In Michael Johnson's world-record 400 m sprint, he ran the first 100 m in 11.20 s; then he reached the 200 m mark after a total time of 21.32 s had elapsed, reached the 300 m mark after 31.76 s,and finished in 43.18 s.

During what 100 m segment was his speed the highest?

a) Between the start and the 100 m mark.

b) Between the 100 m mark and the 200 m mark.

c) Between the 200 m mark and the 300 m mark.

d)Between the 300 m mark and the finish.

Answers

Answer 1

Answer:

b)

Explanation:

Assuming that we are talking about average speed during any segment, we can apply the definition of average speed, as follows:

v(avg) = Δx / Δt = (xf-x₀) / (tfi-t₀)

Using this definition for the 4 segments, we have:

1) v(0-100m) = 100 m / 11.20 sec = 8.93 m/s

2) v(100m-200m) = 100 m / (21.32 s - 11.2 s) = 100 m / 10.12 s = 9.88 m/s

3) v(200m -300m) = 100 m / (31.76 s- 21.32s) = 100 m / 10.44 s = 9.58 m/s

4) v(300m-400m) = 100 m / (43.18 s - 31.76 s) = 100 m / 11.42 s = 8.76 m/s

As we can see, the highest speed was reached between the 100m mark and the 200m mark, so the statement b) is the one that results to be true.

Answer 2

Final answer:

The highest speed in Michael Johnson's world-record 400 m sprint was achieved during the segment between the 100 m mark and the 200 m mark with a speed of 9.88 m/s.

Explanation:

To determine which 100 m segment Michael Johnson had the highest speed, we should calculate his speed for each segment separately. Speed is calculated by dividing the distance by the time taken to cover that distance. We are given the total time at each 100 m interval, so we need to calculate the time for each interval separately and then calculate the speed.

For the first 100 m: Speed = 100 m / 11.20 s = 8.93 m/sFor the second 100 m (100 m to 200 m): Time taken = 21.32 s - 11.20 s = 10.12 s, so Speed = 100 m / 10.12 s = 9.88 m/sFor the third 100 m (200 m to 300 m): Time taken = 31.76 s - 21.32 s = 10.44 s, so Speed = 100 m / 10.44 s = 9.58 m/sFor the fourth 100 m (300 m to 400 m): Time taken = 43.18 s - 31.76 s = 11.42 s, so Speed = 100 m / 11.42 s= 8.76 m/s

Comparing these speeds, we can see that the highest speed was achieved during the segment between the 100 m mark and the 200 m mark.


Related Questions

A baseball has a mass of about 155 g. What is the magnitude of the momentum of a baseball thrown at a speed of 87 miles per hour? (Note that you need to convert mass to kilograms and speed to meters/second. A mile is 1.6 kilometers or 1600 meters.)

Answers

Answer:

5.994 kgm/s

Explanation:

Momentum: This can be defined as the product of the mass of a body to the velocity of that body. The S.I unit of momentum is kgm/s.

Mathematically, momentum can be expressed as,

M = mv.................................... Equation 1

Where M = momentum of the baseball, m = mass of the baseball, v = velocity of the baseball.

Given: m = 155 g = (155/1000) kg = 0.155 kg. v = 87 miles per hour = 87(1600/3600) m/s = 38.67 m/s.

Substituting into equation 1

M = 0.155(38.67)

M = 5.994 kgm/s.

Thus the momentum of the baseball = 5.994 kgm/s

Final answer:

The magnitude of the momentum of a baseball with a mass of 155 g thrown at 87 mph is approximately 6.03 kg·m/s after converting units and applying the momentum equation.

Explanation:

To calculate the magnitude of the momentum of a baseball thrown at a speed of 87 miles per hour, we use the formula p = mv, where p is the momentum, m is the mass in kilograms, and v is the velocity in meters per second. First, we need to convert the mass of the baseball from grams to kilograms (155 g to 0.155 kg) and the speed from miles per hour to meters per second (87 mph to 38.88 m/s, using the conversion factor 1 mile = 1600 meters and 1 hour = 3600 seconds).

Then, we substitute the values into the momentum equation to find the momentum: p = (0.155 kg)(38.88 m/s) = 6.0264 kg·m/s. Therefore, the magnitude of the momentum of the baseball is approximately 6.03 kg·m/s when rounded to two decimal places.

The voltage across a membrane forming a cell wall is 74.0 mV and the membrane is 8.40 nm thick. What is the electric field strength in volts per meter?

Answers

Answer:

Electric field strength=8.81×10⁶V/m

Explanation:

Given Data

voltage v= 74.0 mV

Membrane thickness d=8.40 nm

To find

Electric field strength E=?

