Heat and thermodynamics Numerical ​

Answer:

A)a red giant star

Answer:

-4.7 x 10^-3 J/K-s

Explanation:

The Power generated by metabolizing food = 80 W

The watt W is equivalent to the Joules per sec J/s

therefor power = 80 J/s

20% of this energy is not used for heating, amount available for heating is

==> H = 80% of 80 = 0.8 x 80 = 64 J/s

The inner body temperature = 37 °C = 273 + 37 = 310 K

The entropy of this inner body ΔS = ΔH/T

ΔS = 64/310 = 0.2065 J/K-s

The skin temperature is cooler than the inner body by 7 °C

Temperature of the skin =  37 - 7 = 30 °C = 273 + 30 = 303 K

The entropy of the skin = ΔS = ΔH/T

ΔS = 64/303 = 0.2112 J/K-s

change in entropy of the person's body = (entropy of hot region: inner body) - (entropy of cooler region: skin)

==> 0.2065 - 0.2112 = -4.7 x 10^-3 J/K-s

Answer: d₂ = 170 mGya

Explanation:

the relationship between absonbed 'd' and exposure 'E' is given as;

D(Gv) = F . x (AS/xB)

F is a conversion coefficient depending on medium

so we can simply write

d₁/d₂ = x₁/x₂

Given that;

our x₁ = 60 mAs, x₂ = 120 mAs,  d₁ = 85 mGya, d₂ = ?

from the given formula,

d₂ = (x₂d₁ / x₁)

now we substitute

d₂ = (120 × 85) / 60

d₂ = 170 mGya

∴ if 120 mAa is used,  the new exposure will be 170 mGya

Answer: Astro 1 will have a 10x larger field of view than the Hubble.

Explanation: The hubble will also be extremely light weight that way it can go further into space and the mission will be able to last a longer amount of time.

Explanation:

It is given that,

The velocity of football is 18 m/s

It is projected at an angle of 20 degrees

We need to find the ball's acceleration in the horizontal direction as it flies through the air.

When it is projected with some velocity, it has two rectangular components i.e. horizontal and vertical.

In vertical direction, it will move under the action of gravity. There is no change in velocity in horizontal direction. So, ball's acceleration in the horizontal direction is equal to 0.

Answer:

a)step-down" transformer

b) 1440 turns

Explanation:

There are two types of transformers; step up transformers and step down transformers. A step down transformer converts a higher voltage to a lower voltage.

In a stepdown transformer, there are more turns in the primary coil than in the secondary coil, the turns ratio Ns/Np is less than 1 for a stepdown transformer.

If

Number of turns in primary coil Np= 2760

Number of turns in secondary coil Ns= unknown

Voltage in primary coil Vp= 230 V

Voltage in secondary coil Vs= 120 V

Ns/Np= Vs/VP

NsVp= NpVs

Ns= NpVs/VP = 2760 × 120/230

Ns= 1440 turns

Answer:

0.6 m/s

Explanation:

2Hz = 2^-1 = 2 /s

30cm = .3m

Velocity is in the units m/s, so multiplying wavelength in meters by the frequency will give you the velocity.

(.3m)*(2 /s) = 0.6 m/s

Answer:

24

Explanation:

The mass = 3 kg

at point O all the mechanical energy of the system is due to its potential energy PE. The body is at rest.

PE = mgh

but ME = PE + KE = constant   (law of energy conservation)

KE is the kinetic energy

since KE is zero at this point, then,

ME = mgh

where m is the mass

g is acceleration due to gravity = 9.81 m/s^2

h is the height = O

ME = 3 x 9.81 x O

ME = 29.43-O

At point A the total ME is due to its PE and its kE

PE at this point = mgh = 3 x 9.81 x A = 29.43-A

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

velocity = v at this point, therefore,

KE = [tex]\frac{1}{2}*3*v^{2}[/tex] = [tex]\frac{3}{2} }v^{2}[/tex]

therefore,

ME = 29.43-A + [tex]\frac{3}{2} }v^{2}[/tex]

Equating ME for the points O and A, we have

29.43-O = 29.43-A + [tex]\frac{3}{2} }v^{2}[/tex]

29.43-O - 29.43-A = [tex]\frac{3}{2} }v^{2}[/tex]

(O - A)29.43 = [tex]\frac{3}{2} }v^{2}[/tex]

