You are at a furniture store and notice that a Grandfather clock has its time regulated by a physical pendulum that consists of a rod with a movable weight on it. When the weight is moved downward, the pendulum slows down; when it is moved upward, the pendulum swings faster. If the rod has a mass of 1.23 kg and a length of 1.25 m and the weight has a mass of [10] kg, where should the mass be placed to give the pendulum a period of 2.00 seconds

Answers

Answer 1

Answer:

The distance is 1.026 m.

Explanation:

mass of rod, M = 1.23 kg

Length, L = 1.25 m

mass, m = 10 kg

Time period, T = 2 s

Let the distance is d.

The formula of the time period is given by

[tex]T = 2\pi\sqrt\frac{\frac{1}{3}ML^2+md^2}{(M +m)g}\\\\2\times 2 = 4\pi^2\times \frac{\frac{1}{3}\times1.23\times1.25\times 1.25+ 10d^2}{(1.23 + 10)\times9.8}\\\\11.16 = 0.64 + 10d^2\\\\d= 1.026 m[/tex]


Related Questions

A 2 kg stone is dropped from a height of 100 m. How far does it travel in the third second? take g = 9.8 m/s2​

Answers

Answer:

S = 1/2 gt² = 1/2 × 9.8 × 3² = 4.9×9 = 44.1 m

Explanation:

What is the mass of an object that experiences a gravitational force of 510 N near Earth's surface?

53.0 kg

52.0 kg

51.0 kg

54.1 kg

Answers

Answer:

52.006 Kilograms

.............................................

The mass of an object that experience a gravitional force of 510 N near earths surface in 52.0 kg

Use a variation model to solve for the unknown value. Use as the constant of variation. The stopping distance of a car is directly proportional to the square of the speed of the car. (a) If a car travelling has a stopping distance of , find the stopping distance of a car that is travelling . (b) If it takes for a car to stop, how fast was it travelling before the brakes were applied

Answers

Complete question is;

Use a variation model to solve for the unknown value.

The stopping distance of a car is directly proportional to the square of the speed of the car.

a. If a car traveling 50 mph has a stopping distance of 170 ft, find the stopping distance of a car that is traveling 70 mph.

b. If it takes 244.8 ft for a car to stop, how fast was it traveling before the brakes were applied?

Answer:

A) d = 333.2 ft

B) 60 mph

Explanation:

Let the stopping distance be d

Let the speed of the car be v

We are told that the stopping distance is directly proportional to the square of the speed of the car. Thus;

d ∝ v²

Therefore, d = kv²

Where k is constant of variation.

A) Speed is 50 mph and stopping distance of 170 ft.

v = 50 mph

d = 170 ft = 0.032197 miles

Thus,from d = kv², we have;

0.032197 = k(50²)

0.032197 = 2500k

k = 0.032197/2500

k = 0.0000128788

If the car is now travelling at 70 mph, then;

d = 0.0000128788 × 70²

d = 0.06310612 miles

Converting to ft gives;

d = 333.2 ft

B) stopping distance is now 244.8 ft

Converting to miles = 0.046363636 miles

Thus from d = kv², we have;

0.046363636 = 0.0000128788(v²)

v² = 0.046363636/0.0000128788

v² = 3599.99658

v = √3599.99658

v ≈ 60 mph

two factor of a number are 5 and 6 .what is the number show working​

Answers

Answer:

30

Explanation:

since  [tex]\frac{30}{5}[/tex]=6

         [tex]\frac{30}{6}[/tex]=5

then both 5 and 6 are factors of 30

Have a nice day

Leslie incorrectly balances an equation as 2C4H10 + 12O2 → 8CO2 + 10H2O.

Which coefficient should she change?

Answers

Answer:

13 behind o2

Explanation:

answer is in photo above

Answer:

12

Explanation:

here did you walk? What did you find most enjoyable while walking: listening to music, listening to an audio book, or nothing? How did your body react to this introductory amount of exercise? Was it more exercise or less exercise than you are used to? If you did not walk, what other type of physical movement did you do?

Answers

Working out in the guy

he cheetah is considered the fastest running animal in the world. Cheetahs can accelerate to a speed of 21.8 m/s in 2.50 s and can continue to accelerate to reach a top speed of 28.8 m/s. Assume the acceleration is constant until the top speed is reached and is zero thereafter. 1) Express the cheetah's top speed in mi/h. (Express your answer to three significant figures.) mih 2) Starting from a crouched position, how long does it take a cheetah to reach its top speed

Answers

Answer:

a) the cheetah's top speed is 64.4 miles/hr

b) time taken to reach top speed is 3.3 seconds

Explanation:

Given the data in the question;

Cheetahs can accelerate to a speed of 21.8 m/s in 2.50 s.

