PHYSICS COLLEGE

We showed that the length of the pendulum of period 2.000 seconds on the Earth’s surface was 0.99396 meters. What period would this same pendulum have on the surface of Mars? What length would the pendulum be in order to have a period of 2.000 seconds?

Answers

Answer 1

Answer:

To solve this problem it is necessary to apply the concepts related to the Period based on gravity and length.

Mathematically this concept can be expressed as

$T= 2\pi \sqrt{(l)/(g)}$

Where,

l = Length

g = Gravitational acceleration

First we will find the period that with the characteristics presented can be given on Mars and then we can find the length of the pendulum at the desired time.

The period on Mars with the given length of 0.99396m and the gravity of the moon (approximately $1.62m / s ^ 2)$ will be

$T= 2\pi \sqrt{(l)/(g)}$

$T= 2\pi \sqrt{(0.99396)/(1.62)}$

$T = 4.921seg$

For the second question posed, it would be to find the length so that the period is 2 seconds, that is:

$T= 2\pi \sqrt{(l)/(g)}$

$2= 2\pi \sqrt{(l)/(1.62)}$

$l = 0.16414m$

Therefore, we can observe also that the shorter distance would be the period compared to the first result given.

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The concentration of Biochemical Oxygen Demand (BOD) in a river just downstream of a wastewater treatment plant’s effluent pipe is 75 mg/L. If the BOD is destroyed through a first-order reaction with a rate constant equal to 0.05/day, what is the BOD concentration 50 km downstream? The velocity of the river is 15 km/day.

Answers

Answer:

The BOD concentration 50 km downstream when the velocity of the river is 15 km/day is 63.5 mg/L

Explanation:

Let the initial concentration of the BOD = C₀

Concentration of BOD at any time or point = C

dC/dt = - KC

∫ dC/C = -k ∫ dt

Integrating the left hand side from C₀ to C and the right hand side from 0 to t

In (C/C₀) = -kt + b (b = constant of integration)

At t = 0, C = C₀

In 1 = 0 + b

b = 0

In (C/C₀) = - kt

(C/C₀) = e⁻ᵏᵗ

C = C₀ e⁻ᵏᵗ

C₀ = 75 mg/L

k = 0.05 /day

C = 75 e⁻⁰•⁰⁵ᵗ

So, we need the BOD concentration 50 km downstream when the velocity of the river is 15 km/day

We calculate how many days it takes the river to reach 50 km downstream

Velocity = (displacement/time)

15 = 50/t

t = 50/15 = 3.3333 days

So, we need the C that corresponds to t = 3.3333 days

C = 75 e⁻⁰•⁰⁵ᵗ

0.05 t = 0.05 × 3.333 = 0.167

C = 75 e⁻⁰•¹⁶⁷

C = 63.5 mg/L

Final answer:

The BOD concentration 50 km downstream from the wastewater treatment plant is approximately 15.865 mg/L.

Explanation:

To calculate the BOD concentration 50 km downstream, we need to consider the rate of dilution due to the flow of the river and the first-order reaction that destroys BOD. The concentration of BOD downstream can be calculated using the equation C2 = C1 * exp(-k * d/v), where C1 is the initial concentration, k is the rate constant, d is the distance, and v is the velocity of the river.

Plugging in the given values, we have C2 = 75 * exp(-0.05 * 50/15), which gives us a BOD concentration of approximately 15.865 mg/L 50 km downstream from the wastewater treatment plant.

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How much work would it take to push two protons very slowly from a separation of 2.00×10−10 m (a typical atomic distance) to 3.00×10−15 m (a typical nuclear distance)?

Answers

Electric potential is the amount of work needed to move a unit charge from a point to a specific point against an electric field. The work would it take to push two protons will be 7.7×10⁻¹⁴.

What is electric potential?

Electric potential is the amount of work needed to move a unit charge from a point to a specific point against an electricfield.

The given data in the problem is;

q is the charge= 1.6 ×10⁻¹⁹ C

V is the electric potential

r₁ is the first separation distance= 2.00×10−10 m

r₂ is the second separation distance= 3.00×10−15 m

The electric potential generated by the proton at rest at the two points, using the formula:

Firstly the electric potential at loction 1

$\rm V=(Kq)/(r) \n\n v_i= 9* 10^9 * (1.6*10^(-19))/(2.0*10^(-10))$

The electric potential at loction 2

$V_f = 9 * 10^9 (1.6 * 10^(-19))/(3.0*10^(-15)) \n\n \rm v_f= 4.8 *10^5 \ V$

The product of difference of electric potential and charge is defined as the workdone.

$\rm W= q \triangle V \n\n \rm W= 1.6 * 10^-19 *( 4.8*10^5 -7.2) \n\n \rm W= 7.7 * 10^(-14)$

Hence the work would it take to push two protons will be 7.7×10⁻¹⁴.

