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3 years ago

8

Which law of physics relates electric fields and current

1 answer:

riadik2000 [5.3K]3 years ago

7 0

Answer:

Ohms law

Explanation:

Which states that the current flowing through any cross-section of the conductor is directly proportional to the potential differenceapplied across its end, provided physical conditions like temperature and pressure remain constant.

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You have a set of calipers that can measure thicknesses of a few inches with an uncertainty of ± 0.005 inches. You measure the t

Bond [772]

Answer:

a) x = (0.0114 ± 0.0001) in , b) the number of decks is 5

Explanation:

a) The thickness of the deck of cards (d) is measured and the thickness of a card (x) is calculated

x = d / 52

x = 0.590 / 52

x = 0.011346 in

Let's look for uncertainty

Δx = dx /dd Δd

Δx = 1/52 Δd

Δx = 1/52 0.005

Δx = 0.0001 in

The result of the calculation is

x = (0.0114 ± 0.0001) in

b) You want to reduce the error to Δx = 0.00002, the number of cards to be measured is

#_cards = n 52

The formula for thickness is

x = d / n 52

Uncertainty

Δx = 1 / n 52 Δd

n = 1/52 Δd / Δx

n = 1/52 0.005 / 0.00002

n = 4.8

Since the number of decks must be an integer the number of decks is 5

3 0

3 years ago

.Why does it take double the applied force to move a mass double the size?

Likurg_2 [28]

Answer:

NVIDIA GeForce RTX 3080 10GB GDDR6X PCI Express 4.0 Graphics Card

Explanation:

6 0

2 years ago

The following equation is an example of decay?<br><br> 232/90 TH---4/2 HE +228/88 RA?

DaniilM [7]

Answer:

Alpha decay

Explanation:

Alpha decay is one of the three major types of decays, others being, beta decay and gamma decay.
<em><u>When a radioactive isotope undergoes alpha decay it emits alpha particles. An alpha particle is equivalent to the nucleus of Helium atom.</u></em>
<em><u>Therefore, an atom undergoing decay, its atomic mass is decreased by 4 and its atomic number is decreased by 2. </u></em>
Thus, since 232/90 Th, has undergone alpha decay its mass number is reduced by 4 to 228 and its atomic number by 2 to 88, and becomes 228/88 Ra.

5 0

2 years ago

A piston-cylinder device contains Helium gas initially at 150 kPa, 20 o C, and 0.5m 3 . The helium is now compressed in a polytr

Molodets [167]

Answer:

Explanation:

Given

$P_1=150 kPa$

$T_1=20^{\circ}C$

$V_1=0.5 m^3$

$T_2=140^{\circ}C$

$P_2=400 kPa$

R for Helium $R=2.076$

$c_v=3.115 kJ/kg-K$

mass of gas $m=\frac{P_1V_1}{RT_1}$

$m=\frac{150\times 0.5}{2.076\times 293}$

$m=0.123 kg$

Similarly $V_2$ can be found

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

$V_2=0.264 m^3$

Work done $W=\int_{V_1}^{V_2}PdV$

$W=\frac{P_2V_2-P_1V_1}{n-1}$

$W=\frac{mR(T_2_T_1)}{n-1}$

Since it is a polytropic Process

therefore $PV^n=c$

$P_1V_1^n=P_2V_2^n$

$(\frac{V_1}{V_2})^n=\frac{P_2}{P_1}$

$(\frac{0.5}{0.264})^n=\frac{400}{150}$

$n=\frac{\ln 2.66}{\ln 1.893}$

$n=1.533$

$W=\frac{0.123\times 2.076(140-20)}{1.533-1}$

$W=57.48 kJ$

From Energy balance

$E_{in}-E_{out}=\Delta E_{system}$

Neglecting kinetic and Potential Energy change

$Q_{in}+W_{in}=change\ in\ Internal\ Energy$

Change in Internal Energy $\Delta U=u_2-u_1$

$\Delta U=mc_v(T_2-T_1)$

$\Delta U=0.123\times 3.115(140-20)$

$\Delta U=45.977 kJ$

$Q_{in}+57.48=45.977$

$Q_{in}=-11.50 kJ$

i.e. Heat is being removed

3 0

2 years ago

PLEASE HELP 100 POINTS

tia_tia [17]

Answer:

okay, So Um,

Explanation:

BTW, There are only 50 points, brainly subtracts half of the points

8 0

2 years ago