4. According to Newton’s law of cooling, if an object at temperature T is immersed in a medium having the constant temperature M, then the rate of change of T is proportional to the difference of temperature M − T. This gives the differential equation dT dt = k(M − T). (a) Solve the differential equation for T. (b) A thermometer reading 100◦F is placed in a medium having a constant temperature of 70◦F. After 6 min, the thermometer reads 80◦F. What is the reading after 20 min?

Answers

Answer 1

Answer:

a) The solution of the differential equation is [tex]T(t) = M + (T_{o}-M) \cdot e^{-\frac{t}{\tau} }[/tex].

b) The reading after 20 minutes is approximately 70.770 ºF.

Explanation:

a) Newton's law of cooling is represented by the following ordinary differential equation:

[tex]\frac{dT}{dt} = -\frac{T-M}{\tau}[/tex] (Eq. 1)

Where:

[tex]\frac{dT}{dt}[/tex] - Rate of change of temperature of the object in time, measured in Fahrenheit per minute.

[tex]\tau[/tex] - Time constant, measured in minutes.

[tex]T[/tex] - Temperature of the object, measured in Fahrenheit.

[tex]M[/tex] - Medium temperature, measured in Fahrenheit.

Now we proceed to solve the differential equation:

[tex]\frac{dT}{T-M} = -\frac{t}{\tau}[/tex]

[tex]\int {\frac{dT}{T-M} } = -\frac{1}{\tau} \int \, dt[/tex]

[tex]\ln (T-M) = -\frac{t}{\tau} + C[/tex]

[tex]T(t) -M = (T_{o}-M)\cdot e^{-\frac{t}{\tau} }[/tex]

[tex]T(t) = M + (T_{o}-M) \cdot e^{-\frac{t}{\tau} }[/tex] (Eq. 2)

Where:

[tex]t[/tex] -Time, measured in minutes.

[tex]T_{o}[/tex] - Initial temperature of the object, measured in Fahrenheit.

b) From (Eq. 2) we obtain the time constant of the cooling equation for the object: ([tex]M = 70\,^{\circ}F[/tex], [tex]T_{o} = 100\,^{\circ}F[/tex], [tex]t = 6\,min[/tex], [tex]T(t) = 80\,^{\circ}F[/tex])

[tex]80\,^{\circ}F = 70\,^{\circ}F + (100\,^{\circ}F-70\,^{\circ}F)\cdot e^{-\frac{6\,min}{\tau} }[/tex]

[tex]e^{-\frac{6\,min}{\tau} } = \frac{80\,^{\circ}F-70\,^{\circ}F}{100\,^{\circ}F-70\,^{\circ}F}[/tex]

[tex]e^{-\frac{6\,min}{\tau} } = \frac{1}{3}[/tex]

[tex]-\frac{6\,min}{\tau} = \ln \frac{1}{3}[/tex]

[tex]\tau = -\frac{6\,min}{\ln \frac{1}{3} }[/tex]

[tex]\tau = 5.461\,min[/tex]

The cooling equation of the object is [tex]T(t) = 70 +30\cdot e^{-\frac{t}{5.461} }[/tex] and the temperature of the object after 20 minutes is:

[tex]T(20) = 70+30\cdot e^{-\frac{20}{5.461} }[/tex]

[tex]T(20) \approx 70.770\,^{\circ}F[/tex]

The reading after 20 minutes is approximately 70.770 ºF.


Related Questions

The average speed of an object, S SS, is calculated using the formula S = D T S= T D ​ S, equals, start fraction, D, divided by, T, end fraction, where T TT is the time it takes to travel a distance of D DD units. Rearrange the formula to solve for time ( T ) (T)

Answers

Answer:

T = D/S

Explanation:

Given the formula for calculating the average speed of an object as:

average speed = Distance/Time

Let average speed = S

Distance = D

Time = T

Substitute

S = D/T

Make T the subject of the formula:

From S = D/T

Cross multiply

ST = D

Divide both sides by S

ST/S = D/S

T = D/S

Hence the expression for he time T after rearrangement is T = D/S

What is the maximum speed with which a 1200-kg car can round a turn of radius 92.0 m on a flat road if the coefficient of static friction between tires and road is 0.65

Answers

Answer:

24.22m/s

Explanation:

Given data

mass m= 1200kg

radius r=92m

coefficient of friction μ=0.65

We know that

F=mv2/r

F=μmg  

mv2/r = μsmg  

v^2/r = μsg  

vmax = √(rμg)

Substituting our data we have

vmax=√(92*0.65*9.81)

vmax=√586.638

vmax=24.22m/s

Homeboy Joe is riding his skateboard and playing Among Us. Because he's distracted he doesn't notice he's about to skate right off the 69.0 m cliff he's on. If he lands 22.0 m from the cliff, how fast was he skating?