They can continue to accelerate to reach a top speed of 28.8 m/s.

a) Express the cheetah's top speed in mi/h. (Express your answer to three significant figures.)

The cheetah's top speed = 28.8 m/s = ( 28.8 × 2.237 ) miles/hr

= 64.4256 ≈ 64.4 miles/hr

Therefore, the cheetah's top speed is 64.4 miles/hr

b) Starting from a crouched position, how long does it take a cheetah to reach its top speed.

given that

v₁ = 21.8 m/s   and    t₁ = 2.50 s

let a represent the acceleration of the cheetah

From the First Equation of Motion;:

v = u + at

we substitute

21.8 = 0 + ( a × 2.50 )

21.8 = a × 2.50

a = 21.8 / 2.50

a = 8.72 m/s²

Now, let the time taken by cheetah to reach top speed ( 28.8 m/s ) be t

so from the first equation of motion;

v = u + at

we substitute

28.8 = 0 + ( 8.72 × t )

t = 28.8 / 8.72

t = 3.3 seconds

Therefore, time taken to reach top speed is 3.3 seconds

A projectile is launched at ground level with an initial speed of 49.5 m/s at an angle of 40.0° above the horizontal. It
strikes a target above the ground 3.50 seconds later. What are the x and y distances from where the projectile was
launched to where it lands?
x distance
m
y distance
m

Answers

Answer:

x = 132.7 m

y = 51.34 m

Explanation:

Given :

Initial speed, u = 49.5 m/s²

Angle of projection, θ = 40°

Time, t = 3.50 seconds

The distance, x = horizontal component ;

Distance = speed * time

Distance = uCosθ * 3.50

Distance = 49.5 * Cos40° * 3.50

Distance = 49.5 * Cos40° * 3.50

Horizontal distance = 132.7 m

Vertical distance, y :

Sy = ut + 1/2gt²

Sy = Vertical distance ; g = 9.8 m/s²

Sy = 49.5 * sin40 * 3.5 - (0.5 * 9.8 * 3.5²)

Sy = 111.36295 - 60.025

Sy = 51.33795 m

x = 132.7 m

y = 51.34 m

The melting point of a solid is 90.0C. What is the heat required to change 2.5 kg of this solid at 30.0C to a liquid? The specific heat of the solid is 390 J/kgK and its heat of fusion is 4000 J/kg.





Answer and I will give you brainiliest

Answers

Hey again!

Ok..

Now... The melting Point of this solid is 90°C.

Meaning That as soon as it gets to this temp... It STARTS Melting.

So at that temp... It still has some solid parts in it.

You can say its a Solid Liquid Mixture.

Additional Heat being applied at that point is not raising the temperature;rather its used in breaking the bonds in the solid. This is the Fusion stage.

After Fusion...It'd then Be a Pure Liquid with no solids in it.

So

Q'=MC∆0----- This is the heat needed to take the solid's temp from 30°c - 90°c

Q"=ml ----- This is the heat used in breaking the bonds holding the solids in the solid-liquid phase.

So

Q= Q' + Q"

Q= mc∆0 + ml

∆0 = 90°c - 30°c = 60°c

Q= 2.5(390)(60) + (2.5)(4000)

Q=6.9 x 10⁴Joules

The heat required to change 2.5 kg of the solid at 30.0C to a liquid is 6.9 x 10⁴J.

What is specific heat?

The specific heat is the amount of heat energy required to change the temperature of 1kg of object by 1°C.

The heat needed to change the solid's temperature from 30°C - 90°C is

Q' = mC∆T

The heat used to change the phase solid-liquid phase .i.e.

Q'' =mL where L =latent heat of fusion

The total heat required is

Q= Q' + Q"

Q= mc∆T + ml

Q= 2.5(390)(90 - 30) + (2.5)(4000)

Q=6.9 x 10⁴Joules

Thus, the heat required to change the solid to liquid is 6.9 x 10⁴J.

