Q2/Deceleration of a particle is based on relation a=-3 v² m/s² where v in m/s. If it moves along a straight line and has velocity 10 m/s and position s = 8m when t=0, determine its velocity and position when t= 3 s. Where the velocity become zero. Discuss briefly.​

Answers

Answer 1

Explanation:

Given: a = -3v^2

By definition, the acceleration is the time derivative of velocity v:

[tex]a = \frac{dv}{dt} = - 3 {v}^{2} [/tex]

Re-arranging the expression above, we get

[tex] \frac{dv}{ {v}^{2} } = - 3dt[/tex]

Integrating this expression, we get

[tex] \int \frac{dv}{ {v}^{2} } = \int {v}^{ - 2}dv = - 3\int dt[/tex]

[tex] - \frac{1}{v} = - 3t + k[/tex]

Since v = 10 when t = 0, that gives us k = -1/10. The expression for v can then be written as

[tex] - \frac{1}{v} = - 3t - \frac{1}{10} = - ( \frac{30 + 1}{10} )[/tex]

or

[tex]v = \frac{10}{30t +1 } [/tex]

We also know that

[tex]v = \frac{ds}{dt} [/tex]

or

[tex]ds = vdt = \frac{10 \: dt}{30t + 1} [/tex]

We can integrate this to get s:

[tex]s = \int v \: dt = \int ( \frac{10}{30t + 1}) \: dt = 10 \int \frac{dt}{30t + 1} [/tex]

Let u = 30t +1

du = 30dt

so

[tex] \int \frac{dt}{30t + 1} = \frac{1}{30} \int \frac{du}{u} = \frac{1}{30}\ln |u| + k[/tex]

[tex]= \frac{1}{30}\ln |30t + 1| + k[/tex]

So we can now write s as

[tex]s = \frac{1}{3}\ln |30t + 1| + k[/tex]

We know that when t = 0, s = 8 m, therefore k = 8 m.

[tex]s = \frac{1}{3}\ln |30t + 1| + 8[/tex]

Next, we need to find the position and velocity at t = 3 s. At t = 3 s,

[tex]v = \frac{10}{30(3) +1 } = \frac{10}{91}\frac{m}{s} = 0.11 \: \frac{m}{s} [/tex]

[tex]s = \frac{1}{3}\ln |30(3) + 1| + 8 = 9.5 \: m[/tex]

Note: velocity approaches zero as t --> [tex]\infty [/tex]


Related Questions

As shown in Fig. 4, an ideal gas of monatomic molecules expands from its initial state A to a state B through an isobaric process and then further expands to a volume C. Find the work done by the gas, increase in internal energy, and the energy transferred by heat to the gas over the whole process​

Answers

The picture is very blurry

Why do Hubble Space Telescope images show what objects looked like long ago?

Answers

Answer:

when we get images from the telescope, we're not seeing those object like they are right now. we're seeing them like they looked when that light left those distant objects...... the animation cuts to a visualization of distant galaxies imaged by Hubble.

why are cows is important?​

Answers

Answer:

cause they give u milk

Explanation:

Answer:

Cows are important as they provide humans many things for survival. They provide milk, meat, and leather, all of these are important resources.

A light year is a unit that measures time

Answers

Answer:

Light-year is the distance light travels in one year. Light zips through interstellar space at 186,000 miles (300,000 kilometers) per second and 5.88 trillion miles (9.46 trillion kilometers) per year.

Explanation:

A car moving in a straight line uniformly accelerated speed increased from 3 m / s to 9 m / s in 6 seconds. With what acceleration did the car move?


a.
2 m/s2


b.
1 m/s2


c.
0 m/s2


d.
3 m/s2

Answers

I think it’s 1 . The answer is B

Answer:

b) 1 m/s

I am sure...........

A man weighing 720 N and a woman weighing 500 N have the same momentum. What is the ratio of the man's kinetic energy Km to that of the woman Kw?

Answers

Answer:

Because weight W = M g, the ratio of weights equals the ratio of masses.