Solution

Electric field strength =voltage/Membrane thickness

[tex]E=v/d\\E=\frac{74.0*10^{-3} }{8.40*10^{-9} }\\ E=8.81*10^{6}V/m[/tex]

determine the required values of the tension in ac and ad so that the resultant of the three forces applied at A is verticalFigure:A wire is coupled from three sides

Answers

Answer:

Fac = 21 KN

Fad = 64.29 KN

Explanation:

Step 1: Find the unit vectors in BA , CA, and DA directions

BA = -16 i + 48 j - 12 k

CA = -16 i + 48 j + 24 k

DA = 14 i + 48 j

The following magnitudes

mag (BA) = sqrt ( 16 ^2 + 48^2 + 12^2 ) = 52

mag (CA) = sqrt ( 16 ^2 + 48^2 + 24^2 ) = 56

mag (DA) = sqrt ( 14 ^2 + 48^2 ) = 50

Now compute unit vectors

unit (BA) = (-4 / 13) i + (12 / 13) j - (3 / 13) k

unit (CA) = (-2 / 7) i + (6 / 7) j - (3 / 7) k

unit (DA) = (7 / 25) i + (24 / 25) j

Step 2: Compute Force Vectors F(ml) dot unit( ML )

Fab = (-4*Fab / 13) i + (12*Fab / 13) j - (3*Fab / 13) k

Fac =  (-2*Fac / 7) i + (6*Fac / 7) j - (3*Fac / 7) k

Fad = (7*Fad / 25) i + (24*Fad / 25) j

Step 3: Equilibrium force equations in x and z direction

Fr,x = 0 = 39*( -4 / 13 ) - Fac*( 2 / 7 ) + Fad*( 7 / 25 )

Fr,z = 0 = 39*( -3 / 13 ) + Fad*( 3 / 7 )

Step 4: Solve the above equations for Fca and Fda

Fca = 21 KN

Fda = 64.29 KN

A rocket has a mass 280(103) slugs on earth. Specify its mass in SI units, and its weight in SI units.

Answers

To solve this problem we will apply the concepts related to the conversion of units for which we will have that 1 slug is equal to 14.59kg. At the same time we will use Newton's second law for which weight is defined as the product between mass and acceleration (Due to gravity). This is then

A: Using the conversion ratio of slug to kilogram we have to,

[tex]1 slug = 14.59kg[/tex]

Then

[tex]m = 280*10^3 slugs (\frac{14.59kg}{1slug})[/tex]

[tex]m = 4.09*10^6kg[/tex]

B: Using Newton's second law we have to,

[tex]W = mg[/tex]

[tex]W = (4.09*10^6)(9.8)[/tex]

[tex]W = 40034960N\approx 4*10^7N[/tex]

Organ pipe A, with both ends open, has a fundamental frequency of 270 Hz. The third harmonic of organ pipe B, with one end open, has the same frequency as the second harmonic of pipe A. How long are___________.

(a) pipe A
(b) pipe B

Answers

Answer

Organ Pipe A has both end open.

Organ Pipe B has one end open.

speed of sound, v = 343 m/s

The fundamental frequency = 270 Hz

wavelength of the pipe when both end open

  λ = 2 L

we know,

 [tex]\lambda = \dfrac{v}{f}[/tex]

now,

 [tex]L = \dfrac{v}{2f}[/tex]

inserting all the values

 [tex]L = \dfrac{343}{2\times 270}[/tex]

  L_A = 0.635 m

length of pipe A is equal to 0.635 m

b) since third harmonic of pipe B is equal to second harmonic of pipe A

     [tex]f_B = f_A[/tex]

     [tex]\dfrac{n_BV}{4L_B} = \dfrac{n_AV}{2L_A} [/tex]

     [tex]L_B= \dfrac{2n_BL_A}{4n_A}[/tex]      

     [tex]L_B= \dfrac{2\times 3 \times 0.635}{4\times 2}[/tex]

        L_B = 0.476 m

length of pipe B is equal to 0.476 m.

If Star A is magnitude 1.0 and Star B is magnitude 9.6 , which is brighter and by what factor?

Answers

Answer:

Star A is brighter than Star B by a factor of 2754.22

Explanation:

Lets assume,

the magnitude of star A = m₁ = 1

the magnitude of star B = m₂ = 9.6

the apparent brightness of star A and star B are b₁ and b₂ respectively

Then, relation between the difference of magnitudes and apparent brightness of two stars are related as give below: [tex](m_{2} - m_{1}) = 2.5\log_{10}(b_{1}/b_{2})[/tex]

The current magnitude scale followed was formalized by Sir Norman Pogson in 1856. On this scale a magnitude 1 star is 2.512 times brighter than magnitude 2 star. A magnitude 2 star is 2.512 time brighter than a magnitude 3 star. That means a magnitude 1 star is (2.512x2.512) brighter than magnitude 3 bright star.

We need to find the factor by which star A is brighter than star B. Using the equation given above,

[tex](9.6 - 1) = 2.5\log_{10}(b_{1}/b_{2})[/tex]

[tex]\frac{8.6}{2.5} = \log_{10}(b_{1}/b_{2})[/tex]

[tex]\log_{10}(b_{1}/b_{2}) = 3.44[/tex]

Thus,

[tex](b_{1}/b_{2}) = 2754.22[/tex]

It means star A is 2754.22 time brighter than Star B.

Final answer:

Star A is brighter than Star B by approximately 2512 times.