Answers

Answer: He wasn’t skating he was on Ice ;)

A 7.0 µF capacitor and a 5.0 µF capacitor are connected in series across a 5.0 kV potential difference. The charged capacitors are then disconnected from the source and connected to each other with terminals of like sign together. Find the charge on each capacitor (in mC) and the voltage across each capacitor (in V). (Due to the nature of this problem, do not use rounded intermediate values in your calculations—including answers submitted in WebAssign.)

Answers

Explanation:

Given two capacitors 7.0 µF and 5.0 µF connected in series, let s calculate the effective capacitance.

[tex]\frac{1}{C_T} = \frac{1}{7.0} +\frac{1}{5.0}\\ \frac{1}{C_T} = \frac{5+7}{35} \\C_T = \frac{35}{12}\\ C_T = 2.9166\mu F[/tex]

Given

V = 5.0kV = 5000Volts

To get the total charge in the circuit, we will use the formula:

[tex]Q = C_TV\\Q = 2.9166\times 10^{-6} \times 5000\\Q = 0.014583 C\\Q = 1.4583 \times 10^{-2}C[/tex]

Since the same charge flows in a series connected capacitors, the charge on them in mC is 14.583mC

For the voltage across 7.0 µF:

[tex]V = \frac{Q}{C}\\V = \frac{0.014583}{0.000007} \\V = 2,083.28V[/tex]

Voltage across the 7.0 µF is 2,083.28V

The voltage across the  5.0 µF:

V = 5000 - 2,083.28

V = 2,916.72Volts

Voltage across the  5.0 µF is 2,916.72V

Answer:

The charge on the first capacitor is 14.59 mC

The voltage across the first capacitor is 2084 V

The charge on the second capacitor is 14.59 mC

The voltage across the second capacitor is 2918 V

Explanation:

Given;

first capacitor, C₁ = 7.0 µF

second capacitor, C₂ = 5.0 µF

potential difference, V = 5000 V

The equivalent capacitance is given by;

[tex]\frac{1}{C_{eq}}= \frac{1}{C_1} +\frac{1}{C_2}\\\\\frac{1}{C_{eq}}= \frac{C_1 + C_2}{C_1C_2}\\\\C_{eq} = \frac{C_1C_2}{C_1 +C_2}\\\\ C_{eq} = \frac{(7*10^{-6})(5*10^{-6})}{(7*10^{-6} \ +\ 5*10^{-6})}\\\\C_{eq} = 2.917*10^{-6} \ F[/tex]

The charge on the first capacitor is given by;

[tex]Q_1 = C_{eq} V\\\\Q_1 = 2.917*10^{-6} *5000\\\\Q_1 = 0.01459 \ C\\\\Q_1 = 14.59 \ mC[/tex]

The voltage across the first capacitor is given by;

[tex]V_1 = \frac{Q_1}{C_1} \\\\V_1 = \frac{0.01459}{7*10^{-6}}\\\\ V_1 = 2084.286 \ V[/tex]

V₁ = 2084 V

The charge on the second capacitor is given by;

[tex]Q_2 = C_{eq} V\\\\Q_2 = 2.917*10^{-6} *5000\\\\Q_2 = 0.01459 \ C\\\\Q_2 = 14.59 \ mC[/tex]

The voltage across the second capacitor is given by;

[tex]V_2 = \frac{Q_2}{C_2}\\\\ V_2 = \frac{0.01459}{5*10^{-6}}\\\\V_2 = 2918 \ V[/tex]

is this correct?? or wrong?​

Answers

No isnt it decreasing to the right in a negative acceleration

Regardless of what state a substance is in, it is always that substance

True
False

Answers

Hello,
I think your answer is false.
Hope this helps. Have a great day. :)

1. Consider the point exactly halfway between the two wires. Can you adjust the current so that, with current passing through each of the wires, the net field is zero here? 2. Note the distance between the field lines from one of the wires. Are they equally space? Explain. 3. With currents of different magnitudes passing through the two wires, how do the forces on the two wires compare? What is the reason for this?