Learn more about  specific heat

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The slope at point A of the graph given below is:


WILL MARK BRAINLIEST TO CORRECT ANSWER

Answers

RQ/PQ I think

rise/run

An object with a mass of 5 kg is swung in a vertical circle by a rope with a length of 0.67 m. The tension at the bottom of the circle is 88 Newtons. What is the tension, in Newtons, at the side of the circle, halfway between the top and bottom if the speed of the mass is the same at the bottom and side

Answers

Answer:

[tex]T_2=39.5N[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=5kg[/tex]

Length [tex]L=0.67m[/tex]

Tension [tex]T=88N[/tex]

Generally the equation for Tension is mathematically given by

 [tex]T = m * ( g + v^2 /l)[/tex]

Therefore

 [tex]T_1 = m * ( g + \frac{v^2}{l})[/tex]

 [tex]88 = 5 * ( 9.8 + \frac{v^2}{0.67})[/tex]

 [tex]v^2=5.2[/tex]

 [tex]v=2.4m/s[/tex]

The uniform velocity is

 [tex]v=2.4m/s[/tex]

Therefore

The tension at the side of the circle halfway between the top and bottom is

 [tex]T_2=5*\frac{2.3^2}{0.67}[/tex]

 [tex]T_2=39.5N[/tex]

A train is moving at a constant
speed of 55.0 m/s. After 5.00
seconds, how far has the train
gone?
cara
(Units = m)

Answers

Answer:

Distance = speed * time

55*5

275 meters.

The train would have covered a distance of 275 m

What is distance ?

We can define distance as to how much ground an object has covered despite its starting or ending point.

Distance = speed * time

given

speed= 55 m/s

time = 5 sec

Distance = 55 * 5 = 275 m

The train would have covered a distance of 275 m

learn more about distance

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The temperature of a quantity of an ideal gas is a. one measure of its ability to transfer thermal energy to another body. b. proportional to the average molecular kinetic energy of the molecules. c. proportional to the internal energy of the gas. d. correctly described by all the statements above. e. correctly described only by the first two statements above.

Answers

Answer:

d. correctly described by all the statements above.

Explanation:

Kinetic molecular theory of gases states that gas particles exhibit a perfectly elastic collision and are constantly in motion.

According to the kinetic-molecular theory, the average kinetic energy of gas particles depends on temperature.

This ultimately implies that, the average kinetic energy of gas particles is directly proportional to the absolute temperature of an ideal gas. Thus, an increase in the average kinetic energy of gas particles would cause an increase in the absolute temperature of an ideal gas.

Temperature can be defined as a measure of the degree of coldness or hotness of a physical object. It is measured with a thermometer and its units are Celsius (°C), Kelvin (K) and Fahrenheit (°F).

Generally, the temperature of a quantity of an ideal gas is;

a. a measure of the ability of an ideal gas to transfer thermal energy to another body.

b. the average kinetic energy of gas particles is directly proportional to the absolute temperature of an ideal gas

c. proportional to the internal energy of the gas.

help asap please I will give you 5stars

Answers

Explanation:

In the parallel combination, the equivalent resistance is given by :

[tex]\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+....[/tex]

4. When three 150 ohms resistors are connected in parallel, the equivalent is given by :

[tex]\dfrac{1}{R}=\dfrac{1}{150}+\dfrac{1}{150}+\dfrac{1}{150}\\\\R=50\ \Omega[/tex]

5. Three resistors of 20 ohms, 40 ohms and 100 ohms are connected in parallel, So,

[tex]\dfrac{1}{R}=\dfrac{1}{20}+\dfrac{1}{40}+\dfrac{1}{100}\\\\=11.76\ \Omega[/tex]

Hence, this is the required solution.

Elapsed Time
(s)
Cart Speed
(Low fan speed)
(cm/s)
Cart Speed
(Medium fan speed)
(cm/s)
Cart Speed
(High fan speed)
(cm/s)
0
1
16.4
23.0
31.7
2
31.5
64.0 cm/s
3
54.0 cm/s
36.0
89.8
4
118,81
5
120.0 cm/s
6
128.2
96.0
7
145.8
Table B

Answers

Answer:

cm/s

6

128.2

96.0

7

145.8

Table B

A computer monitor uses 200 W of power. How much energy does it use in
10 seconds?