Explanation:

The brightness of stars is measured using the magnitude scale. The smaller the magnitude, the brighter the star. In this case, Star A has a magnitude of 1.0 and Star B has a magnitude of 9.6. Since Star A has a smaller magnitude, it is brighter than Star B. The difference in magnitude between the two stars can be calculated using the formula:

difference = 2.512^(m₁ - m₂)

where m₁ is the magnitude of Star A and m₂ is the magnitude of Star B. Plugging in the values, we get:

difference = 2.512^(1.0 - 9.6) ≈ 2512

Therefore, Star A is approximately 2512 times brighter than Star B.

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What is the general relationship between earthquake epicenters and volcano positions along the Pacific coast of Mexico and Central America? What sort of plate boundary does this relationship indicate?

Answers

Geographically throughout this area of Mexico, Central America Caribbean is located the Cocos plate. This area is scientifically known as the Central American subduction zone.

In order for a volcano to form, there is usually a clash between the technical plates that generates the elevation of the ground and the connection with ducts that release the magma from the earth. If this entire area is a subduction area, it will also be a land stress release area where volcano lines will be formed, that is, it is a convergent plate boundary area

The relationship between the epicenters of the earthquake and the position of the volcano should be explained below.

What is Central American subduction zone?

When the Mexico area should be geographically located, so the Caribbean of central America should be located on the Cocos plate. So this area should be called as the subduction zone. At the time when the volcano should be created so there is normally clash that lies between the technical plats where it generated the ground elevation and the linked with the ducts due to this, it releases the magma from the earth.

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A +6.00 μC point charge is moving at a constant 9.00 ×10^6m/s in the + y-direction, relative to a reference frame.
At the instant when the point charge is at the origin of this reference frame, what is the magnetic field B it produces at the following points?

A) x=0.500m,y=0, z=0
B) x=0, y=-0.500m, z=0
C) x=0, y=0, z=+0.500m
D) x=0, y=-0.500m, z=+0.500m

Answers

SOLUTION OF EACH CASES ARE GIVEN BELOW.

Explanation:

We know magnetic field due to a charge q moving with a velocity v [tex]=\dfrac{\mu_oq}{4\pi r^3}(v \times r)[/tex].

Case A:

r=0.5i

Putting value of r, q and v .

We get, [tex]B=-2.34\times 10^{-5} \ k.[/tex] [tex](\ j\times i=-k)[/tex]

Case B:

r=0.5 j

B=0 [tex]( j\times j=0)[/tex]

Case C:

We get, [tex]B=2.34\times 10^{-5} \ i\ (Since\ , j \times k=i)[/tex]

Case D:

We get , [tex]B=8.28\times 10^{-6} \ T \ i[/tex].

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When the balls are brought close together, without touching, the following observations are made: Balls B, C, and D are attracted to ball A. Balls B and D have no effect on each other. Balls B is attracted to ball C. What are the charge states (positive, negative, or neutral) of balls A, B, C, and D

Answers

Answer:

B and D is neutral. A and C have opposite charges.

Explanation:

Opposite charges attract each other and similar charges repel each other. Furthermore, a neutral object is attracted to a charged object whether negative or positive.

- If the balls B, C, and D is attracted to A, then their charge is either opposite to that of A or neutral.  

- If the balls B and D have no effect on each other, then both balls are neutral.

- If B is neutral and attracted to C, then ball C have opposite charges than A.

Finally, B and D is neutral. A and C have opposite charges. Whether they are positive or negative cannot be determined by the information given in the question.

A kicker punts a football from the very center of the field to the sideline 39 yards downfield.

What is the magnitude of the net displacement of the ball? (a,football field is 53 yards wide)

What is the angle between the direction of the net displacement of the ball and the central line of the feild.

Answers

Answer:

L = 47.15 yards

∅ = 55.80°

Explanation:

Given

y = 39 yards

x = 53/2 yards

L = ?

We apply the formula

L = √(x²+y²)

L = √((53/2 yards)²+(39 yards)²)

L = 47.15 yards

The angle between the direction of the net displacement of the ball and the central line of the feild is obtained as follows

∅ = tg⁻¹(y/x)

∅ = tg⁻¹(39/(53/2)) = 55.80°

Total lung capacity of a typical adult is approximately 5.0L. Approximately 20% of the air is oxygen, as air is 20% oxygen. At sea level and at anaverage body temperature of 37°C. How many moles ofoxygen do the lungs contain at the end of an inflation?

Answers

Answer:

n=0.03928 moles

Moles of oxygen do the lungs contain at the end of an inflation are 0.03928 moles

Explanation:

The amount of oxygen which lung can have is 20% of 5 L which is the capacity of lungs

Volume of oxygen in lungs =V=5*20%= 1 L=[tex]1*10^{-3} m^3[/tex]

Temperature=T=[tex]37^oC=273+37=310K[/tex]

Pressure at sea level = P= 1 atm=[tex]1.0125*10^5 Pa[/tex]

R is universal Gas Constant =8.314 J/mol.K

Formula:

[tex]n=\frac{PV}{RT}\\n=\frac{(1.0125*10^5) *(1*10^{-3})}{(8.314)*310} \\n=0.03928 mol[/tex]

Moles of oxygen do the lungs contain at the end of an inflation are 0.03928 moles

Two point charges are released at rest in otherwise empty space. As time goes on, the magnitude of the force exerted on each of the particles decreases. Which of the following must be true?a. Both particles are negatively chargedb. Not enough infoc. The particles may be either both positively charged or both negatively chargedd. One particle is positively charged and the other is negatively chargede. Both particles are positively charged

Answers

Answer:

c.)