O - A = 0.051[tex]v^{2}[/tex]   this is the distance between point O and A

For point B

PE = 29.43-B

KE = [tex]\frac{25}{2}*3*v^{2}[/tex] = 37.5[tex]v^{2}[/tex]        (velocity is equal to [tex]5v[/tex] at this point)

therefore,

ME = 29.43-B + 37.5[tex]v^{2}[/tex]

Equating the ME for points A and B, we have

29.43-A + [tex]\frac{3}{2} }v^{2}[/tex] = 29.43-B + 37.5[tex]v^{2}[/tex]

29.43-A - 29.43-B = 37.5[tex]v^{2}[/tex] - [tex]\frac{3}{2} }v^{2}[/tex]

(A - B)29.43 = 36[tex]v^{2}[/tex]

A - B = 1.22[tex]v^{2}[/tex]    this is the distance between points A and B

The distance between points A & B divided by the distance between points 0 & A will be

1.22[tex]v^{2}[/tex]/0.051 = 23.9 ≅ 24

Answer:

Rp = 3.04×10² Ω.

Explanation:

From the question given:

1/Rp = 1/4.5×10² Ω + 1/ 9.4×10² Ω

Rp =?

We can obtain the value of Rp as follow:

1/Rp = 1/4.5×10² + 1/ 9.4×10²

Find the least common multiple (lcm) of 4.5×10² and 9.4×10².

The result is 4.5×10² × 9.4×10²

Next, divide the result of the lcm by each denominator and multiply the result obtained with the numerator as shown below:

1/Rp = (9.4×10² + 4.5×10²) /(4.5×10²) (9.4×10²)

1/Rp = 13.9×10²/4.23×10⁵

Cross multiply

Rp × 13.9×10² = 4.23×10⁵

Divide both side by 13.9×10²

Rp = 4.23×10⁵ / 13.9×10²

Rp = 3.04×10² Ω.

The color that is reflected when a red card is illuminated by red light is white.

The color an object is perceived to have, depends on the frequency of light it reflects.

If white light incidents on a red filter, red is transmitted while blue and green are absorbed.

Consequently, when a red card is illuminated by red light, the red card will  reflect back almost all the incident light on it, causing it to appear brighter which creates an  illusion of white color to the eyes.

Thus, we can conclude the color that is reflected when a red card is illuminated by red light is white.

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Answer:

(A). 1620 watt.

(B).0.8885.

Explanation:

So, we are given the following data or parameters or information which is going to assist or help us in solving this particular Question or problem. So, we have;

Current = 9.35A, direct current at a potential difference of 195 V, frequency of the inverter = 60 Hz alternating current, alternating current with Vmax = 166V and Imax = 19.5A.

(A). The rms power is produced by the inverter = (19.5 /2 ) × 166 = 1620 watt(approximately).

(B). the rms values to find the power efficiency Pout/Pin of the inverter.

P(in) = 195 × 9.35 = 1823.3 watt.

Thus, the rms values to find the power efficiency Pout/Pin of the inverter = 1620/1823.3 = 0.88852324146441793 = 0.8885.

Answer:

The film thickness is 4.32 * 10^-6 m

Explanation:

Here in this question, we are interested in calculating the thickness of the film.

Mathematically;

The number of fringes shifted when we insert a film of refractive index n and thickness L in the Michelson Interferometer is given as;

ΔN = (2L/λ) (n-1)

where λ is the wavelength of the light used

Let’s make L the subject of the formula

(λ * ΔN)/2(n-1) = L

From the question ΔN = 8 , λ = 540 nm, n = 1.5

Plugging these values, we have

L = ((540 * 10^-9 * 8)/2(1.5-1) = (4320 * 10^-9)/1 = 4.32 * 10^-6 m

Answer:

The magnetic field is 0.0857 T.

Explanation:

The electrons orbit the magnetic field with a centripetal force equal to

F = [tex]\frac{mv^{2} }{r}[/tex]

also, the force on an electron in a magnetic field is gotten as

F = Bqv

equating this two equations give

[tex]\frac{mv^{2} }{r}[/tex] = Bqv

mv/r = Bq

where m is the mass of the electron = 9.11 x 10^-31 kg

v is the the linear speed of the electron

B is the magnetic field on the electron

r is the radius of the orbital movement

q is the charge on an electron = 1.602 x 10^-19 C

but, the linear speed v = ωr

where ω is the angular speed of the electron

substituting into equation above, we have

mωr/r = Bq

which reduces to

mω = Bq

finally, w know that the angular speed is related to the frequency of the electron by

ω = 2πf

we then finally have

2mπf = Bq

where f is the frequency emitted by the electron = 2.4 GHz = 2.4 x 10^9 Hz

substituting values into the equation, we have

2 x 9.11 x 10^-31 x 3.142 x 2.4 x 10^9 = B x 1.602 x 10^-19

B = (1.3734 x 10^-20)/(1.602 x 10^-19) = 0.0857 T

= 85.7 mT

Answer:

Explanation:

We shall apply conservation of momentum law in vector form to solve the problem .