Answers

Answer: see below explanation, should be straight forward from there? ;)

Explanation: 1 watt = 1 joule per second

Watt is a measure of energy over time

So 10 seconds... u got this :)

What is the magnitude of a vector that has the following components: x = 32 m y = -59 m

Answers

Answer:

Explanation:

Since the x and y components are given

The vectors Magnitude = √32²+(-59)²

=67.12m

(15 PTS) An observer riding on the platform measures the angle q that the thread supporting the light ball makes with the vertical. There is no friction anywhere. If you can vary m1 and m2, find the largest angle q you could achieve.

Answers

Solution :

 Given :

Angle q = angle between the thread supporting the ball with the vertical.

Let mass [tex]$m_1 >>>m_2$[/tex].

Then [tex]$m_1+m_2=m_1$[/tex]

In this case, acceleration can be found out by applying Newton's law of motion.

Thus,

Acceleration, [tex]$a=\frac{m_1}{m_1+m_2}. g$[/tex]

                      [tex]$a=\frac{m_1}{m_1}. g$[/tex]

                       [tex]$a=g$[/tex]

Therefore, [tex]$\tan \theta =\frac{a}{g}$[/tex]

or                [tex]$\tan \theta =\frac{a}{a}$[/tex]

or                [tex]$\tan \theta =1$[/tex]

                   [tex]$\theta = \tan ^{-1}(1)$[/tex]

                   [tex]$\theta = 45^\circ$[/tex]

Therefore the largest angle q is  [tex]$\theta = 45^\circ$[/tex]

A 285-kg load is lifted 22.0 m vertically with an acceleration a=0.160g by a single cable. Determine

(a) the tension in the cable, (b) the net work done on the load, (c) the work done by the cable on the

load, (d) the work done by gravity on the load, and (e) the final speed of the load assuming it started

from rest.​

Answers

Answer:

a)  T = 2838.6 N,  b)   W = 1003.2 J,  c) W = 6.22 10⁴ J,  d) W = 2.79 10³ J

e) v_f = 2.65  m / s

Explanation:

a) To find the tension of the cable let's use Newton's second law

        T - W = m a

         T = W + ma

        T = m (g + a)

let's calculate

        T = 285 (-9.8 - 0.160)

        T = 2838.6 N

b) net work is stress work minus weight work

        W = F d

        W = (T-W) d

        W = (m a) d

        W = (285 0.160) 22

        W = 1003.2 J

 

c) the work done by the cable

         W = T d cos 0

          W = 2838.6 22.0

          W = 6.22 10⁴ J

d) The work done by the weight

the displacement is upwards and the weight points downwards, so the angle is 180º

        W = F. d

         W = F d cos 180

         W = -285 22.0

         W = 2.79 10³ J

e) the final speed of the load. Let's use the relationship between work and the change in kinetic energy

         W = ΔK

         

as part of rest K₀ = 0

          W = ½ m v_f²

          v_f = [tex]\sqrt{ \frac{2W}{m} }[/tex]

          v_f = [tex]\sqrt{\frac{2 \ 1003.2}{285} }[/tex]

          v_f = 2.65  m / s

A homeowner has a new oil furnace which has an efficiency of 60%. For every 100 barrels of oil used to heat his house, how much (in barrels of oil) goes up the chimney as waste heat?

Answers

Answer:

below

Explanation:

In electronic circuits:______.
a. the power used by a circuit is the resistance times the current squared.
b. electric and magnetic fields are transporting the energy.
c. electrons are transporting the energy.
d. the power used by a circuit is the voltage times the current squared.
e. the power used by a circuit is the current times the voltage.

Answers

Answer:

(a), (c) and (e) s correct.

Explanation:

a. the power used by a circuit is the resistance times the current squared.

The power is given by P = I^2 R, so the statement is correct.  

b. electric and magnetic fields are transporting the energy.

false

c. electrons are transporting the energy.

The energy is transferred by flow of electrons. It is correct.  

d. the power used by a circuit is the voltage times the current squared.

The power is given by P  = V I, the statement is wrong.  

e. the power used by a circuit is the current times the voltage.

The power is given by P  = V I, the statement is correct.  

The number of current paths in a series circuit is:
a. one
b. two
C. three
d. four

Answers

Answer:

One

Explanation:

In series combination, the circuit follows one path whereas in parallel it follows two or more than two path

One
One

The number of circuit paths is one

Part A
Calculate the work done when a force of 6.0 N moves a book 2.0 m
Express your answer with the appropriate units.