(M_m g)/ (M_w g) = [ (p^2 Man )/ (2 K_man)] / [ (p^2 Woman )/ (2 K_woman)

but p's are equal, so

K_m/K_m = (M_w g)/(M_m g) = W_woman / W_man = 450/680 = 0.662Explanation:

What is 3*10^-6 divided by 2.5*10^6 expressed in standard notation?​

Answers

Answer:

1.2 x 10^-12

Explanation:

3/2.5 x 10^-6/10^6

1.2 x 10^-6 x 10^-6

1.2 x 10^-12

The latent heat of vaporization of water is roughly 10 times the latent heat of fusion of water. The amount of heat required to boil away 1 kg of water is __________ the amount of heat required to melt 1 kg of ice.

Answers

Answer:

The amount of heat required to boil away 1 kg of water is 10 times the amount of heat required to melt 1 kg of ice

Explanation:

let the latent heat of fusion of ice  = L

then, the latent heat of vaporization of water = 10L

The heat of fusion of 1 kg of ice = 1 x L = L

The heat of vaporization 1 kg of water = 1 x 10L = 10L

Therefore, the amount of heat required to boil away 1 kg of water is 10 times the amount of heat required to melt 1 kg of ice

Calculate the equivalent resistance

Answers

Not enough information

In young Goodman’s Brown hawthornes reveals his feelings about his Puritan ancestors when

Answers

Answer:

In "Young Goodman Brown," Hawthorne reveals his feelings about his Puritan ancestors when the dark man reveals that he helped Brown's forebears persecute others.

Explanation:

hey mate i know it is right so don't worry this is the correct

Answer:

In "Young Goodman Brown," Hawthorne reveals his feelings about his Puritan ancestors when the dark man reveals that he helped Brown's forebears persecute others. so that means the answer is D.

Explanation:

A. brown strives to resist his dark mission

B. faith expresses her anxieties about young brown's departure

C. brown discovers his catechism teacher is on speaking terms with the devil

D. the dark man reveals that he helped brown's forebears persecute others(correct answer)

The efficiency of a machine can be increased by

Answers

Explanation:

the efficiency of a machine can be increased by reducing the friction

please mark the brainliest

Reducing the friction

A scenario where reaction time is important is when driving on the highway. During the delay between seeing an obstacle and reacting to avoid it (or to slam on the brakes!) you are still moving at full highway speed. Calculate how much distance you cover in meters before you start to put your foot on the brakes if you are travelling 65 miles per hour.

Answers

Answer:

66.83 meters

Explanation:

After a quick online search, it seems that scientists calculate the average reaction time of individuals as 2.3 seconds between seeing an obstacle and putting their foot on the brakes. Now that we have this reaction time we need to turn the miles/hour into meters/second.

1 mile = 1609.34 meters  (multiply these meters by 65)

65 miles = 104,607 meters

1 hour = 3600 seconds

Therefore the car was going 104,607 meters every 3600 seconds. Let's divide these to find the meters per second.

[tex]\frac{104,607}{3600} = \frac{29.0575 meters}{1 second}[/tex]

Now we simply multiply these meters by 2.3 seconds to find out the distance covered before the driver puts his/her foot on the brakes...

29.0575m * 2.3s = 66.83 meters

eididjsmsisijsjsiakaksannnahshsjeejekekekkeie​

Answers

Answer:

ok

Explanation:

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According to the model, when was the universe at its most dense?

A) During the Dark Ages where matter increased in mass.

B) Just before the Big Bang where all matter existed in a singularity.

C) During the nuclear fusion events, as the atoms become more massive.

D) Current day, as the number of galaxies, solar systems, and planets have increased.

Answers

it d cus you look at the gragh

Answer:

The Answer is D

Explanation:

Hope this helps!!!!

If 1.02 ✕ 1020 electrons move through a pocket calculator during a full day's operation, how many coulombs of charge moved through it?