Explanation:

Assuming no other forces acting on the charges, if as time goes on, the magnitude of the force exerted on each of the particles decreases, as the electric force is inversely proportional to the square of the distance between charges, the only explanation is that this distance is increasing, so the force on both charges is repulsive.

When the force is repulsive, all we can say, is that both charges are of the same type, so they can be both positively charged or both negatively charged, as stated by the choice c).

A 1.0-cm-diameter pipe widens to 2.0 cm, then narrows to 0.5 cm. Liquid flows through the first segment at a speed of 9.0 m/s

A:What is the speed in the second segment?m/s
B:What is the speed in the third segment?m/s
C:What is the volume flow rate through the pipe?L/s

Answers

The speed in the second segment is 36 m/s, in the third segment is 144 m/s, and the volume flow rate through the pipe is 3.38 L/s.

A: Using the principle that the volume flow rate remains constant in an incompressible fluid, we can calculate the speed in the second segment to be 36 m/s and the speed in the third segment to be 144 m/s.

B: The volume flow rate through the pipe can be calculated by applying the equation Q = Av, where Q represents the flow rate, A is the cross-sectional area, and v is the velocity of the fluid. Given the diameter changes, the volume flow rate is 3.38 L/s.

C: The primary concept involved in this problem is the relationship between cross-sectional area and fluid velocity as the diameter changes along the pipe, affecting the flow rate.

Another droplet of the same mass falls 8.4 cm from rest in 0.250 s, again moving through a vacuum. Find the charge carried by the droplet.

Answers

Answer:

The charge carried by the droplet is [tex]1.330\times10^{-19}\ C[/tex]

Explanation:

Given that,

Distance =8.4 cm

Time = 0.250 s

Suppose tiny droplets of oil acquire a small negative charge while dropping through a vacuum in an experiment. An electric field of magnitude [tex]5.92\times10^4\ N/C[/tex] points straight down and if the mass of the droplet is [tex]2.93\times10^{-15} kg[/tex]

We need to calculate the acceleration

Using equation of motion

[tex]s=ut+\dfrac{1}{2}at^2[/tex]

Put the value into the formula

[tex]8.4\times10^{-2}=0+\dfrac{1}{2}\times a\times(0.250)^2[/tex]

[tex]a=\dfrac{8.4\times10^{-2}\times2}{(0.250)^2}[/tex]

[tex]a=2.688\ m/s^2[/tex]

We need to calculate the charge carried by the droplet

Using formula of electric filed

[tex]E=\dfrac{F}{q}[/tex]

[tex]q=\dfrac{ma}{E}[/tex]

Put the value into the formula

[tex]q=\dfrac{2.93\times10^{-15}\times2.688}{5.92\times10^4}[/tex]

[tex]q=1.330\times10^{-19}\ C[/tex]

Hence, The charge carried by the droplet is [tex]1.330\times10^{-19}\ C[/tex]

For tax and accounting purposes, corporations depreciate the value of equipment each year. One method used is called "linear depreciation," where the value decreases over time in a linear manner. Suppose that two years after purchase, an industrial milling machine is worth $820,000, and five years after purchase, the machine is worth $430,000. Find a formula for the machine value V.

Answers

Answer:

[tex]V=-130000x+1080000[/tex]

Explanation:

Linear Dependence

Some variables are known or assumed to have linear dependence which means the graph of the ordered pairs (x,V) is a straight line.

If we know two points of the line, we can come up with the exact equation and therefore make predictions for other values of x

The linear depreciation gives us these points (2,820000) and (5,430000)

The general equation of the line is

[tex]V=mx+b[/tex]

Where V is the machine value and x is the  number of years after purchase. We need to find the values of m and b.

Replacing the first point

[tex]820000=m(2)+b[/tex]

[tex]2m+b=820000[/tex]

Replacing the second point

[tex]5m+b=430000[/tex]

Subtracting them  

[tex]-3m=390000[/tex]

[tex]m=-130000[/tex]

Replacing in any of the equations, say, the first one

[tex]2(-130000)+b=820000[/tex]

Solving for b

[tex]b=820000+260000[/tex]

[tex]b=1080000[/tex]

The formula for the machine value V is  

[tex]\boxed{V=-130000x+1080000}[/tex]

Formula for the machine value V in x year is v = -130000(x) + 1,080,000

Liner depreciation based problem:

What information do we have?

