A student carries a backpack for one mile. Another student carries the same backpack for two miles

Compared to the first student, how much work did the second student do?

Answer:

They are equal.

Explanation:

Answer:

i think rocks towards is correct answer

Answer:

It is gently on

Explanation:

I'm am not a 100 percent sure but try this o took that before

Answer:

220 bastanyan sagot ko yawa

Answer:

density = mass/volume

so . . .  

density = (36 g)/(15 cm³) = 2.4 g/cm³

Explanation:

Answer:

c. yes, 7.7 cm from the 5-A wire

Explanation:

Given;

distance between the two wires, r = 20 cm = 0.2 m

first current, I₁ = 5.0 A

second current, I₂ = 8.0 A

The magnetic field due to the two wires occurs in different directions and it can be zero at this region.

Let x be the distance from 5 A wire where the magnetic field is zero.

[tex]B = \frac{\mu_o}{2\pi} [\frac{I_1}{x} -\frac{I_2}{r-x} ] = 0\\\\ \frac{\mu_o}{2\pi} [\frac{I_1}{x} -\frac{I_2}{r-x} ] = 0\\\\ \frac{\mu_o}{2\pi} [\frac{5}{x} -\frac{8}{0.2-x} ] = 0\\\\\frac{5}{x} -\frac{8}{0.2-x} = 0\\\\\frac{5}{x} = \frac{8}{0.2-x}\\\\5(0.2-x) = 8x\\\\1 -5x = 8x\\\\1 = 5x \ + \ 8x\\\\1 = 13x\\\\x = \frac{1}{13} \\\\x = 0.077 \ m\\\\x = 7.7 \ cm[/tex]

Therefore, the correct option is c. yes, 7.7 cm from the 5-A wire

The Answer to your question is:

A.

The Earth's tectonic activity occurs by the movement of the fourteen main tectonic plates of the planet, which move over the mantle continuously, and result in the formation of mountains, earthquakes, tsunamis, volcanic activities, etc.

The decrease in the Earth's tectonic activity would therefore occur if the Earth's mantle became cooler, as the tectonic plates move over the magma, which is a paste formed by silicate of iron and magnesium and whose temperature reaches 600° and 1,200° Celsius , driven by forces from inside the planet.

Therefore the letter C is correct, as with the Earth's mantle cooler, the magma would become more solid which would decrease the Earth's tectonic activity.

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Answer:

please put pic of the questions

Answer:

a) True

Explanation:

There are several possible explanations for this positive charge

* The explanation of the small positive charge in the plant when the bee approaches is like a defense system of the plants,

to prevent the bees from taking the pollen, but the flowers need the bees to transport the pollen for fertilization, so this possibility is not correct

* The air is conductive so the bee indexes a charge in the nearby air, this charge must be negative and this charge induced in the air induces a charge on the flower that must be positive.

When reviewing the different statements we have

a) True, it agrees with the second explanation of the phenomenon

b) False. The earth is a deposit of negative charge

c) false. If this is the case the charge should be negative

d) False. This residual charge from the other bees is quickly neutralized by the charge from the Earth.

Answer:

Explanation:

.

This question is incomplete, the complete question is;

A monatomic gas fills the left end of the cylinder in the following figure. At 300 K , the gas cylinder length is 14.0 cm and the spring is compressed by65.0 cm . How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?

Answer:

the required heat energy is 16 J

Explanation:

Given the data in the question;

Lets consider the ideal gas equation;

PV = nRT

from the image, we calculate initial pressure;

Pi = ( 2000N/M × 0.06m) / 0.0008 m²

Pi = 15 × 10⁴ Pa

next we find Initial velocity

Vi = (0.0008 m²)(0.14) = 1.1 × 10⁻⁴ m²

now we find the number of moles

n = [(15 × 10⁴ Pa)(1.1 × 10⁻⁴ m²)] / 8.31 J/molK × 300K

N = 6.6 × 10⁻³ mol

next we calculate the final temperature;

