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2 years ago

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A proton is located at the point (x = 1.0 nm, y = 0.0 nm) and an electron is located at the point (x = 0.0 nm, y = 4.0 nm). Find

the magnitude of the electrostatic force that each one exerts on the other. (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2, e = 1.6 × 10-19 C) A) 5.3 × 10-18 NB) 5.3 × 108 N C) 1.4 × 10-11 N D) 5.9 × 10-15 N

1 answer:

enot [183]2 years ago

7 0

Answer:

C : 1.4*10^(-11) N.

Explanation:

q_1 = q_2 = 1.6 * 10^(-19)

R^2 = (1)^2 + (4)^2 = 1.7 * 10^(-17) m^2

The coulomb's law is as follows:

F_e = k*q_1*q_2 / R^2

F_e = k*q^2 / R^2

F_e = (9.0*10^9) * (1.6 * 10^(-19))^2 / 1.7 * 10^(-17)

F_e = 1.35 * 10^(-11) N

Hence, answer closes to obtained is option C : 1.4*10^(-11) N.

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A radioactive isotope has a half-life of 2 hours. If a sample of the element contains 600,000 radioactive nuclei at 12 noon, how

storchak [24]

Answer: There will be 75258 nuclei left at 6 pm.

Explanation:

a) half-life of the radioactive substance:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

$t_{\frac{1}{2}}=\frac{0.69}{k}$

$k=\frac{0.69}{t_{\frac{1}{2}}}=\frac{0.693}{2hours}=0.346hours^{-1}$

b) Expression for rate law for first order kinetics is given by:

$A=A_0e^{-kt}$

where,

k = rate constant

t = time for decomposition = 6 hours ( from 12 noon to 6 pm)

A = activity at time t = ?

$A_0$ = initial activity = 600, 000

$A=600000\times e^{-0.346\times 6}$

$A=75258$

Thus there will be 75258 nuclei left at 6 pm.

7 0

2 years ago

A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate

diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

$\Delta S = \frac{Q}{T}$

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

$W = Q_{source}-Q_{sink}$

According to the data given we have to,

$Q_{source} = 200000Btu$

$T_{source} = 1500R$

$Q_{sink} = 100000Btu$

$T_{sink} = 600R$

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

$\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}$

$\Delta S_{sink} = \frac{100000}{600}$

$\Delta S_{sink} = 166.67Btu/R$

On the other hand,

$\Delta S_{source} = \frac{Q_{source}}{T_{source}}$

$\Delta S_{source} = \frac{-200000}{1500}$

$\Delta S_{source} = -133.33Btu/R$

The total change of entropy would be,

$S = \Delta S_{source}+\Delta S_{sink}$

$S = -133.33+166.67$

$S = 33.34Btu/R$

Since $S\neq 0$ the heat engine is not reversible.

PART B)

Work done by heat engine is given by

$W=Q_{source}-Q_{sink}$

$W = 200000-100000$

$W = 100000 Btu$

Therefore the work in the system is 100000Btu

4 0

2 years ago

The weight of a 72.0 kg astronaut on the Moon, where g = 1.63 m/s2 is (5 points) Select one: a. 112 N b. 117 N c. 135 N d. 156 N

kipiarov [429]

Answer: The weight of a 72.0 kg astronaut on the Moon is 117.36 N.

Explanation:

Mass of the astronaut on the moon , m= 72 kg

Acceleration due to gravity on moon,g = 1.63 $m/s^2$

According to Newton second law of motion: F = ma

This will changes to = Weight = mass × g

$Weight=72 kg\times 1.63m/s^2=117.36 N$

The weight of a 72.0 kg astronaut on the Moon is 117.36 N.

7 0

2 years ago

Read 2 more answers

Calculate the applied force of the washers on the car. First, convert the mass you recorded for one, two, three, and four washer

Andrej [43]

The mass of one washer is 0.0049 kg.

The mass of two washers is 0.0098 kg.

The mass of three washers is 0.0147 kg.

The mass of four washers is 0.0196 kg.

3 0

2 years ago

Read 2 more answers

Bill leaves his 60 W desk lamp on every day, including weekends, for eight hours. After one month (30 days), how much total ener

maxonik [38]

' W ' is the symbol for 'Watt' ... the unit of power equal to 1 joule/second.

That's all the physics we need to know to answer this question.
The rest is just arithmetic.

(60 joules/sec) · (30 days) · (8 hours/day) · (3600 sec/hour)

= (60 · 30 · 8 · 3600) (joule · day · hour · sec) / (sec · day · hour)

= 51,840,000 joules
__________________________________

Wait a minute ! Hold up ! Hee haw ! Whoa !
Excuse me. That will never do.
I see they want the answer in units of kilowatt-hours (kWh).
In that case, it's

(60 watts) · (30 days) · (8 hours/day) · (1 kW/1,000 watts)

= (60 · 30 · 8 · 1 / 1,000) (watt · day · hour · kW / day · watt)

= 14.4 kW·hour

Rounded to the nearest whole number:

14 kWh

7 0

2 years ago