Cost of machine after 2 year = $820,000

Cost of machine after 5 year = $430,000

Liner depreciation equation

v = mx + b

After 2 year

820,000 = m(2) + b

820,000 = 2m + b .........   Eq1

After 5 year

430,000 = m(5) + b

430,000 = 5m + b .........   Eq2

Eq2 - Eq1

3m = -390,000

m = -130,000

From Eq

430,000 = 5m + b

430,000 = 5(-130,000) + b

b = 1,080,000

Liner equation of tha cost.

Amount of machine = mx + b

Amount of machine = -130000(x) + 1,080,000

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How do the properties of an electromagnetic wave change as a result of increasing the period of the wave?

Answers

Final answer:

As the period of an electromagnetic wave increases, the wavelength of the wave also increases. However, the frequency of the wave remains the same. The energy carried by the wave is inversely proportional to its wavelength, so as the wavelength increases, the energy of the wave decreases.

Explanation:

Electromagnetic waves have properties such as wavelength and frequency. As the period of an electromagnetic wave increases, the wavelength of the wave also increases. This means that the distance between successive crests or troughs of the wave becomes larger. However, the frequency of the wave remains the same. The energy carried by the wave is inversely proportional to its wavelength, so as the wavelength increases, the energy of the wave decreases.

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To seven significant figures, the mass of a proton is 1.672623 x10^{-27} kg. Which of the following choices demonstrates correct rounding?

Answers

Answer:

[tex]1.67\times 10^{-27}kg[/tex]

Explanation:

We are given that mass of proton

[tex]1.672623\times 10^{-27}kg[/tex]

There are seven significant figures.

We have to round off.

If we round off to three  significant figures

The  thousandth place  of given mass of proton  is less than five therefore, digits on left side of  thousandth  place remains same and digits on right side of thousandth place and thousandth  replace by zero

Therefore, the mass of proton can be written as

[tex]1.67\times 10^{-27}kg[/tex]

Hence, the mass of proton is [tex]1.67\times 10^{-27}kg[/tex]

Final answer:

To round the mass of a proton to seven significant figures, the correct way is to round up the last significant figure if it is 5 or greater, and if it is less than 5, simply drop the remaining digits.

Explanation:

To round the mass of a proton, we look at the digit right after the desired number of significant figures, which in this case is the seventh significant figure. If this digit is 5 or greater, we round up the last significant figure. If it is less than 5, we simply drop the remaining digits.

1.6726230 x 10-27 kg
This demonstrates rounding to seven significant figures, where the last digit is rounded up as it is greater than 5.1.6726 x 10-27 kg
This demonstrates rounding to four significant figures, which is incorrect based on the given number of significant figures.1.672623 x 10-27 kg
This demonstrates the correct rounding to six significant figures, as the last digit is less than 5.

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You are pulling a box so it moves at increasing speed. Compare the work (W1) you need to do to accelerate it from 0 m/s to speed v to the work (W2) needed to accelerate it from speed v to the speed of 2v in the presence of friction. Compare the work () you need to do to accelerate it from 0 to speed to the work () needed to accelerate it from speed to the speed of 2 in the presence of friction. W2 > W1 W1 > W2 W1

Answers

Answer:

Explanation:

In absence of friction

Work done by force is given by change in kinetic energy of box

According work Energy theorem change in kinetic Energy of object is equal to work done by all the forces

[tex]W_1=\frac{1}{2}mv^2-0[/tex]

[tex]W_1=\frac{1}{2}mv^2[/tex]

where m=mass of object

[tex]W_2=\frac{1}{2}m(2v)^2-\frac{1}{2}mv^2[/tex]

[tex]W_2=\frac{3}{2}mv^2[/tex]

[tex]\frac{W_2}{W_1}=\frac{\frac{3}{2}mv^2}{\frac{1}{2}mv^2}[/tex]

[tex]W_2>W_1[/tex]

so work done in second case is three times of first case

(b)In Presence of friction

[tex]W_f+W_1=\frac{1}{2}mv^2[/tex]

suppose [tex]f_r[/tex] is the friction force and d is the displacement

so [tex]W_f=f_r\cdot d\cos 180[/tex]

[tex]W_f=-f_r\cdot d[/tex]

[tex]W_1=\frac{1}{2}mv^2+f_r\cdot d[/tex]

for second case when speed increases v to 2v

[tex]W_2=\frac{3}{2}mv^2+f_r\cdot d[/tex]

thus [tex]W_2>W_1[/tex] when friction is present                  

The work done to accelerate the box from zero is lesser than the work done to accelerate the box from velocity [tex]v[/tex].

From Work- energy theorem,

Work can be defined as the change in the kinetic energy of the object.

So,

In the first Scenario,

[tex]W_1 = \dfrac 12 mv^2- 0\\\\W_1 = \dfrac 12 mv^2[/tex]

In the second scenario,

[tex]W_2 = \dfrac 12 m(2v^2) -\dfrac 12 mv^2\\\\W_2 = \dfrac 32 mv^2[/tex]

Now compare both scenarios,

[tex]\dfrac {W_1}{W_2}=\dfrac{ \dfrac 32 mv^2}{\dfrac12 mv^2}[/tex]

So, [tex]W_1 <W_2[/tex]

Therefore, the work done to accelerate the box from zero is lesser than the work done to accelerate the box from velocity [tex]v[/tex].