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We can visualize the problem in another way, which is equivalent but easier to solve: let's imagine we hold one proton in the same place, and we move the other proton from a distance of 2.00×10−10 m to a distance of 3.00×10−15 m from the first proton. How much work is done?

The work done is equal to the electric potential energy gained by the proton:

$W=q \Delta V$

where $q=1.6 \cdot 10^(-19)C$ is the charge of the proton and $\Delta V$ is the potential difference between the final position and the initial position of the proton. To calculate this $\Delta V$ , we must calculate the electric potential generated by the proton at rest at the two points, using the formula:

$V=k(Q)/(r)$

where $k=9.0 \cdot 10^9 N m^2 C^(-2)$ is the Coulomb constant and Q is the proton charge. Substituting the initial and final distance of the second proton, we find

$V_i = (9.0 \cdot 10^9 )(1.6 \cdot 10^(-19))/(2.0 \cdot 10^(-10))=7.2 V$

$V_f = (9.0 \cdot 10^9 )(1.6 \cdot 10^(-19))/(3.0 \cdot 10^(-15))=4.8 \cdot 10^5 V$

Therefore, the work done is

$W=q \Delta V=(1.6 \cdot 10^(-19)C)(4.8 \cdot 10^5 V-72 V)=7.7 \cdot 10^(-14) J$

In part (a), suppose that the box weighs 128 pounds, that the angle of inclination of the plane is θ = 30°, that the coefficient of sliding friction is μ = 3 /4, and that the acceleration due to air resistance is numerically equal to k m = 1 3 . Solve the differential equation in each of the three cases, assuming that the box starts from rest from the highest point 50 ft above ground. (Assume g = 32 ft/s2 and that the downward velocity is positive.)

Answers

Answer:

v(t) = 21.3t

v(t) = 5.3t

$v(t) = 48 -48 e ^{ (t)/(9)}$

Explanation:

When no sliding friction and no air resistance occurs:

$m(dv)/(dt) = mgsin \theta$

where;

$(dv)/(dt) = gsin \theta , 0 < \theta < ( \pi)/(2)$

Taking m = 3 ; the differential equation is:

$3 (dv)/(dt)= 128*(1)/(2)$

$3 (dv)/(dt)= 64$

$(dv)/(dt)= 21.3$

By Integration;

$v(t) = 21.3 t + C$

since v(0) = 0 ; Then C = 0

v(t) = 21.3t

ii)

When there is sliding friction but no air resistance ;

$m (dv)/(dt)= mg sin \theta - \mu mg cos \theta$

Taking m =3 ; the differential equation is;

$3 (dv)/(dt)=128*(1)/(2) -(√(3) )/(4)*128*(√(3) )/(4)$

$(dv)/(dt)= 5.3$

By integration; we have ;

v(t) = 5.3t

iii)

To find the differential equation for the velocity (t) of the box at time (t) with sliding friction and air resistance :

$m (dv)/(dt)= mg sin \theta - \mu mg cos \theta - kv$

The differential equation is :

= $3 (dv)/(dt)=128*(1)/(2) - ( √( 3))/(4)*128 *( √( 3))/(2)-(1)/(3)v$

= $3 (dv)/(dt)=16 -(1)/(3)v$

By integration

$v(t) = 48 + Ce ^{(t)/(9)$

Since; V(0) = 0 ; Then C = -48

$v(t) = 48 -48 e ^{ (t)/(9)}$

You are traveling on an interstate highway at the posted speed limit of 70 mph when you see that the traffic in front of you has stopped due to an accident up ahead. You step on your brakes to slow down as quickly as possible. Assume that you to slow down to 30 mph in about 5 seconds. A) With this same average acceleration, how much longer would it take you to stop?B) What total distance would you travel from when you first apply the brakes until the car stops?

Answers

A.The time taken for the car to stop is 8.75 s

B.The distance travelled when the brakes were applied till the car stops is 136.89 m

A. Determination of the time taken for the car to stop.

We'llbegin bycalculatingthedecelerationof thecar

Initial velocity (u) = 70 mph = 0.447 × 70 = 31.29 m/s

Final velocity (v) = 30 mph = 0.447 × 30 = 13.41 m/s

Time (t) = 5 s

Deceleration (a) =?

$a \: = (v \: - u)/(t) \n \n a = (13.41 - 31.29)/(5) \n \n a \: = ( - 17.88)/(5) \n \n$

a = –3.576 m/s²

Finally,we shall determine the time taken for the car to stop.

Initial velocity (u) = 31.29 m/s

Final velocity (v) = 0 m/s

Deceleration (a) = –3.576 m/s²

Time (t) =?

$v \: = u \: + at \n 0 \: = 31.29 \: + \: ( - 3.576 * t) \n 0 \: = 31.29 \: - 3.576 * t \n collet \: like \: terms \n 0 - 31.29 \: = - 3.576 * t \n - 31.29 \: = - 3.576 * t \n divide \: both \: side \: by \: - 3.576 \n t \: = (- 31.29)/(- 3.576) \n$

t = 8.75 s

Thus, the time taken for the car to stop is 8.75 s

B.Determination of the total distance travelled when the brakes were applied.