Answers

Answer:

hello your question is incomplete attached below is missing part of the question

answer:

1 )  Magnetic field due to long current carrying wire : [tex]B = \frac{U_{0} I}{2\pi d}[/tex]

Therefore the net magnetic field due the both wires ; B = B[tex]_{1}[/tex] + B[tex]_{2}[/tex] . when we adjust the current I[tex]_{1}[/tex] = I[tex]_{2}[/tex] then the Netfield (B ) = zero

2) The distance between the field lines are not equally spaced and this is because the separation between field lines increases with the increase in the distance between the wires

3) Increase in current through the wire will lead to increase in force and this can be explained via this equation

[tex]F = \frac{U_{0}I_{1}I_{2} }{2\pi d }[/tex]

Explanation:

1 )  Magnetic field due to long current carrying wire : [tex]B = \frac{U_{0} I}{2\pi d}[/tex]

Therefore the net magnetic field due the both wires ; B = B[tex]_{1}[/tex] + B[tex]_{2}[/tex] . when we adjust the current I[tex]_{1}[/tex] = I[tex]_{2}[/tex] then the Netfield (B ) = zero

2) The distance between the field lines are not equally spaced and this is because the separation between field lines increases with the increase in the distance between the wires

3) Increase in current through the wire will lead to increase in force and this can be explained via this equation

[tex]F = \frac{U_{0}I_{1}I_{2} }{2\pi d }[/tex]

I NEED HELP PLEASEE ITS AN ECONOMICS QUESTION ABOVE

Answers

Answer:

I believe the answer is Property taxes

Explanation:

Answer: I'm pretty sure property taxes

Explanation:

A Navy Seal of mass 80 kg parachuted directly down into an enemy harbor. At one point while she was falling, the resistive force that air exerted on her was 520 N upward. What can you determine about her motion at this point in time

Answers

Answer:

The Navy Seal is accelerating downwards at the rate of 3.3 m/s²

Explanation:

Given;

mass of the Navy Seal, m = 80 kg

the upward resistive force on her, F = 520 N

Her net downward force is given by;

[tex]F_{net} = F_{down} - F_{up}\\\\F_{net} = (80*9.8) - 520\\\\F_{net} = 264 \ N[/tex]

Her downward acceleration at this time is given by;

F = ma

a = F / m

a = 264 / 80

a = 3.3 m/s²

Therefore, the Navy Seal is accelerating downwards at the rate of 3.3 m/s²

Consider the force field and circle defined below. F(x, y) = x2 i + xy j x2 + y2 = 121 (a) Find the work done by the force field on a particle that moves once around the circle oriented in the clockwise direction.

Answers

Answer: the work done by the force is 0

Explanation:

F (x², xy)

121 = 11²

so R = x² + y² = 11²

p = x². Q = xy

Δp/Δy = 0, ΔQ/Δx

using Green's theorem

woek = c_∫F.Δr = R_∫∫ ΔQ/Δx - Δp/Δy) ΔA

=  (x² + y² = 121)_∫∫ yΔA

now let x = rcosФ, y = rsinФ

ΔA = rΔrΔФ

so r from 0 to 11

and Ф from 0 to 2π

= 0_∫^2π   0_∫^11  rsinФ × rΔrΔФ

= 0_∫^2π SinФΔФ   0_∫^11  r²Δr

= [ -cosФ]^2π_0 [r³/3]₀¹¹ = ( -cos2π + cos0) (11³/3) = 0

therefore the work done by the force is 0

KINEMATICS: MOTION ALONG STRAIGHT LINE
The nearest grocery is 60m, east from your house. You are walking at 1.2 m's for 15.Os toward
the grocery when it started to rain and you ran back to your house to get an umbrella. It took
you another 5.0s to go back to your house. You start walking again at 1.2m's until you reach
the grocery. (a) What is your average speed? (b) What is your average velocity?

Answers

Explanation:

hope this helps, cheers!

