A 2.5 kg rock is released from rest at the surface of a pond 1.8 m deep. As the rock falls, a constant upward force of 5.0 N is exerted on it by water resistance. Let y=0 be at the bottom of the pond.

Required:
a. Calculate the nonconservative work, Wnc, done by water resistance on the rock, the gravitational potential energy of the system, U, the kinetic energy of the rock, K, and the total mechanical energy of the system, E, when the depth of the rock below the water's surface is 0 m.
b. Calculate the nonconservative work, Wnc, done by water resistance on the rock, the gravitational potential energy of the system, U, the kinetic energy of the rock, K, and the total mechanical energy of the system, E, when the depth of the rock below the water's surface is 0.50 m.
c. Calculate the nonconservative work, Wnc, done by water resistance on the rock, the gravitational potential energy of the system, U, the kinetic energy of the rock, K, and the total mechanical energy of the system, E, when the depth of the rock below the water's surface is 1.0 m.

Answers

Answer 1

Answer:

a) Nonconservative Work

[tex]W_{disp} = 9\,J[/tex]

Final Gravitational Potential Energy

[tex]U_{f} = 0\,J[/tex]

Final Translational Energy

[tex]K_{f} = 35.131\,J[/tex]

b) Nonconservative Work

[tex]W_{disp} = 6.5\,J[/tex]

Final Gravitational Potential Energy

[tex]U_{f} = 12.259\,J[/tex]

Final Translational Energy

[tex]K_{f} = 25.373\,J[/tex]

c) Nonconservative Work

[tex]W_{disp} = 4\,J[/tex]

Final Gravitational Potential Energy

[tex]U_{f} = 24.518\,J[/tex]

Final Translational Energy

[tex]K_{f} = 15.614\,J[/tex]

Explanation:

The nonconservative work due to water resistance is defined by definition of work:

[tex]W_{disp} = F\cdot (y_{o}-y_{f})[/tex] (1)

Where:

[tex]W_{disp}[/tex] - Dissipate work, in joules.

[tex]F[/tex] - Resistance force, in newtons.

[tex]y_{o}[/tex] - Initial height, in meters.

[tex]y_{f}[/tex] - Final height, in meters.

The final gravitational potential energy ([tex]U_{f}[/tex]), in joules, is calculated by means of the definition of gravitational potential energy:

[tex]U_{f} = m\cdot g\cdot y_{f}[/tex] (2)

Where:

[tex]m[/tex] - Mass of the rock, in kilograms.

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

The final translational kinetic energy ([tex]K_{f}[/tex]), in joules, is obtained by means of the Principle of Energy Conservation, Work-Energy Theorem and definitions of gravitational potential energy and translational kinetic energy:

[tex]m\cdot g\cdot y_{o} = U_{f} + K_{f} + W_{disp}[/tex] (3)

[tex]K_{f} = m\cdot g\cdot y_{o} - U_{f} - W_{disp}[/tex]

Lastly, the mechanical energy of the system ([tex]E[/tex]), in joules, is the sum of final gravitational potential energy, translational kinetic energy and dissipated work due to water resistance:

[tex]E = U_{f} + K_{f} + W_{disp}[/tex] (4)

Now we proceed to solve the exercise in each case:

a) Nonconservative Work ([tex]F = 5\,N[/tex], [tex]y_{o} = 1.8\,m[/tex], [tex]y_{f} = 0\,m[/tex])

[tex]W_{disp} = (5\,N)\cdot (1.8\,m - 0\,m)[/tex]

[tex]W_{disp} = 9\,J[/tex]

Final Gravitational Potential Energy ([tex]m = 2.5\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{f} = 0\,m[/tex])

[tex]U_{f} = (2.5\,kg) \cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (0\,m)[/tex]

[tex]U_{f} = 0\,J[/tex]

Final Translational Energy ([tex]m = 2.5\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{o} = 1.8\,m[/tex], [tex]U_{f} = 0\,J[/tex], [tex]W_{disp} = 9\,J[/tex])

[tex]K_{f} = (2.5\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.8\,m) -0\,J-9\,J[/tex]

[tex]K_{f} = 35.131\,J[/tex]

b) Nonconservative Work ([tex]F = 5\,N[/tex], [tex]y_{o} = 1.8\,m[/tex], [tex]y_{f} = 0.50\,m[/tex])

