PHYSICS COLLEGE

You received a shipment 20 days ago of 13 I for treatment of hyperthyroidism. What fraction of the original shipment would you still have with a half-life of 8.040 days for 31?

Answers

Answer 1

Answer:

Answer:

we will have 17.8 % of the original value

Explanation:

As we know that by radioactive decay the total number of nuclei present at any instant of time is given as

$N = N_o e^(-\lambda t)$

here we need to find the fraction of total number of nuclei present

so we will have

$(N)/(N_o) = e^(-\lambda t)$

so we have

$\lambda = (ln 2)/(8.040)$

now we have

$(N)/(N_o) = e^{-(ln 2)/(8.040)(20)}$

$(N)/(N_o) = 0.178$

so we will have 17.8 % of the original value

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By using a 2-meter stick (like the one in lab) marked in millimeters and a stopwatch that measures to 1/100h of a second, you decide to measure the speed of a motorized toy car that travels at a constant velocity. You measure out a 162.0cm interval with the 2-meter stick and time how long it takes the car to travel that distance using the stopwatch. Repeating the ex 2.95 s Calculate the average speed of the toy car What are the absolute and relative uncertainties of the distance and time measurements? Which measurement is more uncertain? Use the weakest link rule to determine the relative and absolute uncertainty in your speed estimation. Explain why it is necessary to calculate relative uncertainties. Why is absolute uncertainty not enougn ent 5 times you get the following time data: 3.11 s 3.15 s 2.84 s 2.97 s
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As a delivery truck travels along a level stretch of road with constant speed, most of the power developed by the engine is used to compensate for the energy transformations due to friction forces exerted on the delivery truck by the air and the road. If the power developed by the engine is 2.43 hp, calculate the total friction force acting on the delivery truck (in N) when it is moving at a speed of 24 m/s. One horsepower equals 746 W.

A steel cable lying flat on the floor drags a 20 kg block across a horizontal, frictionless floor. A 100 N force applied to the cable causes the block to reach a speed of 4.2 m/s in a distance of 2.0 m.What is the mass of the cable?

Answers

Answer:

m_cable = 2,676 kg

Explanation:

For this exercise we must look for the acceleration with the kinematic ce relations

v² = v₀² + 2 a x

since the block starts from rest, its initial velocity is vo = 0

a = v² / 2x

a = 4.2² /(2 2.0)

a = 4.41 m / s²

now we can use Newton's second law

Note that the mass that the extreme force has to drag is the mass of the block plus the mass of the cable.

F = (m + m_cable) a

m_cable = F / a -m

m_cable = 100 / 4.41 - 20

m_cable = 2,676 kg

Final answer:

Unfortunately, the information given does not provide enough data to determine the mass of the steel cable. This is because the force, acceleration, and distance information given only involve the mass of the block, not the cable.

Explanation:

The question is requesting the mass of the steel cable. However, given the information in the question, we don't actually have enough data to determine this. The application of the force, the acceleration of the block, and the distance it covers are all connected through Newton's second law (F = ma) and the equations of motion, but these only involve the mass of the block, not the mass of the cable. Even if we assumed the cable applies the entire 100 N force to the block, this would only allow us to solve for the acceleration of the block, not the mass of the cable. Therefore, the mass of the steel cable cannot be determined with the information provided in the question.

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Consider a single turn of a coil of wire that has radius 6.00 cm and carries the current I = 1.50 A . Estimate the magnetic flux through this coil as the product of the magnetic field at the center of the coil and the area of the coil. Use this magnetic flux to estimate the self-inductance L of the coil.

Answers

Answer:

$\phi = 1.78 *10^(-7) \ Weber$

$L = 1.183 *10^(-7) \ H$

Explanation:

From the question we are told that

The radius is $r = 6 \ cm = (6)/(100) = 0.06 \ m$

The current it carries is $I = 1.50 \ A$

The magnetic flux of the coil is mathematically represented as

$\phi = B * A$

Where B is the magnetic field which is mathematically represented as

$B = (\mu_o * I)/(2 * r)$

Where $\mu_o$ is the magnetic field with a constant value $\mu_o = 4\pi * 10^(-7) N/A^2$

substituting value

$B = (4\pi * 10^(-7) * 1.50 )/(2 * 0.06)$

$B = 1.571 *10^(-5) \ T$

The area A is mathematically evaluated as

$A = \pi r ^2$

substituting values

$A = 3.142 * (0.06)^2$

$A = 0.0113 m^2$

the magnetic flux is mathematically evaluated as

$\phi = 1.571 *10^(-5) * 0.0113$

$\phi = 1.78 *10^(-7) \ Weber$

The self-inductance is evaluated as

$L = (\phi )/(I)$

substituting values

$L = (1.78 *10^(-7) )/(1.50 )$

$L = 1.183 *10^(-7) \ H$

A 60.0-kg boy is surfing and catches a wave which gives him an initial speed of 1.60 m/s. He then drops through a height of 1.57 m, and ends with a speed of 8.50 m/s. How much nonconservative work was done on the boy

Answers

Answer:

Work = 1167.54 J

Explanation:

The amount of non-conservative work here can be given by the difference in kinetic energy and the potential energy. From Law of conservation of energy, we can write that:

Gain in K.E = Loss in P.E + Work

(0.5)(m)(Vf² - Vi²) - mgh = Work

where,

m = mass of boy = 60 kg

Vf = Final Speed = 8.5 m/s

Vi = Initial Speed = 1.6 m/s

g = 9.8 m/s²

h = height drop = 1.57 m

Therefore,

(0.5)(60 kg)[(8.5 m/s)² - (1.6 m/s)²] - (60 kg)(9.8 m/s²)(1.57 m) = Work

Work = 2090.7 J - 923.16 J

Work = 1167.54 J

Calculate the de Broglie wavelength of an electron and a one-ton car, both moving with speed of 100 km/hour. Based on your calculation could you predict which will behave like a "quantum particle" and why. Please explain each step in words and detail.

Answers

Answer :

(a). The wavelength of electron is 26.22 μm.

(b).The wavelength of car is $2.38*10^(-38)\ m$

Explanation :

Given that,

Speed = 100 km/hr

Mass of car = 1 ton

(a). We need to calculate the wavelength of electron

Using formula of wavelength

$\lambda_(e)=(h)/(p)$

$\lambda_(e)=(h)/(mv)$

Put the value into the formula

$\lambda_(e)=(6.63*10^(-34))/(9.1*10^(-31)*100*(5)/(18))$

$\lambda=0.00002622$

$\lambda=26.22*10^(-6)\ m$

$\lambda=26.22\ \mu m$

(II). We need to calculate the wavelength of car

Using formula of wavelength again

$\lambda_(e)=(6.63*10^(-34))/(1000*100*(5)/(18))$

$\lambda=2.38*10^(-38)\ m$

The wavelength of the electron is greater than the dimension of electron and the wavelength of car is less than the dimension of car.

Therefore, electron is quantum particle and car is classical.

Hence, (a). The wavelength of electron is 26.22 μm.

(b).The wavelength of car is $2.38*10^(-38)\ m$ .

the period of the satellite is exact 42.391 hours, the earth's mass is 5.98 kg and the radius of th earth is 3958.8 miles, what is the distance of the satellite from the surface of the earth in miles?

Answers

Answer:

As the mass is not written well, i will use the equation in terms of the gravitational acceleration:

The equation for the period of a satellite is:

$T = 2*pi*\sqrt{(r^3)/(g*R^2) }$

We want to find r, so isolating r we get:

$r = \sqrt[3]{x ((T)/(2*pi) )^2*g*R^2}$

Where:

T = period.

r = radius of the satellite.

R = radius of the planet.

g = gravitational acceleration of the planet.

pi = 3.14159...

g = 78999.64 mi/h^2 (value of a table)

T = 42.391 h.

R = 3958.8 miles

We can replace those values in the equation and get:

$r = \sqrt[3]{ ((42.391)/(2*3.14159) )^2*78999.64*(3958.8)^2} = 38,339.5 mi$

Now this value is measured from the center of the Earth, then the altitude of the satellite measured from the surface of the Earth will be:

H = r - R = 38,339.1mi - 3958.8mi = 34,380.3 mi

2. Turn on Show current and select Electron flow. The moving dots represent a current of electrons—tiny, negatively charged particles—moving through the wire. Voltage is a measure of how much more potential energy an electron at one end of a battery has than an electron at the other end of the battery. A. How does changing the battery’s voltage affect the current? It makes the current get more electrons which causes more movement

Answers

Answer:

Changing the battery's voltage will also change the flow of electrons through the circuit. An increase in the voltage will produce more electron movement, and a reduction in the voltage will produce less electron movement.

Explanation:

The voltage is the potential energy between two points in an electric circuit. It is also the work done per unit charge to move a charge between these two points, this work is done against the resistance (analogous to frictional forces in the wire) of the wire. The potential energy is like the push required to move an electron through an electric circuit, and negatively charged particle (electron in the case of the wire) are pulled towards the higher potential, which is conventionally at the positive terminal. Current (flow of electrons) will not occur without any potential between two points.

Changing the battery's voltage will change the push that is experienced by the electrons. If the potential is increased, the electrons will experience more push, which means there will be more movement or flow of electrons through the circuit. Reducing the battery's voltage reduces the push experienced by the electrons; meaning a reduced flow or movement of these electrons through the circuit.

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Hotel rooms in Smalltown go for $100, and 1,000 rooms are rented on atypical day.a. To raise revenue, the mayor decides to charge hotels a tax of $10 perrented room. After the tax is imposed, the going rate for hotel roomsrises to $108, and the number of rooms rented falls to 900. Calculate theamount of revenue this tax raises for Smalltown and the deadweightloss of the tax. (Hint: The area of a triangle is l/2Xbase X height.)h. The mayor now doubles the tax to $20. The price rises to $116, andthe number of rooms rented falls to 800. Calculate tax revenue anddeadweight loss with this larger tax. Are they double, more thandouble, or less than double? Explain.