The values for the average speed and average velocity are;

(a) [tex]Average \ speed = 1.37\overline{142857} \ m/s[/tex]

(b) [tex]The \ average \ velocity = 0.\overline{857142} \ m/s[/tex]

The reason the above values are correct is given as follows;

The given parameters are;

Distance of the nearest grocery store to the house, d = 60 m

Direction of the grocery store = East from the house

Walking speed, s = 1.2 m/s

Time of walking before the rain started, t₁ = 15.0 s

Time it takes to go back to the house, t₂ = 5.0 s

Speed at which the start walking again, s = 1.2 m/s

(a) The formula for average speed, [tex]s_{avg}[/tex], is given as follows;

[tex]s_{avg} = \dfrac{Total \ distance }{Sum \ of \ time \ taken}[/tex]

The distance walked before the rain, d₁ = 1.2 m/s × 15.0 s = 18 meters

Distance covered on returning to the house = d₂ = d₁ = 18 meters

The total distance covered, ∑Distance = d₁ + d₂ + 60

∴ ∑Distance = 18 m + 18 m + 60 m = 96 m

The time, t₃, it takes to get to the grocery store on the second attempt after getting the umbrella is given as follows;

[tex]t_3 = \dfrac{d}{v}[/tex]

[tex]t_3 = \dfrac{60 \ m}{1.2 \ m/s} = 50.0 \, s[/tex]

The total time, ∑Time = t₁ + t₂ + t₃

∴ ∑Time = 15.0 s + 5.0 s + 50.0 s = 70.0 s

[tex]Average \ speed = \dfrac{\sum Distance}{\sum Time}[/tex]

[tex]Average \ speed = \dfrac{96 \, m}{70 \, s} = \dfrac{48}{35} \ m/s= 1.37\overline{142857} \ m/s[/tex]

[tex]Average \ speed = 1.37\overline{142857} \ m/s[/tex]

(b) The average velocity, [tex]v_{avg}[/tex], is given as follows;

[tex]v_{avg}= \dfrac{Total \ displacement}{Total \ time}[/tex]

The total displacement = The total change in location = The change in location from the house to the grocery =  60 m

The total time is the same as for the average speed, ∑Time = 70 s

[tex]\therefore v_{avg}= \dfrac{60 \, m}{70 \, s} = \dfrac{6}{7} \ m/s = 0.\overline{857142} \ m/s[/tex]

[tex]The \ average \ velocity = 0.\overline{857142} \ m/s[/tex]

Learn more about average speed and average velocity here:

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When a parachute opens, the air exerts a large drag force on it. This upward force is initially greater than the weight of the sky diver and, thus, slows him down. Suppose the weight of the sky diver is 915 N and the drag force has a magnitude of 1061 N. The mass of the sky diver is 93.4 kg. Take upward to be the positive direction. What is his acceleration, including sign

Answers

Explanation:

According to newton's second law of motion.

[tex]\sum Fx = ma\\\\\sum Fx = 1061 - 915\\\\\sum Fx = 146N[/tex]

m is the mas of the sky diver = 93.4kg

a is the acceleration of the skydiver

From the formula above;

[tex]a = \frac{\sum Fx}{m}\\ \\a = \frac{146}{93.4}\\\\a = 1.563m/s^2[/tex]

Hence the acceleration of the sky diver is 1.563m/s²

While making some observations at the top of the 66 m tall Astronomy tower, Ron
accidently knocks a 0.5 kg stone over the edge. How long will a student at the bottom
have to get out of the way before being hit?

Answers

Analysing the question:

Since the stone was dropped, there was no initial velocity applied on it and hence it's initial velocity of the stone is 0 m/s

We are given:

height of the tower (h) = 66 m

mass of the stone (m) = 0.5 kg

initial velocity of the stone (u) = 0 m/s

time taken by the stone to reach the ground (t) = t seconds

acceleration due to gravity = 10 m/s²

** Neglecting air resistance**

Finding the time taken by the stone to reach the ground:

from the second equation of motion

h = ut + 1/2at²

replacing the variables

66 = (0)(t) + 1/2 (10)(t)²

66 = 5t²

t² = 13.2

t = 3.6 seconds

I initially wanted to subtract the height of the student from the height of the tower since the time i calculated is the time taken by the stone to reach the ground and that means that the stone has already hit the student before 3.6 seconds

but since we were NOT given the height of a student, the person who posed this question wants the time taken by the stone to reach the ground and that is what we solved

Average velocity of Mike Phelps swimming 100 m race in the 50 m long pool (2 laps) is approximately equal to *

A. 0 m/s
B. 1 m/s
C. 2 m/s
D. 4 m/s

Answers

Answer:

2

Explanation:

Question 1 of 15

All digits shown on the measuring device, plus one estimated digit, are

considered

Answer here

SUBMIT

Answers

Answer:

significant

Explanation:

The digits in a measurement that are considered significant are all of those digits that represent marked calibrations on the measuring device plus one additional digit to represent the estimated digit (tenths of the smallest calibration).

