (mcΔt)water

m is the mass

c is specific heat capacity

Δt is the change in temperature

Q = (3(900)(90-5)) + (1.5(4200)(90-5))

Q = 2700*85 + 6300*85

Q = (2700+6300)85

Q = 9000*85

Q = 765,000

Hence the amount of energy needed is 765,000Joules or 765kJ

8 0

3 years ago

Air temperature in a desert can reach 58.0°C (about 136°F). What is the speed of sound (in m/s) in air at that temperature?

Veronika [31]

The approximate speed of sound in dry (0% humidity) air, in meters per second, at temperatures near 0 °C, can be calculated from

$c_{air} = (331.3+0.606 \upsilon)$

Here

$\upsilon =$ Temperature in Celsius

Replacing with our values we have that

$\upsilon=58\° C$

$c_{air} = (331.3+0.606*58)$

$c_{air} = 366.1m/s$

Therefore the speed of sound in air at that temperature is 366.1m/s

3 0

3 years ago

Is it possible to produce a continuous and oriented aramid fiber–epoxy matrix composite having longitudinalandtransverse moduli

Masja [62]

Answer:

Not possible

Explanation:

$E_{cl}$ = longitudinal modulus of elasticity = 35 Gpa

$E_{ct}$ = transverse modulus of elasticity = 5.17 Gpa

$E_m$ = Epoxy modulus of elasticity = 3.4 Gpa

$V_{\rho l}$ = Volume fraction of fibre (longitudinal)

$V_{\rho t}$ = Volume fraction of fibre (transvers)

$E_f$ = Modulus of elasticity of aramid fibers = 131 Gpa

Longitudinal modulus of elasticity is given by

$E_{cl}=E_m(1-V_{\rho l})+E_fV_{\rho l}\\\Rightarrow 35=3.4(1-V_{\rho l})+131V_{\rho l}\\\Rightarrow 35=3.4-3.4V_{\rho l}+131V_{\rho l}\\\Rightarrow V_{\rho l}=\frac{35-3.4}{131-3.4}\\\Rightarrow V_{\rho l}=0.24764$

Transverse modulus of elasticity is given by

$E_{ct}=\frac{E_mE_f}{(1-V_{\rho t})E_f+V_{\rho t}E_m}\\\Rightarrow 5.17=\frac{3.4\times 131}{(1-V_{\rho t})131+V_{\rho t}3.4}\\\Rightarrow \frac{3.4\times 131}{5.17}-131=-127.6V_{\rho t}\\\Rightarrow V_{\rho t}=\frac{\frac{3.4\times 131}{5.17}-131}{-127.6}\\\Rightarrow V_{\rho t}=0.35148$

$V_{\rho l}\neq V_{\rho t}$

Hence, it is not possible to produce a continuous and oriented aramid fiber.

5 0

3 years ago

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

Most energy obtained from water is converted from _____.

Snowcat [4.5K]

Potential energy behind dams

7 0

3 years ago

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