Pf = ( 2000N/m × 0.08) / 0.0008 m²

Pf = 2 × 10⁵ Pa

Calculate the final Volume

Vf = (0.0008 m² × 0.16 m = 1.28 × 10⁻⁴ m³

we also determine the final temperature

[tex]T_{f}[/tex] =  (2 × 10⁵ Pa × 1.28 × 10⁻⁴ m³) / 6.6 × 10⁻³ × 8.31 J/molK

[tex]T_{f}[/tex]  = 466.8 K

so change in temperature ΔT

ΔT =  466.8 K - 300K = 166.8 K

we then calculate the change in thermal energy

ΔU = nCΔT

ΔU = ( 6.6 × 10⁻³ mol ) × 12.5 × 166.8K

ΔU = 13.761 J

C is the isochoric molar specific heat which is equal to 3R/2 for monoatomic

now we calculate the work done;

W = 1/2 × K( x[tex]_{i\\}[/tex]² - x[tex]_{f\\}[/tex]² )

W = 1/2 × ( 2000 N/m) ( 0.06² - 0.08² )

= - 2.8 J

and we then calculate the heat energy using the following expression;

Q = ΔU - W

we substitute

Q = 13.761 - (- 2.8 J)

Q = 13.761 + 2.8 J)

Q =  16 J

Therefore, the required heat energy is 16 J

Answer: gravitational potential energy

Explanation:

Answer:

8 units

Explanation:

Answer:

$ 5.93

Explanation:

If we are to put into consideration the following factors even though it was not all stated in the question.

What is the cost of heating a hot tub containing 1500 kg of water from 10.0°C to 40.0°C, assuming 75.0% efficiency to take heat loss to surroundings into account? The cost of electricity is ¢8.50/kW·h.

Q = mc∆T .........1

Where m is the mass of water = 1500kg

C is the specific heat capacity of water = 4.184

∆T = temperature change

= 40° - 10°

= 30°

Now, substitute the values into equation 1

= 1500×4.184×30

= 188280

Q in kw/s

= 188280/3600

= 52.3kw/hr

Efficiency = 75%

= 75/100

= 0.75

Hence we have

52.3/0.75

= 69.73kw/hr

C = Q × cost per kw/hr

C = 69.73× $8.5/100

C = 5.92705

C =$ 5.93

Hence, the cost of heating the hot tub is $ 5.93

Solution :

Given :

Rectangular wingspan

Length,L = 17.5 m

Chord, c = 3 m

Free stream velocity of flow, [tex]$V_{\infty}$[/tex] = 200 m/s

Given that the flow is laminar.

[tex]$Re_L=\frac{\rho V L}{\mu _{\infty}}$[/tex]

      [tex]$=\frac{1.225 \times 200 \times 3}{1.789 \times 10^{-5}}$[/tex]

    [tex]$= 4.10 \times 10^7$[/tex]

So boundary layer thickness,

[tex]$\delta_{L} = \frac{5.2 L}{\sqrt{Re_L}}$[/tex]

[tex]$\delta_{L} = \frac{5.2 \times 3}{\sqrt{4.1 \times 10^7}}$[/tex]

    = 0.0024 m

The dynamic pressure, [tex]$q_{\infty} =\frac{1}{2} \rho V^2_{\infty}$[/tex]

                                           [tex]$ =\frac{1}{2} \times 1.225 \times 200^2$[/tex]

                                          [tex]$=2.45 \times 10^4 \ N/m^2$[/tex]

The skin friction drag co-efficient is given by

[tex]$C_f = \frac{1.328}{\sqrt{Re_L}}$[/tex]

     [tex]$=\frac{1.328}{\sqrt{4.1 \times 10^7}}$[/tex]

     = 0.00021

[tex]$D_{skinfriction} = \frac{1}{2} \rho V^2_{\infty}S C_f$[/tex]

                  [tex]$=\frac{1}{2} \times 1.225 \times 200^2 \times 17.5 \times 3 \times 0.00021$[/tex]

                  = 270 N

Therefore the net drag = 270 x 2

                                      = 540 N

Answer:

Explanation:

If objects stick together, then a collision is inelastic. When objects don't stick together, If the kinetic energy is the same, then the collision is elastic.