Initial velocity (u) = 31.29 m/s

Final velocity (v) = 0 m/s

Deceleration (a) = –3.576 m/s²

Distance (s) =?

${v}^(2) = {u}^(2) + 2as \n {0}^(2) = {31.29}^(2) + (2 * - 3.576 * s) \n 0 = 979.0641 - 7.152 s \n collect \: like \: terms \n 0 - 979.0641 = - 7.152 s \n - 979.0641 = - 7.152 s \n divide \: both \: side \: by \: - 7.152 \n s = (- 979.0641)/(- 7.152) \n \n$

s = 136.89 m

Therefore, the total distance travelled by the car when the brakes were applied is 136.89 m

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Answer:8.75 s,

136.89 m

Explanation:

Given

Initial velocity $=70 mph\approx 31.29 m/s$

velocity after 5 s is $30 mph\approx 13.41 m/s$

Therefore acceleration during these 5 s

$a=(v-u)/(t)$

$a=(13.41-31.29)/(5)=-3.576 m/s^2$

therefore time required to stop

v=u+at

here v=final velocity =0 m/s

initial velocity =31.29 m/s

$0=31.29-3.576* t$

$t=(31.29)/(3.576)=8.75 s$

(b)total distance traveled before stoppage

$v^2-u^2=2as$

$0^2-31.29^2=2* (-3.576)\cdot s$

s=136.89 m

A white blood cell has a diameter of approximately 12 micrometers or 0.012 um a model represents its diameter as 24 um what ratio of model size

Answers

Answer:

The ratio of the model size is 1 : 2000

Explanation:

Given

Real Diameter = 0.012 um

Scale Diameter = 24 um

Required

Determine the scale ratio

The scale ratio is calculated as follows;

$Scale = (Real\ Measurement)/(Scale\ Measurement)$

Substitute values for real and scale measurements

$Scale = (0.012\ um)/(24\ um)$

Divide the numerator and the denominator by 0012um

$Scale = (1)/(2000)$

Represent as ratio

$Scale = 1 : 2000$

Hence, the ratio of the model size is 1 : 2000

The ratio of the model size to the actual size is 1 : 2000. This means the model represents the white blood cell's diameter 2000 times larger than its actual size.

The ratio of the model size to the actual size can be calculated using the given measurements:

Actual Diameter = 0.012 um

Model Diameter = 24 um

Ratio = Model Diameter / Actual Diameter

Ratio = 24 um / 0.012 um

Ratio = 2000

So, the ratio of the model size to the actual size is 1 : 2000. This means the model represents the white blood cell's diameter 2000 times larger than its actual size.

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The girlfriend of a PHS1282 student proposes to him and offers him a "golden" ring in the process. However, his girlfriend is a student in the Department of Accountancy and Finance, and would struggle to afford a gold ring. Additionally she is known to have a slightly dodgy character. After initially being flattered by the marriage proposal he therefore gets suspicious about whether the ring is actually made of gold. He therefore bring the ring to the PHS「282 lab and measure its mass to be 2.80± 0.02 g and its volume to be 0.16 ± 0.03 cm3 . Gold has a mass density of 19.3 g/cm3. Could the ring be made of gold? (Explain your answer) . Brass has a mass density of 8.4 g/cm3 to 8.7 g/cm3 (depending on the composition of the alloy). Could the ring be made of brass?

Answers

Answer:

The density of the ring is:

$\rho=17.5\pm 3 \, g/cm^3$

This means the ring could very well be made of gold, but it is very unlikely that it is made of brass.

Explanation:

For a quantity f(x,y) that depends on other quantities (in this case two) x and y, the error is given by:

$\sigma_f=\sqrt{\left((\partial f)/(\partial x)\right)^2\sigma_x^2+\left((\partial f)/(\partial y)\right)^2\sigma_y^2 }$

where $\sigma_x$ and $\sigma_y$ are the standard deviations on errors of the variables $x$ and $y$ .

In our case $\rho=f(m,V)=(m)/(V)$ where $m$ is the mass and $V$ is the volume.

Knowing that $\sigma_m=0.02$ and $\sigma_V=0.03$ we can estimate the error on the density

$\sigma_(\rho)=\sqrt{\left((1)/(V)\left)^2\sigma_m^2+\left((m)/(V^2)\right)^2\sigma_V^2}$ $\approx 3$ (values were directly plugged)

The density is by using the given values

$\rho=(m)/(V)=(2.80)/(0.16)=17.5 \, g/cm^3$

The density with error is given by

$\rho=(m)/(V)\pm \sigma_(\rho)=17.5\pm 3 \, g/cm^3$

Which means it could go as high as 20.5 or as low as 14.5, Meaning that the ring could very well be made of gold, but it is very unlikely that it is made of brass.

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