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On a summer day in Narragansett, Rhode Island, the air temperature is 74 F and the barometric pressure is 14.5 lbf/in . Estimate the air density in kg/m.

Answers

There are some mistakes in the question as units pressure and air density are not written correctly.The correct question is here

On a summer day in Narragansett, Rhode Island, the air temperature is 74°F and the barometric pressure is 14.5 lbf/in². Estimate the air density in kg/m³.

Answer:

p=1.175 kg/m³

Explanation:

Given data

Temperature =74 F

Barometric pressure=14.5 lbf/in²

To find

Air density

Solution

From ideal gas law

[tex]pV=mRT\\p=(m/V)RT\\[/tex]

As mass/volume is pressure So

[tex]p=\frac{P}{RT}[/tex]

First we need to convert barometric pressure lbf/in² to N/m²

So

[tex]Pressure=(14.5lbf/in^{2} )*(\frac{6894.76N/m^{2} }{lbf/in^{2} } )\\Pressure=99974.02N/m^{2}[/tex]

Now Substitute the given value and pressure to find air density

[tex]p=\frac{P}{RT}\\ p=\frac{99974.02N/m^{2} }{(287J/kg.K)(\frac{74^{o}F-32}{1.8}+273.15 )}\\ p=\frac{99974.02N/m^{2} }{(287J/kg.K)*296.48K}\\ p=1.175kg/m^{3}[/tex]  

If the x-component of a force vector is 2.16 newtons and its y-component is 4.11 newtons, then what is its magnitude? F = newtons

Answers

Answer:

Magnitude of resultant vector will be 4.643 N

Explanation:

We have given x-component of the force vector is given [tex]F_{X}=2.16N[/tex]

And y component of the force vector is given [tex]F_{y}=4.11N[/tex]

We have to find the magnitude of resultant force F

Resultant force will be equal to vector sum of magnitude of x competent and y component of force vector

So F = [tex]\sqrt{F_X^2+F_Y^2}=\sqrt{2.16^2+4.11^2}=4.643N[/tex]

So magnitude of resultant vector will be 4.643 N

given that a car's distance from a stop sign t seconds since it started moving is represented by d=t2-2t determine the car's average speed over each interval of elapsed time

Answers

Answer:

Average speed =t-2t

Explanation:

Equation given for Average speed = Total distance/Elapsed time

Given: Total distance=d=t^2-2t

Elapsed time= t

Substituting into the equation

Average speed = t^2-2t/t

Dividing both numerator an denomination with t

Average speed=t-2t

A car starts from rest at a stop sign. It accelerates at 4.6 m/s^2 for 6.2 s , coasts for 2.1s , and then slows down at a rate of 3.3 m/s^2 for the next stop sign. How far apart the are the stop signs?

Answers

Answer:

D = 271.54 m

Explanation:

given,

1. car accelerates at 4.6 m/s² for 6.2 s

2. constant speed for 2.1 s

3. slows down at 3.3 m/s²

distance travel for case 1

using equation of motion

 [tex]d_1 = u t +\dfrac{1}{2}at^2[/tex]

 [tex]d_1 =\dfrac{1}{2}\times 4.6\times 6.2^2[/tex]

      d₁ = 88.41 m

case 2

constant speed for 2.1 s now, we have to find velocity

v = u  + at

v = 0 + 4.6 x 6.2

v = 28.52 m/s

distance travel in case 2

d₂ = v x t

d₂ = 28.52 x 2.1 = 59.89 m

for case 3

distance travel by the car

v² = u² + 2 a s

final velocity if the car is zero

0² = 28.52² + 2 x (-3.3) x d₃

6.6 d₃ = 813.39

 d₃ = 123.24 m

total distance travel by the car

D = d₁ + d₂ + d₃

D = 88.41 + 59.89 + 123.24

D = 271.54 m

A charge Q is located inside a rectangular box. Theelectric flux through each of the six surfaces of the box is: electric flux 1 = +1500 N*m^2/C, electric flux 2 = +2200N*m^2/C, electric flux 3 = +4600 N*m^2/C, electric flux 4 = -1800N*m^2/C, electric flux 5 = -3500 N*m^2/C, and electric flux 6 =-5400 N*m^2/C. What is Q?

Answers

To solve this problem we will apply the laws of gaus that relate the Electric Flow as the charge of the object on the permittivity constant of free space. Mathematically this is

[tex]\phi = \frac{Q}{\epsilon_0}[/tex]

Rearranging to find the charge,

[tex]Q = \phi \epsilon_0[/tex]

Here

Q = Charge

[tex]\phi =[/tex] Electric Flux

[tex]\epsilon_0 =[/tex] Permittivity of free space

The total flux would be

[tex]\phi_T = \phi_1+\phi_2+\phi_3+\phi_4+...+\phi_{\infty}[/tex]

[tex]\phi = ( 1500+2200+4600-1800-3500-5400 ) N\cdot m^2 / C[/tex]

[tex]\phi = - 2400 N\cdot m^2 / C[/tex]

Replacing we have that,

[tex]Q = (-2400 N\cdot m^2/C)( 8.85*10^{-12} C^2 / N \cdot m^2)[/tex]

[tex]Q = -21240 * 10^{-12} C[/tex]

[tex]Q = - 21.24 nC[/tex]

Therefore the charge Q inside a rectangular box is -21.24nC

Final answer:

Using Gauss's Law, the charge Q inside the box can be calculated by summing the electric flux values across all six surfaces and then multiplying by the permittivity of free space, resulting in a charge of approximately -1.24 x 10^-8 C.