Consider the balloon and air inside the flask to be in a closed system. Using the First Law of Thermodynamics to explain what happened to the balloon as heat was added by the environment.

Balloon will burst, due to more heat and in another flask air gets heated.

"The first law of thermodynamics states that energy can neither be created nor destroyed, only altered in form. For any system, energy transfer is associated with mass crossing the control boundary, external work, or heat transfer across the boundary. These produce a change of stored energy within the control volume."

"Heat is the energy that is transferred from one body to another as the result of a difference in temperature. If two bodies at different temperatures are brought together, energy is transferred i.e, heat flows—from the hotter body to the colder."

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Answer:

Explanation:

Water.

Carbohydrates.

Protein-Amino acids.

Fat.

Vitamins.

Minerals.

Omega-3 fatty acids.

Answer:

16.6 m

Explanation:

Let d be the distance the other person is ahead of you. Since the other person is walking at a speed, v = 1.17 m/s, after picking the wallet, the other person moves a distance , vt in time, t = 10.5 s, the total distance covered by you till catch up is D = d + vt.

Also, you moves with a speed of v' = 2.75 m/s in time t = 10.5 s as you pick up the wallet, you covers a distance d' = v't at catch up.

At catch up, D = d'

d + vt = v't

d = v't - vt

d = (v' - v)t

Substituting the values of the variables into d, we have

d = (2.75 m/s - 1.17 m/s)10.5 s

d = (1.58 m/s)10.5 s

d = 16.59 m

d ≅ 16.6 m

So, the other person was 16.6 m ahead of you when you started running.

Here is a copy+pasted definition...

Transactive memory is a psychological hypothesis first proposed by Daniel Wegner in 1985 as a response to earlier theories of "group mind" such as groupthink.[1] A transactive memory system is a mechanism through which groups collectively encode, store, and retrieve knowledge. Transactive memory was initially studied in couples and families where individuals had close relationships but was later extended to teams, larger groups, and organizations to explain how they develop a "group mind",[1] a memory system that is more complex and potentially more effective than that of any of its individual constituents. A transactive memory system includes memory stored in each individual, the interactions between memory within the individuals, as well as the processes that update this memory. Transactive memory, then, is the shared store of knowledge.

In simple terms...

A transactive memory is a psychological hypothesis. It is a system in the brain which allows human's to encode, store, or get information.

For example...

let's say I learn 5 colors when I'm younger. My "transactive memory" will store it and will allow my brain to "retrieve" the information to allow me to remember colors I previously "stored".

Please mark brainliest!

Answer:

[tex]0.35598[/tex]

Explanation:

r = Radius = [tex]\dfrac{0.9}{2}=0.45\ \text{cm}[/tex]

R = Resistance = [tex]198\ \Omega[/tex]

A = Area = [tex]\pi r^2[/tex]

l = Length of blood in cylinder = 1 cm

h = Hematocrit of the blood

Resistivity is given by

[tex]\rho=\dfrac{1.32}{1-h}-0.79[/tex]

Resistance is given by

[tex]R=(\dfrac{1.32}{1-h}-0.79)\dfrac{l}{\pi r^2}\\\Rightarrow h=1-\dfrac{1.32}{\dfrac{R\pi r^2}{l}+0.79}\\\Rightarrow h=1-\dfrac{1.32}{\dfrac{198\times \pi\times (0.45\times 10^{-2})^2}{0.01}+0.79}\\\Rightarrow h=0.35598[/tex]

The hematocrit of the blood is [tex]0.35598[/tex].