Answers

Answer:

Explanation:

one electron has [tex]1.60217662*10^{-19}~coulombs~then\\\\1.02*10^{20}~electrons------->1.02*10^{20}*1.60217662*10^{-19}~coulombs= 16.3422~coulombs[/tex]

Calculate the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m

Answers

Answer: The period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.

Explanation:

Given: Mass = 5 kg

Spring constant = 6 N/m

Formula used to calculate period is as follows.

[tex]T = 2 \pi \sqrt\frac{m}{k}[/tex]

where,

T = period

m = mass

k = spring constant

Substitute the values into above formula as follows.

[tex]T = 2 \pi \sqrt\frac{m}{k}\\= 2 \times 3.14 \times \sqrt\frac{5}{6}\\= 5.73 s[/tex]

Thus, we can conclude that the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.

A ship is flying away from Earth at 0.9c (where c is the speed of light). A missile is fired that moves toward the Earth at a speed of 0.5c relative to the ship. How fast does the missile move relative to the Earth

Answers

Answer:

the required speed with which the missile move relative to the Earth is -0.727c

Explanation:

Given the data in the question;

relative velocity relation;

u' = u-v / 1 - [tex]\frac{uv}{c^2}[/tex]

so let V[tex]_B[/tex] represent the velocity as seen by an external reference frame; u=V[tex]_B[/tex]

and let V[tex]_A[/tex] represent the speed of the secondary reference frame; v=V[tex]_A[/tex]

hence, u' is the speed of B as seen by A

so

u' = V[tex]_B[/tex]-V[tex]_A[/tex] / 1 - [tex]\frac{V_BV_A}{c^2}[/tex]

now, given that; V[tex]_A[/tex] = 0.9c  and V[tex]_B[/tex]  = 0.5c

we substitute

u' = ( 0.5c - 0.9c ) / 1 - [tex]\frac{(0.5c)(0.9c)}{c^2}[/tex]

u' = ( 0.5c - 0.9c ) / 1 - [tex]\frac{c^2(0.5)(0.9)}{c^2}[/tex]

u' = ( 0.5c - 0.9c ) / 1 - (0.5 × 0.9)

u' = ( -0.4c ) / 1 - 0.45

u' = -0.4c / 0.55

u' = -0.727c

Therefore, the required speed with which the missile move relative to the Earth is -0.727c

If the pressure of a gas is really due to the random collisions of molecules with the walls of the container, why do pressure gauges – even very sensitive ones – give perfectly steady readings? Shouldn't the gauge be continually jiggling and fluctuating? Explain.

Answers

Answer:

there is no fluctuation in the measurement because the quantity of molecule is too large and a quantity of some molecules is imperceptible.

Explanation:

The pressure measurement is carried out by calibrating the force exerted by the air on a surface of known area, suppose a small area 1 mm² = 0.01 cm²

To find out if the random movement of air molecules affects the pressure reading, let's calculate the number of molecules that reaches the pressure gauge.

In a system at atmospheric pressure and in a volume of 1 m³ (walls of 1 m each) there is one mole of air molecules, this mole is evenly distributed, so how many molecules fall on our surface

           # _molecule = 6.02 10²³ 0.01 10⁻⁴ / 1

           #_molecular = 6.02 10¹⁷ molecules per second

therefore the variation of the number of molecules is not very important

Consequently there is no fluctuation in the measurement because the quantity of molecule is too large and a quantity of some molecules is imperceptible.

A strong trough in a Rossby wave occurs when the jet stream A. bends towards the Equator. B. bends toward the poles. C. does not bend but maintains an east to west flow. D. does not bend but maintains a west to east flow.

Answers

Answer:

A. bends towards the Equator.

Explanation:

Rossby waves are inertial waves that are naturally occurring in a rotating fluids. These waves are also called as the planetary waves.

The Rossby waves are undulated that occur in the polar front jet stream when there is a significant differences in the temperatures between the polar and the tropical air masses.

It occurs when the polar air masses moves towards the equator and when the tropical air masses moves towards the pole.  It is formed when the air bends away from the poles and bends towards the equator.

