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3 years ago

A current of 1.8 A delivers 2.5 C of charge. How much time was required?

Physics

2 answers:

alexira [117]3 years ago

8 0

Answer: C

Explanation:

fgiga [73]3 years ago

4 0

The current is defined as the amount of charge that flows through a given point in a certain time interval:
$I= \frac{Q}{\Delta t}$
where Q is the charge and $\Delta t$ the time interval.

In our problem, I=1.8 A and Q=2.5 C, therefore we can find the time interval by re-arranging the previous formula and plugging these numbers into it:
$\Delta t= \frac{Q}{I}= \frac{2.5 C}{1.8 A}=1.4 s$

Therefore, C) is the correct answer.

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Heike Onnes noticed that pure, solid _____ could be cooled to 4.2 K and have no resistance.

Marrrta [24]

Answer: Mercury

Explanation:

Did it on edg

6 0

2 years ago

Thermopane window is constructed, using two layers of glass 4.0 mm thick, separated by an air space of 5.0 mm.

Bond [772]

To solve this problem it is necessary to apply the concepts related to rate of thermal conduction

$\frac{Q}{t} = \frac{kA\Delta T}{d}$

The letter Q represents the amount of heat transferred in a time t, k is the thermal conductivity constant for the material, A is the cross sectional area of the material transferring heat, $\Delta T$ , T is the difference in temperature between one side of the material and the other, and d is the thickness of the material.

The change made between glass and air would be determined by:

$(\frac{Q}{t})_{glass} = (\frac{Q}{t})_{air}$

$k_{glass}(\frac{A}{L})_{glass} \Delta T_{glass} = k_{air}(A/L)_{air} \Delta T_{air}$

$\Delta T_{air} = (\frac{k_{glass}}{k_{air}})(\frac{L_{air}}{L_{glass}}) \Delta T_{glass}$

$\Delta T_{air} = (\frac{0.84}{0.0234})(\frac{5}{4}) \Delta T_{glass}$

$\Delta T_{air} = 44.9 \Delta T_{glass}$

There are two layers of Glass and one layer of Air so the total temperature would be given as,

$\Delta T = \Delta T_{glass} +\Delta T_{air} +\Delta T_{glass}$

$\Delta T = 2\Delta T_{glass} +\Delta T_{air}$

$20\°C = 46.9\Delta T_{glass}$

$\Delta T_{glass} = 0.426\°C$

Finally the rate of heat flow through this windows is given as,

$\Delta {Q}{t} = k_{glass}\frac{A}{L_{glass}}\Delta T_{glass}$

$\Delta {Q}{t} = 0.84*24*10 -3*0.426$

$\Delta {Q}{t} = 179W$

Therefore the correct answer is D. 180W.

3 0

2 years ago

For heat transfer purposes, an egg can be considered to be a 5.5-cm-diameter sphere having the properties of water. An egg that

Harrizon [31]

Answer:

The time taken is $t = 40007 sec$

Explanation:

From the question we are told that

The diameter of the egg is $d_e = 5.5cm = \frac{5.5}{100} = 5.5*10^{-2}m$

The initial temperature of egg the $T_e = 4.3^{o}C$

The temperature of the boiling water $T_b = 100^oC$

The heat transfer coefficient is $H = 800 W/m^2 \cdot K$

The final temperature is $T_e_f = 74^oC$

The thermal conductivity of water is $k = 0.607 W/m^oC$

The diffusivity of the egg $\alpha = 0.146 * 10^{-6} m^2 /s$

Using one term approximation

We have the

$\frac{T_e_f - T_b}{T_e - T_b} = Ae^{-\lambda ^2 \tau}$

The radius is $r = \frac{5.5*10^ {-2}}{2} =2.75*10^{-2}m$ Note that this radius is approximation to that of a real egg

Now we need to obtain the Biot number which help indicate the value of $A \ and \ \lambda$ to use in the above equation

The Biot number is mathematically represented as

$Bi = \frac{H r}{k}$

Substituting values

$Bi = \frac{800 * 2.75 *10^{-2}}{0.607}$

$= 36.24$

So for this value which greater than 0.1 the coefficient $\lambda_1 \ and \ A_1$ is

$\lambda = 3.06632$

$A = 1.9942$

Substituting this into equation 1 we have

$\frac{74- 100}{4.3 - 100} = 1.9942 e^{-(3.0632^2) \tau}$

$0.2717= 1.9942 e^{-(3.0632^2) \tau}$

$0.2717= 1.9942 e^{-9.383 \tau}$

$0.13624 = e^{-9.383 \tau}$

Taking natural log of both sides

$-1.993 = -9.383\ \tau$

$\tau = 0.2124$

The time required for the egg to be cooked is mathematically represented as

$t = \frac{\tau r^2}{\alpha }$

substituting value is

$= \frac{0.2124 * 2.75 *10^{-2}}{0.146 *10^{-6}}$

$t = 40007 sec$

8 0

2 years ago

An electron moves through a uniform electric field vector E = (2.80î + 5.20ĵ) V/m and a uniform magnetic field vector B = 0.400k

alina1380 [7]

Answer:

1.758820×10^11(-2.5i-0.8j) m/s^2

Explanation:

From the question, the parameters given are; E=(2.80i+ 5.20j) v/m, a uniform magnetic field,B= 0.400K T, acceleration, a= ??? and velocity vector, v= 11.0i metre per seconds (m/s)...

We can solve this problem using the formula below;

Ma= q[E+V × B] ---------------(1).

Note: q is negative, m= mass of electron.

Making acceleration,a the subject of the formula and substituting the parameters into equation (1);

a= -e/m × (2.5i + 5.2j +11.0i × 0.400K)

a= -e/m × (2.5i+5.2j-4.4j)

a= e/m × (-2.5i - 0.8j)

e/m= 1.758820×10^11 c/kg

Therefore, slotting in the value of charge to mass(e/m) ratio;

a= 1.7588×10^11×(-2.5i-0.8j) m/s^2

7 0

2 years ago

PLEASE HELP I WILL GIVE MAX POINTS

Finger [1]

you can collect water and shine a light though it and record your finings

4 0

3 years ago

Read 2 more answers