PHYSICS COLLEGE

You are using a Geiger counter to measure the activity of a radioactive substance over the course of several minutes. If the reading of 400. counts has diminished to 100. counts after 90.3 minutes, what is the half-life of this substance?

Answers

Answer 1

Answer:

Answer : The half-life of this substance will be, 45 minutes.

Explanation :

First we have to calculate the value of rate constant.

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

$k=(2.303)/(t)\log(a)/(a-x)$

where,

k = rate constant = ?

t = time passed by the sample = 90.3 min

a = initial amount of the reactant = 400

a - x = amount left after decay process = 100

Now put all the given values in above equation, we get

$k=(2.303)/(90.3min)\log(400)/(100)$

$k=1.54* 10^(-2)\text{ min}^(-1)$

Now we have to calculate the half-life of substance, we use the formula :

$k=(0.693)/(t_(1/2))$

$1.54* 10^(-2)\text{ min}^(-1)=(0.693)/(t_(1/2))$

$t_(1/2)=45min$

Therefore, the half-life of this substance will be, 45 minutes.

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The minimum charge on any object cannot be less than

Answers

Answer:

1.6 x 10^{-19} Coulombs

Explanation:

In Physics, the standard unit of measurement of a charge is Coulombs and it's denoted by C. Also, the symbol for denoting a charge is Q.

In Chemistry, electrons can be defined as subatomic particles that are negatively charged and as such has a magnitude of -1.

The minimum charge on any object such as an electron cannot be less than 1.6 x 10^{-19} Coulombs and it's usually referred to as the fundamental unit of charge.

What frequency corresponds to a period of 4.31s.
T =1/f = 1/4.31s = 0.232hz correct?

Answers

Answer:correct

Explanation: Period T is the reciprocal of frequency (i.e T=1/f)

Frequency is the reciprocal of period (i.e F= 1/T)

Therefore if T=4.31s

Frequency F= 1/4.31s=0.232hz

You are an electrician installing the wiring in a new home. The homeowner desires that a ceiling fan with light kits be installed in five different rooms. Each fan contains a light kit that can accommodate four 60-watt lamps. Each fan motor draws a current of 1.8 amperes when operated on high speed. It is assumed that each fan can operate more than three hours at a time and therefore must be considered a continuous-duty device. The fans are to be connected to a 15-ampere circuit. Because the devices are continuous-duty, the circuit current must be limited to 80% of the continuous connected load. How many fans can be connected to a single 15-ampere circuit

Answers

Answer:

3 fans per 15 A circuit

Explanation:

From the question and the data given, the light load let fan would have been

(60 * 4)/120 = 240/120 = 2 A.

Next, we add the current of the fan motor to it, so,

2 A + 1.8 A = 3.8 A.

Since the devices are continuos duty and the circuit current must be limited to 80%, then the Breaker load max would be

0.8 * 15 A = 12 A.

Now, we can get the number if fans, which will be

12 A/ 3.8 A = 3.16 fans, or approximately, 3 fans per 15 A circuit.

Final answer:

The total power draw of each fan is 3.8 amperes. Thus, considering a limit of 80% usage of 15 amperes, only 3 fans can be connected to a single circuit to keep the total power draw below 12 amperes.

Explanation:

The question is asking how many ceiling fans, each with a certain power draw, can be connected on a single 15-ampere circuit, considering that each fan is a continuous-duty device. The power draw of each fan when the motor is operated at high speed and the light kit is fully loaded is the sum of the power draw of the motor and the light kit. As the power draw of each motor is 1.8 amperes and the light kit is 240 watts or 2 amperes (calculated using the formula Power = Voltage x Current; assuming a voltage of 120 volts), the total power draw of each fan is 3.8 amperes. Considering the limit of 80% of the continuous load, only 12 amperes (80% of 15) can be used. Thus, 3 fans can be connected to the circuit as it reaches 11.4 amperes, close enough to the 12 amperes limit.

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What kind of exercise should you do when you're cooling down after an
intense workout?

Answers

Answer:

Planks?

Explanation:

It's kinda resting

Answer:

weights

walking

stretching

etc.

A small segment of wire contains 10 nC of charge. The segment is shrunk to one-third of its original length. A proton is very far from the wire. What is the ratio Ff/Fi of the electric force on the proton after the segment is shrunk to the force before the segment was shrunk?

Answers

The ratio of the electric force on the proton after the wire segment is shrunk to three times its original length to the force before the segment was shrunk is 3.

The electric force between a point charge and a segment of wire with a distributed charge is given by Coulomb's law.

The formula for the electric force on a point charge q due to a segment of wire with charge Q distributed along its length L is:

$F=(k.q.Q)/(L)$

where:

F is the electric force on the point charge,

k is Coulomb's constant ( 8.988 × 1 0⁹ Nm²/ C²),

q is the charge of the point charge,

Q is the charge distributed along the wire segment, and

L is the length of the wire segment.

When the wire segment is shrunk to one-third of its original length, the new length becomes 1/3 L.

The chargedistribution remains the same, only the length changes.

So, the new electric force $F_f$ on the proton after the segment is shrunk becomes:

$F_f=(k.q.Q)/((1)/(3)L)$

The original electric force $F_i$ on the proton before the segment was shrunk is:

$F_i = (k.q.Q)/(L)$

let's find the ratio $(F_f)/(F_i)$ :

$(F_f)/(F_i)=((k.q.Q)/((1)/(3)L))/((k.q.Q)/(L))$

$(F_f)/(F_i)=3$

Hence, the ratio of the electric force on the proton after the wire segment is shrunk to the force before the segment was shrunk is 3.

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Final answer:

The ratio of the electric force on the proton after the wire segment is shrunk is equal to the ratio of their charges.

Explanation:

The ratio of the electric force on the proton after the wire segment is shrunk to the force before the segment was shrunk can be found using Coulomb's law. Coulomb's law states that the electric force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

In this case, the charges involved are the charge of the wire segment and the charge of the proton. Since the wire segment contains 10 nC of charge, we can consider it as one of the charged objects. The proton is very far from the wire, so we can assume that the distance between them remains the same before and after the wire segment is shrunk. Therefore, the ratio of the electric force on the proton after the segment is shrunk to the force before the segment was shrunk is equal to the ratio of their charges.

Let's assume that the initial force on the proton is Fi and the final force on the proton is Ff. Using the given information, we have:

Fi = k(q1 * q2) / r^2

where k is the electrostatic constant, q1 and q2 are the charges of the wire segment and the proton respectively, and r is the distance between them.

After the wire segment is shrunk to one-third of its original length, the charge of the wire segment remains the same and the distance between the wire segment and the proton also remains the same. Therefore, the ratio Ff/Fi can be calculated as:

Ff/Fi = (q1 * q2) / (q1 * q2) = 1

A ball is dropped from a 19m high cliff. The acceleration on the ball was 9.8m/s². What was the ball's final velocity before hitting the ground?