Hence the correct option is (A).

15. A car travelling towards the right has a mass of 1332 kg and has a speed of 25 m/s. A truck is
travelling towards the left with a mass of 3000 kg and a speed of 15 m/s. They collide head
on with each other. What is the total momentum after the crash? In which direction will the
vehicles travel after the collision?

Answers

Explanation:

Given that,

The mass of a car, m₁ = 1332 kg

The speed of the car, u₁ = 25 m/s (right)

The mass of a truck, m₂ = 3000 kg

The speed of the truck, u₂ = -15 m/s

The total momentum after the crash is given by :

p=m₁u₁ + m₂u₂

Put all the values,

P = 1332(25) + 3000(-15)

= −11700 kg-m/s

So, the total momentum after the crash is equal to 11700 kg-m/s and it is in the left direction.

A digital signal differs from an analog signal because it a.consists of a current that changes smoothly. b. consists of a current that changes in pulses. c.carries information. d. is used in electronic devices.​

Answers

Answer:

d.it is used in electronic devices

why the walls of tyres becomes warm as the car moves​

Answers

Answer:

the particles vibrate inside the tyre

Explanation:

as the car moves kinetic energy is transfered in the tyres which causes the particles to vibrate inside the tyre so the kinetic store is. transferred into thermal

A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Teflon frying pan is only 0.150 N. Knowing the coefficient of kinetic friction between the two materials (0.04), he quickly calculates the normal force. What is it (in N)

Answers

Answer:

[tex]f_n=3.75N[/tex]

Explanation:

From the question we are told that:

Frictional force [tex]F=0.150N[/tex]

Coefficient of kinetic friction [tex]\mu=0.04[/tex]

Generally the equation for Normal for is mathematically given by

 [tex]f_n=\frac{F}{\mu}[/tex]

Therefore

 [tex]f_n=\frac{0.150}{0.04}[/tex]

 [tex]f_n=3.75N[/tex]

When hydrogen peroxide antiseptic is tightly capped and left on the shelf for a long period of time, the bottle is observed to swell. When opened, it “hisses” noticeably. Explain these observations. Is the antiseptic likely to be as effective?

Answers

Answer:

The hydrogen peroxide decomposes as: [tex]H_2O_2\rightarrow 2H_2O+O_2[/tex]

Explanation:

Hydrogen peroxide antiseptic although tightly capped but left on the shelf for a long time leads to swollen bottle and opens with a hissing sound because hydrogen peroxide reacts with air in presence of light to dissociate into water and oxygen molecules that leads to high volume of content in the tightly closed bottle and formation of water molecules from hydrogen peroxide.

Once the hydrogen peroxide has turned into water molecules it is no longer effective as an antiseptic.

A car moves at a constant speed of 90km/h from a starting point. Another car moves at 70km/h after 2hours from the same starting point. if both cars moves in the same direction, after how many hours will the distance between the first car and the second car to be 40 km.​

Answers

Answer:

400

Explanation:

N 4. Which of the following can cause a short circuit?

Answers

Answer:

pretty sure its A

Explanation:

please give brainliest if i'm correct

Answer:

A.

Explanation:

I have done this and that was correct

hope this helps

the unit of area is called a derived unit.why?​

Answers

Explanation:

the unit of area is called a derived unit because it is made of two fundamental unit metre and metre.

A 2 kg object traveling at 5 m s on a frictionless horizontal surface collides head-on with and sticks to a 3 kg object initially at rest. Which of the following correctly identifies the change in total kinetic energy and the resulting speed of the objects after the collision? Kinetic Energy Speed
(A) Increases 2 m/s 3.2 m/s
(B) Increases Soold 2 m/s
(C) Decreases 3.2 m/s
(D) Decreases

Answers

Answer: (d)

Explanation:

Given

Mass of object [tex]m=2\ kg[/tex]

Speed of object [tex]u=5\ m/s[/tex]

Mass of object at rest [tex]M=3\ kg[/tex]