Initial momentum = 0

momentum of 12 g piece

= .012 x 37 i since it moves along x axis .

= .444 i

momentum of 22 g

= .022 x 34 j

= .748 j

Let momentum of third piece = p

total momentum

= p + .444 i + .748 j

so

applying conservation law of momentum

p + .444 i + .748 j  = 0

p = - .444 i -  .748 j  

magnitude of p

= √ ( .444² + .748² )

= .87 kg m /s

mass of third piece = 58 - ( 12 + 22 )

= 24 g = .024 kg

if v be its velocity

.024 v = .87

v = 36.25 m / s .

Answer:

Answer in explanation

Explanation:

Water is mainly used as coolant in heating systems like hot-water radiators. The main function of water in such systems, is to absorb as much heat as possible, in order to decrease the temperature of the system and as a result cool it.

The specific heat capacity is the measure of heat energy that is required to raise the temperature of unit mass of a substance through 1 °C. In other words, specific heat capacity quantifies the amount of heat that can be stored by a unit mass of a substance having a degree rise in temperature.

Thus, the more specific heat a substance has, the more heat it can absorb from the hot system. Hence, the specific heat capacity of a coolant must be high.

This is the reason why water, with its high specific heat capacity, is utilized for heating systems, such as radiators.

The question is incomplete. Here is the complete question.

The isotope of Plutonium 238Pu is used to make thermoeletric power sources for spacecraft. Suppose that a space probe was launched in 2012 with 3.5 kg of 238Pu.

(a) If the half-life of 238Pu is 87.7 yr, write a function of the form [tex]Q(t)=Q_{0}e^{-kt}[/tex] to model the quantity Q(t) of 238Pu  left after t years. Round ythe value of k to 3 decimal places. Do not round intermediate calculations.

(b) If 1.7kg of 238Pu is required to power the spacecraft's data transmitter, for low long after launch would scientists be able to receive data? Round to the nearest year. Do not round intermediate calculations.

Answer: (a) [tex]Q(t)=3.5e^{-0.0079t}[/tex]

              (b) 91 years.

Explanation:

(a) Half-life is time it takes a substance to decrease to half of itself, i.e.:

Q(t) = [tex]0.5Q_{0}[/tex]

[tex]0.5Q_{0}=Q_{0}e^{-87.7k}[/tex]

[tex]0.5=e^{-87.7k}[/tex]

[tex]ln(0.5)=ln(e^{-87.7k})[/tex]

[tex]ln(0.5)=-87.7k[/tex]

[tex]k = \frac{ln(0.5)}{-87.7}[/tex]

k = 0.0079

Knowing k and [tex]Q_{0}[/tex]=3.5kg, function is [tex]Q(t)=3.5e^{-0.0079t}[/tex]

(b) Using function:

[tex]Q(t)=3.5e^{-0.0079t}[/tex]

[tex]1.7=3.5e^{-0.0079t}[/tex]

[tex]e^{-0.0079t}=\frac{1.7}{3.5}[/tex]

[tex]e^{-0.0079t}=0.4857[/tex]

[tex]ln(e^{-0.0079t})=ln(0.4857)[/tex]

[tex]-0.0079t=-0.7221[/tex]

[tex]t = \frac{-0.7221}{-0.0079}[/tex]

t = 91.41

t ≈ 91 years

Scientists will be able to receive data for approximately 91 years.