Answer:

Hello your question is poorly written below is the complete question

Suppose the battery in a clock wears out after moving Ten thousand coulombs of charge through the clock at a rate of 0.5 Ma how long did the clock run on does battery and how many electrons per second slowed?

answer :

a) 231.48 days

b) n = 3.125 * 10^15

Explanation:

Battery moved 10,000 coulombs

current rate = 0.5 mA

A) Determine how long the clock run on the battery. use the relation below

q = i * t ----- ( 1 )

q = charge , i = current , t = time

10000 = 0.5 * 10^-3 * t

hence  t = 2 * 10^7 secs

hence the time = 231.48  days

B) Determine how many electrons per second flowed

q = n*e ------ ( 2 )

n = number of electrons

e = 1.6 * 10^-19

q = 0.5 * 10^-3 coulomb ( charge flowing per electron )

back to equation 2

n ( number of electrons ) = q / e = ( 0.5 * 10^-3 ) / ( 1.6 * 10^-19 )

hence : n = 3.125 * 10^15

Answer:

Explanation:

Yes , it is possible .

When a negative charge moves towards a positive charge , it is moving in the direction of increasing electrical potential . In the whole process , its electrical potential energy decreases and its kinetic energy increases .

Actually the potential energy of a negative charge near a positive charge is negative and it is inversely proportional to distance .

V = - Qq / R , When R decreases , the negative value of potential increases . That means potential energy decreases .

Answer:

R' = R/4

Explanation:

The resistance of a metal rod is R. It is given by the relation as follows :

[tex]R=\rho\dfrac{l}{A}[/tex]

Where

l is the length and A is the area of cross-section

[tex]A=\pi r^2=\pi (\dfrac{d}{2})^2[/tex]

If both its length and its diameter are quadrupled, it means,

l' = 4l

and d'= 4d

It means,

[tex]A'=\pi (\dfrac{4d}{2})^2[/tex]

Let new resistance be R'. So,

[tex]R'=\rho\dfrac{l'}{A'}\\\\R'=\rho\dfrac{4l}{\pi (\dfrac{4d}{2})^2}\\\\=\rho \dfrac{4l}{\pi \dfrac{16d^2}{2}}\\\\=\dfrac{4}{16}\times \dfrac{\rho l}{\pi \dfrac{d^2}{2}}\\\\=\dfrac{1}{4}\times \dfrac{\rho l}{\pi \dfrac{d^2}{2}}\\\\R'=\dfrac{R}{4}[/tex]

So, the correct option is (B) "R/4".

Answer:

Hello your question is incomplete attached below is the complete question

answer :

Total charge enclosed within the sphere : [tex]\frac{q_{r1} }{4\pi e_{0}R^3 } . r[/tex]

Total charge enclosed outside the sphere : [tex]\frac{q}{4\pi e_{0}r^2 } .r[/tex]

Explanation:

Given data:

Total charge of a uniformly charged sphere = Q

radius = R

first step : find the electric field inside and outside the uniformly charged sphere

2nd step : determine the total charge enclosed within and outside the sphere

make a sketch of the uniformly charged sphere

Attached below is a detailed solution

Answer:

Kim was adopted as a baby and raised by loving parents in an enriched environment. Studies have shown that the correlation between IQ

( intelligence quotient) scores of adopted children and those of their biologically unrelated family members reduce to zero as they grow into adulthood.

Explanation:

Kim was adopted as a baby and raised by loving parents in an enriched environment. Studies have shown that the correlation between IQ

( intelligence quotient) scores of adopted children and those of their biologically unrelated family members reduce to zero as they grow into adulthood.

Also, a loving environment is equally important for a child's upbringing.

Answer:

The distance (in miles) by the bus up the hill when its speed decreased to 50 mph is approximately 1.353 miles

Explanation:

The parameters of the motion of the driver are;

The upgrade of the road, θ = 10°

The rate of constant acceleration of the bus driver = 5 ft./s²

The speed of the bus as it begins to go up the hill, v₁ = 80 mph = 117.3228 ft./s

The speed of the driver at a point on the hill, v₂ = 50 mph ≈ 73.32677 ft./s

The acceleration due to gravity, g ≈ 32.1740 ft./s²

Therefore, we have;

The acceleration due to gravity down the incline plane, gₓ = g·sinθ

∴ gₓ = g·sin(θ) ≈ 32.1740 ft./s² × sin(10°) ≈ 5.587 ft/s²

The net acceleration of the bus, on the incline plane, [tex]a_{Net}[/tex] = gₓ - a = 5.587 ft./s² -5 ft./s² = 0.587 ft./s²