A microwave oven operates at 2.50 GHzGHz . What is the wavelength of the radiation produced by this appliance? Express the wavelength numerically in nanometers.

Answers

Answer:

The wavelength is [tex]\lambda  =  1.2  * 10^8 nm[/tex]

Explanation:

From the question we are told that

   The frequency of operation of the microwave is  [tex]f =  2.50 GHz  =  2.50 *10^{9} \ Hz[/tex]

     Generally the wavelength is mathematically represented as

          [tex]\lambda  =  \frac{c}{f}[/tex]

Here c is the speed of light with value [tex]c =  3.0 *10^{8} \  m/s[/tex]

So  

         [tex]\lambda  =  \frac{3.0 *10^{8}}{  2.50 *10^{9}}[/tex]

=>       [tex]\lambda  =  0.12 \  m [/tex]

converting to nanometer

           [tex]\lambda  =  1.2  * 10^8 nm[/tex]

The earth has a net electric charge that causes a field at points near its surface equal to 150 N/C and directed in toward the center of the earth. (a) What magnitude and sign of charge would a 60-kg human have to acquire to overcome his or her weight by the force exerted by the earth’s electric field? (b) What would be the force of repulsion between two people each with the charge calculated in part (a) and separated by a distance of 100 m? Is use of the earth’s electric field a feasible means of flight? Why or why not?

Answers

Answer:

a) The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The force of repulsion between two people is [tex]13.851\times 10^{6}[/tex] newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.

Explanation:

a) From Second Newton's Law, we form this equation of equilibrium:

[tex]\Sigma F = F_{E}-W = 0[/tex] (Eq. 1)

Where:

[tex]F_{E}[/tex] - Electrostatic force exerted on human, measured in Newton.

[tex]W[/tex] - Weight of the human, measured in Newton.

If we consider that human can be represented as a particle and make use of definitions of electric field and weight, the previous equation is expanded and electric charge is cleared afterwards:

[tex]q\cdot E-m\cdot g = 0[/tex]

[tex]q = \frac{m\cdot g}{E}[/tex] (Eq. 2)

[tex]E[/tex] - Electric field, measured in Newtons per Coloumb.

[tex]m[/tex] - Mass, measured in kilograms.

[tex]g[/tex] - Gravity acceleration, measured in meters per square second.

[tex]q[/tex] - Electric charge, measured in Coulomb.

As electric field of the Earth is directed in toward the center of the planet, that is, in the same direction of gravity, electric field must be a negative value. If we know that [tex]m = 60\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]E = -150\,\frac{N}{C}[/tex], the charge that a 60-kg human must have to overcome weight is:

[tex]q = \frac{(60\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{-150\,\frac{N}{C} }[/tex]

[tex]q = -3.923\,C[/tex]

The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The electric force of repulsion between two people with the same charge calculated in part (a) is determined by Coulomb's Law, whose definition we proceed to use:

[tex]F = \kappa \cdot \frac{q^{2}}{r^{2}}[/tex] (Eq. 3)

Where:

[tex]\kappa[/tex] - Electrostatic constant, measured in Newton-square meter per square Coulomb.

[tex]q[/tex] - Electric charge, measured in Coulomb.

[tex]r[/tex] - Distance between two people, measured in meters.

If we know that [tex]\kappa = 9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}[/tex], [tex]q = -3.923\,C[/tex] and [tex]r = 100\,m[/tex], then the force of repulsion between two people is:

[tex]F = \left(9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot \left[\frac{(-3.923\,C)^{2}}{(100\,m)^{2}} \right][/tex]

[tex]F = 13.851\times 10^{6}\,N[/tex]

The force of repulsion between two people is [tex]13.851\times 10^{6}[/tex] newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.