Answer:

93.3x10^-3s

Explanation:

If

Resistance = 1.8 kΩ

Current = 50 mA

Capacitor = 120 μF

Voltage = 140 V

to calculate the discharge current

Applying the formula of discharge current

io=vo/R

io= 140/ 1.8x 10³

= 0.078A

to calculate the time

Applying the formula of current

io= vo/R e-t/RC

50= 140/1800e-t/RC

0.649= e-t/RC

-t/RC= ln( 0.649)

t = 0.432x 120x10^-6x 1800

t=93.3 x 10^-3seconds

Answer:

In the beginning, I was not familiar to assess assessments of the other students. Ifelt a little bit weird that is it possible to check assignments while having an instructor.I was also a bit frustrated, to be honest, that why do we have to assess thoseassessments. It was kind of extra burden for me. But after few weeks assessingmore assignments, my feeling had changed because I was learning lots of thingsthat were changing my perspectives. I was gaining extra knowledge from my peersin the form of assessments. Yes, I am comfortable with assessing assessments,because I got to learn many vocabularies and making structures of the sentencecorrectly by improving grammatically as I am not a native English speaker. Thus, inthis way, I was learning something new in each and every assessment.

Answer:

Explanation:

Given data:

power rating of the heater P= 1010 W

voltage of the heater V= 115 volts

current taken by the heater I= ?

We can apply the power formula to solve for the current in the heater

i.e P= IV

Making I the current subject of formula we have

I= P/V

Substituting our given data into the expression for I we have

I=1010/115= 8.78 A

Answer:

The AC generator has one unbroken slip ring

Explanation:

In physics, the application of electromagnetic induction can be seen in generators and dynamos. Electromagnetic induction is the process of generating electricity using magnets. It found applications in generators and the types of generator they found application is in AC and DC generator.

An AC generator is also called a Dynamo. A DC generator contains what is called a SPLIT RING fixed to the end of the coil which can be separated and coupled back according to the name "split". An AC generator also called a Dynamo makes use of a SLIP ring which cannot be divided into two. It comes as an entity. The presence of this rings is what differentiates a DC generator from an AC generator.

We can replace split rings with slip rings when converting a DC generator to an AC generator and vice versa.

It can therefore be concluded that the difference between a DC and an AC generator is that the AC generator has one unbroken slip ring.

Answer:

Explanation:

The phase angle of an RLC circuit  ϕ is expressed as shoen below;

ϕ = [tex]tan^{-1} \dfrac{X_l-X_c}{R}[/tex]

Xc is the capacitive reactance = 1/2πfC

Xl is the inductive reactance = 2πfL

R is the resistance = 25.0Ω

Given C = 35.5 μF, L = 0.0940 H, and frequency f = 70.0Hz

Xl = 2π * 70*0.0940

Xl = 41.32Ω

For the capacitive reactance;

Xc = 1/2π * 70*35.5*10⁻⁶

Xc = 1/0.0156058

Xc = 64.08Ω

Phase angle ϕ = [tex]tan^{-1} \frac{41.32-64.08}{25} \\\\[/tex]

ϕ = [tex]tan^{-1} \frac{-22.76}{25} \\\\\\\\[/tex]

[tex]\phi = tan^{-1} -0.9104\\\\\phi = -42.31^0[/tex]

Since tan is negative in the 2nd quadrant;

[tex]\phi = 180-42.31^0\\\\\phi = 137.69^0[/tex]

Hence the phase angle ϕ of the circuit in degrees is 137.69°

The phase angle ϕ of the series RLC circuit that is driven at a frequency of 70.0 Hz is ϕ = 137.69°

Given that:

capacitance C = 35.5 μF,

Inductance L = 0.0940 H,

The resistance R = 25.0Ω

and frequency f = 70.0Hz

The capacitive reactance is given by:

Xc = 1/2πfC

Xc = 1/2π × 70 × 35.5× 10⁻⁶

Xc = 1/0.0156058

Xc = 64.08Ω

The inductive reactance is given by:

Xl = 2πfL

Xl = 2π × 70 × 0.0940

Xl = 41.32Ω

The phase angle of an RLC circuit ϕ  is given by:

[tex]\phi=tan^{-1}\frac{X_l-X_c}{R}\\\\\phi=tan^{-1}\frac{41.32-64.08}{25}[/tex]

Ф = -42.31°

Since tan is negative in the 2nd quadrant, thus:

ϕ = 180° - 42.31°

ϕ = 137.69°

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Answer:

Explanation:

Equation of refraction of a lens is expression according to the formula given below;

[tex]\dfrac{n_2}{v} = \dfrac{n_1}{u}= \dfrac{n_2-n_1}{R}[/tex]