The vertical component of the velocity, [tex]v_y[/tex] = v × sin(θ)

∴ [tex]v_y[/tex] = 117.3228 ft./s × sin(10°) ≈ 20.37289 ft./s

vₓ = 117.3228 ft./s × cos(10°) ≈ 115.5404 ft./s

The velocity of the car, v₂, on the inclined plane is given as follows;

v₂ = v₁ - [tex]a_{Net}[/tex] × t

∴ t = (v₁ - v₂)/[tex]a_{Net}[/tex]  = (117.3228 ft./s - 73.32677 ft./s)/(0.587 ft./s²) ≈ 74.95 s

The distance covered, 's', is given as follows;

s = v₁·t - 1/2·[tex]a_{Net}[/tex]·t²

∴ s = 117.3228 × 74.95 - 1/2 × 0.587 × 74.95² ≈ 7144.6069 ft.

The distance travelled up the hill, s ≈ 7144.6069 ft. ≈ 1.3531452 miles ≈ 1.353 miles

Answer:

5.7375 seconds

Explanation:

The computation of the time required to reach that maximum velocity is shown below:

Given that

Mass = m =  5100 kg

Net upward force F = 8 × times 10^5 N

Initial speed = V_i = 0

Maximum speed = V = 900 m.s

Based on the above information

Impluse J = m(V - V_i)

= 5100 (900 - 0)

= 459 × 10^4 kg m.s

As we know that

J = FT

So

T = J ÷ F

= (459 × 10^4) ÷ (8 × 10^5)

= 5.7375 seconds

Answer:

g ±Δg = (9.8 ± 0.2) m / s²

Explanation:

For the calculation of the acceleration of gravity they indicate the equation of the simple pendulum to use

          T = [tex]2\pi \sqrt{ \frac{L}{g} }[/tex]

          T² =  [tex]4\pi ^2 \frac{L}{g}[/tex]4pi2 L / g

          g = [tex]4\pi ^2 \frac{L}{T^2}[/tex]

They indicate the average time of 20 measurements 1,823 s, each with an oscillation

let's calculate the magnitude

           g = [tex]4\pi ^2 \frac{0.823}{1.823^2}[/tex]4 pi2 0.823 / 1.823 2

            g = 9.7766 m / s²

now let's look for the uncertainty of gravity, as it was obtained from an equation we can use the following error propagation

for the period

             T = t / n

             ΔT = [tex]\frac{dT}{dt}[/tex] Δt + [tex]\frac{dT}{dn}[/tex] ΔDn

In general, the number of oscillations is small, so we can assume that there are no errors, in this case the number of oscillations of n = 1, consequently

              ΔT = Δt / n

              ΔT = Δt

now let's look for the uncertainty of g

             Δg = [tex]\frac{dg}{dL}[/tex] ΔL + [tex]\frac{dg}{dT}[/tex]  ΔT

             Δg = [tex]4\pi ^2 \frac{1}{T2}[/tex]   ΔL + 4π²L  (-2  T⁻³) ΔT

a more manageable way is with the relative error

             [tex]\frac{\Delta g}{g} = \frac{\Delta L }{L} + \frac{1}{2} \frac{\Delta T}{T}[/tex]

we substitute

              Δg = g ( \frac{\Delta L }{L} + \frac{1}{2}  \frac{\Delta T}{T}DL / L + ½ Dt / T)

the error in time give us the stanndard deviation  

let's calculate

               Δg = 9.7766 ([tex]\frac{0.001}{0.823} + \frac{1}{2} \ \frac{0.671}{1.823}[/tex])

               Δg = 9.7766 (0.001215 + 0.0184)

               Δg = 0.19 m / s²

the absolute uncertainty must be true to a significant figure

                Δg = 0.2 m / s2

therefore the correct result is

               g ±Δg = (9.8 ± 0.2) m / s²

Answer:

42,250

Explanation:

It goes inside=

Displacemt

It does work=

Work done

To find efficiency of jule we do=

Dicplacement × Work done

650 × 65

42,250

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