You are driving your car at 45 m/s, when a raccoon runs into the street in front of you. You slam on the brakes and come to a stop in 5 seconds. What is the acceleration of your car?

Answers

Answer:

-9m/s²

Explanation:

Given parameters:

Initial velocity = 45m/s

Final velocity  = 0

duration  = 5s

Unknown:

acceleration = ?

Solution:

Acceleration is the rate of change of velocity with time;

        Acceleration  = [tex]\frac{v- u}{t}[/tex]

v is the final velocity

u is the initial velocity

t is the time taken

 Input the parameters and solve;

     Acceleration  = [tex]\frac{0 - 45}{5}[/tex]   = -9m/s²

The car accelerates at a rate of -9m/s² which is a deceleration

What is a personified statement?​

Answers

Answer:

A statement about you

Explanation:

Answer:

Personification is when you give an animal or object qualities or abilities that only a human can have. This creative literary tool adds interest and fun to poems or stories. Personification is what writers use to bring non-human things to life. It helps us better understand the writer's message.

Explanation:

Matter is anything that takes up space and has mass. O A. True O B. False​

Answers

Answer:True

Explanation:Matter is everything around you. Atoms and compounds are all made of very small parts of matter. Those atoms go on to build the things you see and touch every day. Matter is defined as anything that has mass and takes up space (it has volume).



When driving at slower speeds you need to use what type of steering

wheel movements compared to when driving at faster speeds? *

Answers

Answer:

slower speeds = larger and faster steering wheel movements

faster speeds = small and slow steering wheel movements

Explanation:

When driving at slower speeds you need to use larger and faster steering wheel movements. This is because at slow speeds the car does not have enough momentum to make certain maneuvers with small steering wheel movements in a given amount of time, therefore making large and faster steering wheel movements gives the car enough time with the momentum it has to make the desired maneuver. At faster speeds only small and slow steering wheel movements are needed and while cause the car to quickly change to the desired direction due to the increased momentum of the car.

What is the effect of applying an unbalanced force on an object?
A. Object moves
B. Object remains stationary
C. Mass of object increases
D. Inertia of object increases

Answers

A object move that the answer

Object will start moving in the direction where comparatively high force  is directing to move

What is an unbalance force ?

When the resultant force acting on a body is not equal to zero , the force acting on the body are known as unbalance forces . The body acted upon by unbalance forces changes its state of motion.

According to Newton if a body is experiencing net force =0 (which means that all horizontal and vertical forces are cancelling each other ), then the body will acquire an equilibrium state and will not move because forces are balancing each other and making net force on the body zero .

But unbalance force can tend the body to move in that direction in which a comparatively high force is acting due to which cancellation will not occur ( equilibrium state will not occur )

hence correct option A)Object moves

learn more about unbalance forces

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A projectile is fired straight up with an initial velocity of 40.0 m/s . Approximately how high will the projectile ?

Answers

Answer:

it depends on the wind and any other conditions but if you have a controlled environment it should take 1 second to get 40 meters but it could go higher in which it could take about 5 seconds to go 200 meters

Explanation:

hope it helped

:)

Protons and neutrons grouped in a specific pattern
Ob
Protons and electrons spread around randomly
Ос
А group of protons and neutrons that are surrounded by electrons
Od
A ball of electrons and neutrons surrounded by protons

Answers

Answer b protons and electrons

Which state of matter is most similar to solids

Answers

Answer:

liquids

Explanation

A 0.5 kg basketball moving 5 m/s to the right collides with a 0.05 kg tennis
ball moving 30 m/s to the left. After the collision, the tennis ball is moving 34
m/s to the right. What is the velocity of the basketball after the collision?
Assume an elastic collision occurred.
A. 1.4 m/s to the left
B. 1.4 m/s to the right
C. 11.4 m/s to the right
D. 11.4 m/s to the left

SOMEBODY HELP ME

Answers

Answer:

A. 1.4 m/s to the left

Explanation:

To solve this problem we must use the principle of conservation of momentum. Let's define the velocity signs according to the direction, if the velocity is to the right, a positive sign will be introduced into the equation, if the velocity is to the left, a negative sign will be introduced into the equation. Two moments will be analyzed in this equation. The moment before the collision and the moment after the collision. The moment before the collision is taken to the left of the equation and the moment after the collision to the right, so we have:

[tex]M_{before} = M_{after}[/tex]

where:

M = momentum [kg*m/s]

M = m*v

where:

m = mass [kg]

v = velocity [m/s]

[tex](m_{1} *v_{1} )-(m_{2} *v_{2})=(m_{1} *v_{3} )+(m_{2} *v_{4})[/tex]

where:

m1 = mass of the basketball = 0.5 [kg]

v1 = velocity of the basketball before the collision = 5 [m/s]

m2 = mass of the tennis ball = 0.05 [kg]

v2 = velocity of the tennis ball before the collision = - 30 [m/s]

v3 =  velocity of the basketball after the collision [m/s]

v4 = velocity of the tennis ball after the collision = 34 [m/s]

Now replacing and solving:

(0.5*5) - (0.05*30) = (0.5*v3) + (0.05*34)

1 - (0.05*34) = 0.5*v3

- 0.7 = 0.5*v

v = - 1.4 [m/s]

The negative sign means that the movement is towards left

Answer:

A. 1.4 m/s to the left

Explanation:

If we throw a body upwards at time t=0 with an initial speed s=25 m/s, what would the body's speed, position and velocity?

Answers

Answer:

The speed is     v = 25 - 9.8 t

the velocity is   v = (25 - 9.8 t) j^

the position is   y = 25 t - 4.9 t²

Explanation:

This is a vertical throwing exercise.

The first thing we must do is set a coordinate system, in this case we will make the upward direction positive.

Let's write the kinematics equations

        v = v₀ - g t

        v² = v₀² - g (y-y₀)

        y = y₀ + v₀ t - ½ g t²

in our case the initial velocity is

         v₀ = 25 m / s

and we zero the system at the launch point

        y₀ = 0

the equations remain

      v = 25 - 9.8 t

        v² = 25² - 9.8 y

        y = 0 + 25 t - ½ 9.8 t²

         

The speed is     v = 25 - 9.8 t

the velocity is   v = (25 - 9.8 t) j^

the position is   y = 25 t - 4.9 t²

find the vector sum of the 3 vectors. A= -3i - 2j + 7k, B= i + 3j + 3k, B= -5k​

Answers

Answer:

34k+B plus 9 :)

Explanation:

The volume of 2.0 kg of helium in a piston-cylinder device is initially 7 m3. Now the helium is compressed to 5 m3 while its pressure is maintained constant at 160 kPa. Determine: (a) The initial and final temperatures of helium. (b) The work required to compress it, in kJ.

Answers

Answer:

A) T1 = 269.63 K

T2 = 192.59 K

B) W = -320 KJ

Explanation:

We are given;

Initial volume: V1 = 7 m³

Final Volume; V2 = 5 m³

Constant Pressure; P = 160 KPa

Mass; m = 2 kg

To find the initial and final temperatures, we will use the ideal gas formula;

T = PV/mR

Where R is gas constant of helium = R = 2.0769 kPa.m/kg

Thus;

Initial temperature; T1 = (160 × 7)/(2 × 2.0769) = 269.63 K

Final temperature; T2 = (160 × 5)/(2 × 2.0769) = 192.59 K

B) world one is given by the formula;

W = P(V2 - V1)

W = 160(5 - 7)

W = -320 KJ

In a railroad switchyard, a rail car of mass 41,700 kg starts from rest and rolls down a 2.65-m-high incline and onto a level stretch of track. It then hits a spring bumper, whose spring compresses 79.4 cm. Find the spring constant.

Answers

Answer:

The spring constant is 3.44x10⁶ kg/s².

Explanation:

We cand find the spring constant by conservation of energy:

[tex] E_{i} = E_{f} [/tex]

[tex] mgh = \frac{1}{2}kx^{2} [/tex]   (1)

Where:

m is the mass =  41700 kg

g is the gravity = 9.81 m/s²

h is the height = 2.65 m

x is the distance of spring compression = 79.4 cm

k is the spring constant =?

Solving equation (1) for k:

[tex]k = \frac{2mgh}{x^{2}} = \frac{2*41700 kg*9.81 m/s^{2}*2.65 m}{(0.794 m)^{2}} = 3.44 \cdot 10^{6} kg/s^{2}[/tex]              

Therefore, the spring constant is 3.44x10⁶ kg/s².

I hope it helps you!

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