Answers

Answer:

Work done applied = 12 newton-meter

Explanation:

Given examples:

Force applied = 6 newton

Distance of book = 2 meter

Find from the given data:

Work done

Computation:

The equation can be used to compute work.

Work done applied = Force applied x Distance of book

Work done applied = Force x Distance

Work done applied = 6 x 2

Work done applied = 12 newton-meter

The length of a cylindrical axon is 8 cm and its radius of 8μm,and the thickness of the membrane is 0.01μm,dielectric constant ( ε=24.78x10-12 F/m),the capacitance of nerve cell is

Answers

Answer:

9.965 nF

Explanation:

The capacitance of the axon C = εA/d where ε = dielectric constant = 24.78 × 10⁻¹² F/m, A = surface area of axon = 2πrL where r = radius of axon = 8 μm = 8 × 10⁻⁶ m and L = length of axon = 8 cm = 8 × 10⁻² m and d = thickness of membrane = 0.01 μm = 0.01 × 10⁻⁶ m

So, C = εA/d

C = ε2πrL/d

Substituting the of the values variables into the equation, we have

C = ε2πrL/d

C =  24.78 × 10⁻¹² F/m × 2π × 8 × 10⁻⁶ m × 8 × 10⁻² m/0.01 × 10⁻⁶ m

C =  9964.63 × 10⁻²⁰ Fm/0.01 × 10⁻⁶ m

C =  996463 × 10⁻¹⁴ F

C = 9.96463 × 10⁻⁹ F

C = 9.96463 nF

C ≅ 9.965 nF

A 9V battery is connected to two light bulbs. If the current through the circuit is 1.5 A, what is the resistance of each light bulb (assume both light bulbs are identical)?

Answers

Answer:

3 ohm

Explanation:

Given :

V=9V

And according to given question same current is flowing in both resistance that means resistance will connected in series

So,

R= R+R=2R

Now,

Applying ohm's law

[tex]V=IR\\9=1.5*2R\\9=3R\\R=\frac{9}{3} \\R= 3ohm[/tex]

Therefore, answer is 3 ohm

Its Acceleration during the upward Journey ? ​

Answers

Acceleration will be 9.81 if it goes downwards. If it accelerates upwards it will be -9.81m/s^2

Hạt mang điện q > 0 chuyển động trong từ trường của một dòng điện thẳng dài có cường độ I = 10A như hình. Hạt mang điện chuyển động song song với dây dẫn và cách dây một khoảng 5cm. Vẽ hình và:
a. Xác định cảm ứng từ do dòng điện gây ra tại điểm mà hạt mang điện đi qua.
b. Hạt mang điện chuyển động với tốc độ 104m/s, lực Lorentz tác dụng lên hạt là 8.10-4N. Tính độ lớn của điện tích.

Answers

Answer:

I dnt know that language

Explanation:

dòng điện là gì ?/???????

Answers

Answer:

Dòng điện là một dòng các hạt mang điện, chẳng hạn như electron hoặc ion, di chuyển qua vật dẫn điện hoặc không gian. Nó được đo bằng tốc độ thực của dòng điện tích qua một bề mặt hoặc vào một thể tích điều khiển.

Xin lưu ý rằng tôi đã sử dụng một trình dịch để nhập nội dung này, vì vậy có thể có một số từ không hợp lý.

Explanation:

If we will be conducting the experiments in tanks with dimensions of roughly 10 in by 10 in. Estimate the volume of oil you would need to cover half the total area. Again assume a molecule is 10 nm in size. Show your work.

Answers

Answer:

1. 10³ m²

2. 32.4 ml

3. 1.91 × 10²¹ molecules

Explanation:

Here is the complete question

1. Estimate the size of a one-molecule-thick oil film formed by spreading 1 ml of oil on the surface of the water. Assume that an oil molecule is roughly 10 nm in size. Show your work.

2. If we will be conducting the experiments in tanks with dimensions of roughly 10 in by 10 in. Estimate the volume of oil you would need to cover half the total area. Again assume a molecule is 10 nm in size. Show your work.

3. Estimate the number of molecules in 1 ml of oil assuming they’re 10 nm in size. What assumptions do you have to make? Show your work.

Solution

1. Since the film would cover an area, A, and would have a height which is the thickness of the molecule, h = 10 nm = 1 × 10⁻⁹ m, its volume is V = Ah. This volume also equals the volume of the oil. Since the volume of oil is 1 ml = 1 × 10⁻⁶ m³, V = 1 × 10⁻⁶ m³.