R is the radius of curvature of the convex refracting surface = 12cm

v is the image distance from the refracting surface

u  is the object distance from the refracting surface

n₁ and n₂ are the refractive indices of air and the medium respectively

Given parameters

R = 12 cm

u = [tex]\infty[/tex] (since light incident is parallel to the axis)

n₁  = 1

n₂  = 1.5

Required

focus point of the light that is incident and parallel to the central axis (v)

Substituting this values into the given formula we will have;

[tex]\dfrac{1.5}{v} - \dfrac{1}{\infty}= \dfrac{1.5-1}{12}\\\\\dfrac{1.5}{v} -0= \dfrac{0.5}{12}\\\\\dfrac{1.5}{v}= \dfrac{0.5}{12}\\\\[/tex]

Cross multiply

[tex]1.5*12 = 0.5*v\\ \\18 = 0.5v\\\\v = \frac{18}{0.5}\\ \\v = 36cm[/tex]

Hence  Light incident parallel to the central axis is focused at a point 36cm from the surface

Answer:

Ok, the question is incomplete buy ill try to answer this in a general way.

Suppose that you have no-polarized light.

When that light hits one polaroid, the light becomes polarized along some line, and has an intensity I0.

Now, when polarized light hits a polaroid which axis is at an angle θ with respect to the polarization of the light, the intensity of the resulting beam is given by the Malus's law:

I(θ) = I0*cos^2(θ)

For example, if the axis of the polaroid is exactly the same as the one of the polarized light, then we have θ = 0°

and:

I(0°) = I0*cos^2(0°) = I0

So the intensity does not change.

Now, knowing the initial intensity, you can find the angle needed to get a given intensity.

For example, if the question was:

"At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to A"

We should solve:

I(θ) = A = I0*cos^2(θ)

(A/i0) = cos^2(θ)

√(A/I0) = cos(θ)

Acos(√(A/I0)) = θ

Answer:

The complete question is

NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, low-mass sail and the energy and momentum of sunlight for propulsion. (a) Should the sail be absorbing or reflective? Why? (b) The total power output of the sun is 3.9 x 10^26  W. How large a sail is necessary to propel a 10,000-kg spacecraft against the gravitational force of the sun? Express your result in square kilometers. (c) Explain why your answer to part (b) is independent of the distance from the sun.

a) The sail should be reflective because, an incident electromagnetic wave, in this case, light wave, impacts twice the energy density on a reflective sail, and hence twice the force on a totally reflective sail as would be impacted on a sail that is totally absorbing.

For totally reflective, F = (2I/c)A    ....1

for totally reflective, F = (I/c)A       ....2

where I is the intensity of the light

c is the speed of light = 3 x 10^8 m/s

A is the area the sail

b) The intensity of the light from the sun = power/area

==> I = [tex]\frac{3.9*10^{26}}{4\pi r^{2} }[/tex]

where r is the distance from the sun and the sail

The Force from the sail from equation 1  is therefore

[tex]F[/tex] = [tex]\frac{2*3.9*10^{26}*A}{4\pi r^{2} *3*10^{8}}[/tex] = [tex]2.069*10^{17}\frac{A}{r^{2}}[/tex]

gravitational force between the sail and the sun [tex]F_{g}[/tex] = [tex]\frac{GMm}{r^{2}}[/tex]

where

G is the gravitational constant = 6.67 x 10^−11 m^3⋅s−2⋅kg−1.

m is the mass of the sail = 10000 kg

M is the mass of the sun = 1.99 x 10^30 kg.

==> [tex]F_{g}[/tex] = [tex]\frac{6.67*10^{-11}*1.99*10^{30}*10000}{r^{2}}[/tex] = [tex]\frac{1.33*10^{24}}{r^{2}}[/tex]

Equating the forces, we have

[tex]2.069*10^{17}\frac{A}{r^{2}}[/tex]  =  [tex]\frac{1.33*10^{24}}{r^{2}}[/tex]

the distance cancels out

A = (1.33 x 10^24)/(2.069 x 10^17) = 6428226.196 m^2

==> 6428.2 km^2

c) The force of the solar radiation is proportional to the intensity of the sun from the light, and the intensity is inversely proportional to the square of the distance from the source. Also, the force of gravitation  is inversely proportional to the square of the distance, so they both cancel out.