Suppose after collision, speed is v

conserving momentum

[tex]\Rightarrow mu+0=(m+M)v\\\\\Rightarrow v=\dfrac{2\times 5}{2+3}\\\\\Rightarrow v=2\ m/s[/tex]

Initial kinetic energy

[tex]k_1=\dfrac{1}{2}\times 2\times 5^2\\\\k_1=25\ J[/tex]

Final kinetic energy

[tex]k_2=\dfrac{1}{2}\times (2+3)\times 2^2\\\\k_2=10\ J[/tex]

So, it is clear there is decrease in kinetic energy . Thus, energy decreases and velocity becomes 2 m/s.

The tendon from the flexor hallucis longus muscle in the hindlimb of a turkey (Meleagris gallopavo) has a resilience of 0.93. If 9.00 J of work are done on the tendon to stretch it out, how many Joules of work does the tendon do as it is relaxing?

Answers

Answer:

8.37 Joules

Explanation:

The amount of energy that a substance, such as animal tissue or rubber can take and yet return to its original condition is known as resilience. When we stress such a substance and then let it restore to its previous state, we are referring to the substance to maintain its elastic area of the stress-strain curve.

From the given information:

the resilence is 0.93

The amount of work done during stretching of the tendon = 9.00 J

Thus,

the work done when relaxing = 0.93 × 9.00 J

= 8.37 Joules

Two identical satellites orbit the earth in stable orbits. Onesatellite orbits with a speed vat a distance rfrom the center of the earth. The second satellite travels at aspeed that is less than v.At what distance from the center of the earth does the secondsatellite orbit?At a distance that is less than r.At a distance equal to r.At a distance greater than r.Now assume that a satellite of mass m is orbiting the earth at a distance r from the center of the earth with speed v_e. An identical satellite is orbiting the moon at thesame distance with a speed v_m. How does the time T_m it takes the satellite circling the moon to make onerevolution compare to the time T_e it takes the satellite orbiting the earth to make onerevolution?T_m is less than T_e.T_m is equal to T_e.T_m is greater than T_e.

Answers

Answer:

a. At a distance greater than r

b. T_m is greater than T_e.

Explanation:

a. Two identical satellites orbit the earth in stable orbits. One satellite orbits with a speed vat a distance r from the center of the earth. The second satellite travels at a speed that is less than v. At what distance from the center of the earth does the second satellite orbit?

Since the centripetal force on any satellite, F equals the gravitational force F' at r,

and F = mv²/r and F' = GMm/r² where m = mass of satellite, v = speed of satellite, G = universal gravitational constant, M = mass of earth and r = distance of satellite from center of earth.

Now, F = F'

mv²/r = GMm/r²

v² = GM/r

v = √GM/r

Since G and M are constant,

v ∝ 1/√r

So, if the speed decreases, the radius of the orbit increases.

Since the second satellite travels at a speed less than v, its radius, r increases since v ∝ 1/√r.

So, the distance the second satellite orbits is at a distance greater than r

b. An identical satellite is orbiting the moon at the same distance with a speed v_m. How does the time T_m it takes the satellite circling the moon to make one revolution compare to the time T_e it takes the satellite orbiting the earth to make one revolution?

Since the speed of the satellite, v = √GM/r where M = mass of planet

Since the satellite is orbiting at the same distance, r is constant

So, v ∝ √M

Since mass of earth M' is greater than mass of moon, M", the speed of satellite circling moon, v_m is less than v the speed of satellite circling earth at the same distance, r

Now, period T = 2πr/v where r = radius of orbit and v = speed of satellite

Since r is constant for both orbits, T ∝ 1/v

Now, since the speed of the speed of the satellite on earth orbit v  is greater than the speed of the satellite orbiting the moon, v_m, and T ∝ 1/v, it implies that the period of the satellite orbiting the earth, T_e is less than the period of the satellite orbiting the moon, T_m since there is an inverse relationship between T and v. T_e is less T_m implies T_m is greater than T_e

So, T_m is greater than T_e.

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