Stress is a factor that contributes to heart disease risk.true or false

What electrical force dies a Uranium nucleus exert on one of its inner electrons, located at a distance of 175 picometers (= 1.7

AlekseyPX

Answer:

correct option is d) 7.0 x 10^-7 N

Explanation:

given data

distance = 175 picometers = 1.75 × $10^{-10}$ m

to find out

electrical force

solution

we know atomic no of uranium is 92

and charge on electron is = 1.6 × $10^{-19}$ C

and electrical force is express as

electrical force = $\frac{1}{4 \pi \epsilon _o} \frac{q1q2}{r^2}$ .............1

put here value we get

electrical force = $9*10^9 \frac{92*(1.6*10^{-19})^2}{(1.75*10^{-10})^2}$

electrical force = 6.921 × $10^{-7}$ N

so correct option is d) 7.0 x 10^-7 N

5 0

2 years ago

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How do I find the speed of sound when given Frequency and displacement?

IceJOKER [234]

Answer: Taking into account sound is a wave, we can use the information of the displacement (generally given as a graph) to find the wavelength and frequency, then we can calculate the speed with the formula of the speed of a wave.

Explanation:

If we have the displacement graph of the sound wave, we can find its amplitude, its wavelength and period (which is the inverse of frequency).

Now, if we additionally have the frequency as data, we can use the equation of the speed of a wave:

$s=\lambda f$

Where:

$s$ is the speed of the sound wave

$\lambda$ is the wavelength

$f$ is the frequency

3 0

2 years ago

Problem 9: Suppose you wanted to charge an initially uncharged 85 pF capacitor through a 75 MΩ resistor to 90.0% of its final vo

Bingel [31]

Answer:

$t=14.678\times 10^{-3}s$

Explanation:

Given:

Capacitance, C = 85 pF = 85 × 10⁻¹² F

Resistance, R = 75 MΩ = 75×10⁶Ω

Charge in capacitor at any time 't' is given as:

$Q=Q_o(1-e^{-\frac{t}{RC}})$

where,

Q₀ = Maximum charge = CE

E = Initial voltage

t = time

also, Q = CV

V= Final voltage = 90% of E = 0.9E

thus, we have

$C\times 0.9E=CE(1-e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}})$

or

$0.9=1-e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}}$

or

$e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}}=1-0.9=0.1$

taking log both sides, we get

$ln(e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}})=ln(0.1)=ln(\frac{1}{10})$

or

$-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}=-ln(10)$

or

$t=75\times 10^6\ \times\ 85\times 10^{-12}\times ln{10}$

or

$t=14.678\times 10^{-3}s$

3 0

3 years ago

Can someone help me :)

zhenek [66]

Answer:

undergone a chemical change

Explanation:

6 0

2 years ago

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An ideal monatomic gas initially has a temperature of 300 K and a pressure of 5.79 atm. It is to expand from volume 420 cm3 to v

maxonik [38]

Answer:

a) The final pressure is 1.68 atm.

b) The work done by the gas is 305.3 J.

Explanation:

a) The final pressure of an isothermal expansion is given by:

$T = \frac{PV}{nR}$

$T_{i} = T_{f}$

$\frac{P_{i}V_{i}}{nR} = \frac{P_{f}V_{f}}{nR}$

Where:

$P_{i}$ : is the initial pressure = 5.79 atm

$P_{f}$ : is the final pressure =?

$V_{i}$ : is the initial volume = 420 cm³

$V_{f}$ : is the final volume = 1450 cm³

n: is the number of moles of the gas

R: is the gas constant

$P_{f} = \frac{P_{i}V_{i}}{V_{f}} = \frac{5.79 atm*420 cm^{3}}{1450 cm^{3}} = 1.68 atm$

Hence, the final pressure is 1.68 atm.

b) The work done by the isothermal expansion is:

$W = P_{i}V_{i}ln(\frac{V_{f}}{V_{i}}) = 5.79 atm*\frac{101325 Pa}{1 atm}*420 cm^{3}*\frac{1 m^{3}}{(100 cm)^{3}}ln(\frac{1450 cm^{3}}{420 cm^{3}}) = 305.3 J$

Therefore, the work done by the gas is 305.3 J.

I hope it helps you!

3 0

3 years ago