Answer:

2)The Cell with OER 7

Explanation:

OER is the acronym for Oxygen Enhancement Ratio. It is the measure of the  enhancement of the effect of ionizing radiation due to the presence of oxygen. The ionization effect can be detrimental or therapeutic (use in cancer treatment). OER is the ratio of radiation dose during the lack of oxygen (hypoxia), to the dosage in the presence of oxygen (air is used as a reference). From the definition, one can see that the higher the OER the higher the sensitivity of the cell.

Answer:

t = T / 2 all energy is stored in the inductor

Explanation:

The circuit described is an oscillating circuit where the charge of the condensation stops the inductor and vice versa, in this system the angular velocity of the oscillation is

          w = √1/LC

          2π / T =√1 / LC

          T = 2π  √LC

The energy is constant and for the initial instant it is completely stored in the capacitor

         Uc = Q₀² / 2C

In the process, the capacitor is discharging and the energy is stored in the inductor until when the charge in the capacitors zero, all the energy is stored in the inductor

        U = L I² / 2

in the intermediate instant the energy is stored in the two elements.

Since the period of the system is T for time t = 0 all energy is stored in the capacitor and for t = T / 2 all energy is stored in the inductor

After t = 0 the maximum energy stored in the magnetic field of the inductor is equal to [tex]U'=\dfrac{L I^{2}}{2}[/tex] for the time period, half of period of oscillation  (t = T/2).

The given problem is based on the charging and discharging concepts of capacitor. An oscillating circuit is a circuit where the charge of the capacitor stops the inductor and vice versa, in this system the angular frequency of the oscillation is given as,

[tex]\omega =\dfrac{1}{\sqrt{LC}}\\\\\\\dfrac{2 \pi}{T} =\dfrac{1}{\sqrt{LC}}\\\\\\T = 2\pi \times \sqrt{LC}[/tex]

here, T is the period of oscillation.

Also, the energy stored in the capacitor is constant and for the initial instant it is completely stored in the capacitor. So, the energy stored is given as,

[tex]U =\dfrac{Q^{2}}{2C}[/tex]

here, C is the capacitance.

In the process, the capacitor is discharging and the energy is stored in the inductor until when the charge in the capacitors zero, all the energy is stored in the inductor. So, the expression for the energy stored in the inductor is,

[tex]U'=\dfrac{L I^{2}}{2}[/tex]

here, L is the inductance and I is the current.

Note :- The period of the system is T for time t = 0 all energy is stored in the capacitor and for t = T / 2 all energy is stored in the inductor.

Thus, we conclude that after t = 0 the maximum energy stored in the magnetic field of the inductor is equal to [tex]U'=\dfrac{L I^{2}}{2}[/tex] for the time period, half of period of oscillation  (t = T/2).

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Answer:

1. the sphere of the radius a

Explanation:

Because the charge distribution for each case is spherically symmetric, we can choose a spher- ical Gaussian surface of radius r , concentric with the sphere in question.

So E = k (Q /r 2) (for r ≥ R ) , where R is the radius of the sphere being considered, either a or b .

With the choice of potential at r = ∞ being zero, the electric potential at any distance r from the center of the sphere can be expressed as V = - integraldisplay r ∞ E dr = k /Q r

(for r ≥ R ) .

On the spheres of radii a and b , we have V a = k (Q/ a)and V b = k (Q/ b), respectively.

So Since b > a , the sphere of radius a will have the higher potential.

Also recall Because E = 0 inside a conductor, the potential

Answer:

Distance = 13 units

Explanation:

The overall path covered by an object during its journey is called distance covered.

In this problem, a person starts at position zero, walks to position 8, then walks to position 5.

We need to find the person's distance traveled. It can be calculated simply by adding all the positions i.e.

Distance = 0+8+5

Distance = 13

Hence, the distance covered by the person is 13 units.

Answer:

No.

Explanation:

Given the following :

Velocity (V) of ball = 5m/s

Radius = 1m

Can the ball reach the highest point of the circular track

of radius 1.0 m?

The highest point in the track could be considered as the diameter of the circle :

Radius = diameter / 2;

Diameter = (2 * Radius) = (2*1) = 2

Maximum height which the ball can reach :

Using the relation :

Kinetic Energy = Potential Energy

0.5mv^2 = mgh

0.5v^2 = gh

0.5(5^2) = 9.8h

0.5 * 25 = 9.8h

12.5 = 9.8h

h = 12.5 / 9.8

h = 1.2755

h = 1.26m

Therefore maximum height which can be reached is 1.26m.

Since h < Diameter

Answer:middle

Explanation:

Because it will make the seasaw balanced