The size of the oil drop is its area. So, A = V/h

= 1 × 10⁻⁶ m³ ÷ 1 × 10⁻⁹ m

= 10³ m²

2. The area of the tank is 10 in by 10 in = 100 in². Since we want to cover half the area, we require 100 in²/2 = 50 in² = 50 in² × (0.0254)² m²/in² = 0.0324 m².

If the thickness of oil is one molecule thick which is 10nm = 1 × 10⁻⁹ m, the volume of oil is then thickness × area = 1 × 10⁻⁹ m × 0.0324 m²

= 0.0324 × 10⁻⁹ m³

= 32.4 × 10⁻⁶ m³

= 32.4 ml since 1 × 10⁻⁶ m³ = 1 ml

3. Since the volume of oil is 1 ml = 1 × 10⁻⁶ m³, we need to find the volume of one molecule. Since it is assumed to be a sphere, its volume is V' = πd³/6 where d = size of oil molecule = 10 nm = 1 × 10⁻⁹ m.

Let n be the number of molecules present in 1 ml, then nV' = 1 ml = 1 × 10⁻⁶ m³. So, n = 1 × 10⁻⁶ m³/V' = 1 × 10⁻⁶ m³ ÷ πd³/6 = 6 × 10⁻⁶ m³/πd³

Substituting d into the equation, we have

n = 6 × 10⁻⁶ m³/π(1 × 10⁻⁹ m)³

n = 6 × 10⁻⁶ m³/π × 10⁻²⁷ m³

n = 1.91 × 10²¹ molecules

Calculate the sample standard deviation and sample variance for the following frequency distribution of hourly wages for a sample of pharmacy assistants.

Lower Bound Upper Bound
6.51 8.50
50 8.51
10.50 18
10.51 12.50
42 12.51
14.50 20
14.51 16.50

Answers

Answer:

(a) standard deviation = σ = 4.9996

(b) variance = σ² = 24.996

Explanation:

Given frequency table (find attached as Table 1);

(a) To find the sample standard deviation and sample variance, follow these steps;

i. Calculate the mid-point c for each group by using the mid-point formula;

c = (lower bound + upper bound) / 2

=> c = (6.51 + 8.50) / 2 = 7.505

=> c = (8.51 + 10.50) / 2 = 9.505

=> c = (10.51 + 12.50) / 2 = 11.505

=> c = (12.51 + 14.50) / 2 = 13.505

=> c = (14.51 + 16.50) / 2 = 15.505

So the new table becomes (find attached as Table 2);

ii. Calculate the total number of samples (n) which is the sum of all the frequencies.

n = 50+18+42+20+46

n = 176

iii. Calculate the mean (M)

This is done by first multiplying the midpoints by the corresponding frequencies and then dividing the result by the total number of samples (n).

M = [(7.505 x 50) + (9.505 x 18) + (11.505 x 42) + (13.505 x 20) + (15.505 x 46)] / 176

M = [375.25 + 171.09 + 483.21 + 270.1 + 713.23] / 176

M = [2012.88] / 176

M = 11.44

iv. Find the variance (σ²);

The variance is calculated using the following formula

σ² = [Σ(f x c²) - (n x M²)] / (n - 1)                ------------(i)

Where;

f = frequency of each boundary data point

=>  Let's first calculate Σ(f x c²).

This is done by finding the sum of the product of the frequency (f) of each boundary point and the square of their corresponding mid-points(c)

Σ(f x c²) = [(50 x 7.505²) + (18 x 9.505²) + (42 x 11.505²) + (20 x 13.505²) + (46 x 15.505²)]

Σ(f x c²) = [(2816.25125) + (1626.21045) + (5559.33105) + (3647.7005) + (11058.63115)]

Σ(f x c²) = 24708.1244

=> Now calculate (n x M²)

n x M² = 176 x 11.44²

n x M² = 23033.7536

=> Now substitute these values into equation (i) to calculate the variance

σ² = [Σ(f x c²) - (n x M²)] / (n - 1)

σ² = [24708.1244 - 23033.7536] / (176 - 1)

σ² = [4374.3708] / (175)

σ² = 24.996

Therefore, the variance is 24.996

v. Find the standard deviation (σ)

The standard deviation is the square root of the variance. i.e

σ = √σ²

σ = √24.996

σ = 4.9996

Therefore, the standard deviation is 4.9996

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