Explanation:

To determine the charge Q inside the rectangular box, we need to use Gauss's Law, which relates the electric flux through a closed surface to the charge enclosed within that surface. The net electric flux (Φnet) through a closed surface is equal to the charge inside (Q) divided by the permittivity of free space (ε0). Mathematically, this is expressed as Φnet = Q/ε0.

The net flux is the algebraic sum of the fluxes through each surface, so we have:

Φnet = (electric flux 1) + (electric flux 2) + (electric flux 3) + (electric flux 4) + (electric flux 5) + (electric flux 6)

Substituting the given values, we get:

Φnet = (+1500 N·m2/C) + (+2200 N·m2/C) + (+4600 N·m2/C) + (-1800 N·m2/C) + (-3500 N·m2/C) + (-5400 N·m2/C)

Summing these yields:

Φnet = -1400 N·m2/C

Assuming ε0 is the permittivity of free space (approximately 8.854 x 10-12 C2/N·m2), we can find Q:

Q = Φnet ε0

Q = (-1400 N·m2/C)(8.854 x 10-12 C2/N·m2)

Q ≈ -1.24 x 10-8 C

Thus, the charge Q inside the box is approximately -1.24 x 10-8 C.

A sandbag is dropped from a balloon which is ascending vertically at a constant speed of 5 m/s. The bag is released with the same upward velocity of 5 m/s when t = 0 and hits the ground when t = 8 s.

(a) Determine the speed of the bag as it hits the ground.

(b) Determine the altitude of the balloon when the bag hits the ground

Answers

Answer:

a) 73.48 m/s

b) 313.92 m

Explanation:

Data provided in the question:

ascending velocity = - 5 m/s      [ negative sign depicts upward movement]

Time taken by bag to hit the ground, t = 8 s

a) from the Newton's equation of motion

we have  

[tex]s=ut+\frac{1}{2}at^2[/tex]

where,  

u is the initial speed  

a is the acceleration = 9.81 m/s²   (since it is a case of free fall )

s is the distance

thus,

[tex]s=(-5)(8)+\frac{1}{2}(9.81)(8)^2[/tex]

s = - 40 + 313.92

s = 273.92 m

from

v = u + at

v is the final speed

v = -5 + (9.81)(8)

or

v = 73.48 m/s

b) Distance traveled by balloon  = Speed × Time

= 5 × 8

= 40 m

Therefore,

Altitude of the balloon

= Distance traveled by bag + Distance traveled by balloon

= 273.92 m + 40 m

= 313.92 m

Final answer:

The speed of the bag as it hits the ground is 5 m/s. The altitude of the balloon when the bag hits the ground is 40 meters.

Explanation:

To determine the speed of the bag as it hits the ground, we can use the equation for velocity:

v = u + at

Since the bag is dropped with an initial velocity of 5 m/s and there is no acceleration in the vertical direction, the final velocity of the bag as it hits the ground is also 5 m/s.

To determine the altitude of the balloon when the bag hits the ground, we can calculate the distance traveled by the bag using the equation for distance:

s = ut + (1/2)at^2

Plugging in the values, we get:

s = 5(8) + (1/2)(0)(8^2) = 40 meters

Learn more about Calculating velocity and distance of a falling object here:

https://brainly.com/question/32413086

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Suppose the ring rotates once every 4.30 ss . If a rider's mass is 58.0 kgkg , with how much force does the ring push on her at the top of the ride? Express your answer with the appropriate units.

Answers

Answer:

422.36 N

Explanation:

given,

time of rotation = 4.30 s

T = 4.30 s

Assuming the diameter of the ring equal to 16 m

radius, R = 8 m

[tex]v = \dfrac{2\pi R}{T}[/tex]

[tex]v = \dfrac{2\pi\times 8}{4.30}[/tex]

  v = 11.69 m/s

now, Force does the ring push on her at the top

[tex]- N - m g = \dfrac{-mv^2}{R}[/tex]

[tex] N + m g = \dfrac{mv^2}{R}[/tex]

[tex] N = \dfrac{mv^2}{R}- m g[/tex]

[tex] N = m(\dfrac{v^2}{R}- g)[/tex]

[tex] N = 58\times (\dfrac{11.69^2}{8}- 9.8)[/tex]

N = 422.36 N

The force exerted by the ring to push her is equal to 422.36 N.

The force does the ring push on her at the top of the ride will be [tex]N=422.36\ Newton[/tex]

What will be the force does the ring push on her at the top of the ride?

It is Given that

Time rotation  T= 4.30 s

Mass m= 58 kg

Now the Velocity will be calculated as

[tex]V=\dfrac{2\pi r}{T} =\dfrac{2\pi 8}{4.30} =11.69 \frac{m}{s}[/tex]

Now by balancing the forces

[tex]N=\dfrac{mv^2}{R} -mg[/tex]

[tex]N=m(\dfrac{v^2}{R} -g)[/tex]

[tex]N=58\times (\dfrac{11.69^2}{8} -9.8)[/tex]

[tex]N=422.36 \ Newton[/tex]

Thus the force does the ring push on her at the top of the ride will be [tex]N=422.36\ Newton[/tex]

To know more about the balancing of forces follow

https://brainly.com/question/517289

Why is a frequency distribution useful? It allows researchers to see the "shape" of the data. It tells researchers how often the mean occurs in a set of data. It can visually identify the mean. It ignores outliers.

Answers

Answer: Frequency distribution helps to understand data. some complex data needs to be converted into intervals to see its frequency of occurrence. it helps to find the mean median and mode of the data. It helps in the analysis of data and its practical implementation.

Explanation:

A circular-shaped object of mass 10 kg has an inner radius of 13 cm and an outer radius of 26 cm. Three forces (acting perpendicular to the axis of rotation) of magnitudes 12 N, 26 N, and 16 N act on the object, as shown. The force of magnitude 26 N acts 30❦ below the horizontal. 12 N 16 N 26 N 30❦ ϝ Find the magnitude of the net torque on the wheel about the axle through the center of the object. Answer in units of N · m

Answers

The net torque on the circular object, considering the three forces acting on it, is calculated to be 1.46 N × m in an anti-clockwise direction.

To find the net torque on the circular object about its axle, we will first consider each of the three forces acting on it separately and determine the torque produced by each force. Torque (Τ) is defined by the equation Τ = r  ×  F  × sin(θ), where r is the radius at which the force is applied, F is the magnitude of the force, and θ is the angle between the force and the direction of the radius.

The 12 N force acts on the outer radius, so r = 0.26 m (26 cm converted to meters). Since the force is perpendicular to the radius, θ = 90 degrees, and the sin(90) = 1. Therefore, the torque from this force is Τ = 0.26 m × 12 N × 1 = 3.12 N × m.

The 16 N force also acts on the outer radius at 90 degrees to the radius, so its torque is Τ = 0.26 m × 16 N × 1 = 4.16 N   × m.

The 26 N force acts at an angle 30 degrees below the horizontal, so it makes an angle of 60 degrees with the radius. Hence, the torque from this force is Τ = 0.26 m × 26 N  × sin(60) = 5.82 N × m (rounded to two decimal places).

To find the net torque, we need to consider the direction of each torque. Both the 12 N and 16 N forces produce torque in the same direction (let's say clockwise), whereas the 30 degree component of the 26 N force produces torque in the opposite direction (counter-clockwise).

Net torque = Torque from 12 N + Torque from 16 N - Torque from 26 N = 3.12 N × m + 4.16 N × m - 5.82 N × m = 1.46 N   × m (anti-clockwise).

The 9-m boom AB has a fixed end A. A steel cable is stretched from the free end B of the boom to a point Clocated on the vertical wall. If the tension in the cable is 3 kN, determine the moment about A of the force exertedby the cable at B.

Answers

Answer:

27000 Nm

Explanation:

The boom end at A is fixed and end at B is subjected to a 3kN force. Boom AB has length of 9m. The moment about A is the product of force 3kN (3000 N) at B and the moment arm of 9m

M = FL = 3000 * 9 = 27000 Nm

So the moment about A is 27000 Nm

A simple harmonic oscillator completes 1550 cycles in 30 min. (a) Calculate the period. s (b) Calculate the frequency of the motion. Hz

Answers

Answer:

(a) 1.16 s

(b)0.861 Hz

Explanation:

(a) Period : The period of a simple harmonic motion is the time in seconds, required for a object undergoing oscillation to complete one cycle.

From the question,

If 1550 cycles is completed in (30×60) seconds,

1 cycle is completed in x seconds

x = 30×60/1550

x = 1.16 s

Hence the period is 1.16 seconds.

(b) Frequency : This can be defined as the number of cycles that is completed in one seconds, by an oscillating body. The S.I unit of frequency is Hertz (Hz).

Mathematically, Frequency is given as

F = 1/T ........................... Equation 1

Where F = frequency, T = period.

Given: T = 1.16 s.

Substitute into equation 1

F = 1/1.16

F = 0.862 Hz

Hence thee frequency = 0.862 Hz

Water is pumped from a lake to a storage tank 15 m above at a rate of 90 L/s while consuming 15.4 kW of electric power. Disregarding any frictional losses in the pipes and any changes in kinetic energy, answer the questions that follow.
a) the overall efficiency of the pump motor unit.
b) the pressure difference between the inlet and the exit of the pump.

Answers

Answer:

B

Explanation:

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