[tex]W_{disp} = (5\,N)\cdot (1.8\,m - 0.5\,m)[/tex]

[tex]W_{disp} = 6.5\,J[/tex]

Final Gravitational Potential Energy ([tex]m = 2.5\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{f} = 0.5\,m[/tex])

[tex]U_{f} = (2.5\,kg) \cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (0.5\,m)[/tex]

[tex]U_{f} = 12.259\,J[/tex]

Final Translational Energy ([tex]m = 2.5\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{o} = 1.8\,m[/tex], [tex]U_{f} = 12.259\,J[/tex], [tex]W_{disp} = 6.5\,J[/tex])

[tex]K_{f} = (2.5\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.8\,m) -12.259\,J-6.5\,J[/tex]

[tex]K_{f} = 25.373\,J[/tex]

c) Nonconservative Work ([tex]F = 5\,N[/tex], [tex]y_{o} = 1.8\,m[/tex], [tex]y_{f} = 1\,m[/tex])

[tex]W_{disp} = (5\,N)\cdot (1.8\,m - 1\,m)[/tex]

[tex]W_{disp} = 4\,J[/tex]

Final Gravitational Potential Energy ([tex]m = 2.5\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{f} = 1\,m[/tex])

[tex]U_{f} = (2.5\,kg) \cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (1\,m)[/tex]

[tex]U_{f} = 24.518\,J[/tex]

Final Translational Energy ([tex]m = 2.5\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{o} = 1.8\,m[/tex], [tex]U_{f} = 24.518\,J[/tex], [tex]W_{disp} = 4\,J[/tex])

[tex]K_{f} = (2.5\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.8\,m) -24.518\,J-4\,J[/tex]

[tex]K_{f} = 15.614\,J[/tex]


Related Questions

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Answers

Direct relation is present between distance and time which means one has direct impact on the value of other.

There is direct relationship between distance and time. Direct relation means if one increases the other automatically increases while on the other hand, if one decreases the other will also decreases. If there is a large amount of distance so more time is required to pass it while on the other hand, if there is less distance so less time is required to complete it so we can conclude that there is direct relationship between distance and time.

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A 59.0 kg bungee jumper jumps off a bridge and undergoes simple harmonic motion. If the period of oscillation is 0.250 mins, what is the spring constant (in N/m) of the bungee cord, assuming it has negligible mass compared to that of the jumper

Answers

Answer:

The spring constant of the spring is 10.3 N/m.

Explanation:

Given that,

Mass of a bungee jumper, m = 59 kg

The period of oscillation, T = 0.25 min = 15 sec

We need to find the spring constant of the bungee cord. We know that the period of oscillation is given by :

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

Where

k is the spring constant

[tex]T^2=4\pi^2\times \dfrac{m}{k}\\\\k=4\pi^2\times \dfrac{m}{T^2}\\\\k=4\pi^2\times \dfrac{59}{(15)^2}\\\\k=10.3\ N/m[/tex]

So, the spring constant of the spring is 10.3 N/m.

A fast moving vehicle travelling at a speed of 25.4 m/s comes up behind another vehicle which is
travelling at a slower constant speed of 13.6 m/s. If the faster vehicle does not begin braking until it
is 11.4 meters away from the car in front of it, what is the minimum acceleration that the faster car
must exhibit if it is to avoid colliding with the car in front? Assume that both cars are travelling in the
positive direction

Answers

Answer:

     a = 6.1 m / s²

Explanation:

For this kinematics exercise, to solve the exercise we must set a reference system, we place it in the initial position of the fastest vehicle

Let's find the relative initial velocity of the two vehicles

          v₀ = v₀₂ - v₀₁

          v₀ = 25.4 - 13.6

          v₀ = 11.8 m / s

 

the fastest vehicle

          x = v₀ t + ½ a t²

The faster vehicle has an initial speed relative to the slower vehicle, therefore it is as if the slower vehicle were stopped, so the distance that must be traveled in a fast vehicle to reach this position is

          x = 11.4 m

         

let's use the expression

           v² = v₀² - 2 a x

how the vehicle stops v = 0

          a = v₀² / 2x

          a = [tex]\frac{11.8^2}{2 \ 11.4}[/tex]

          a = 6.1 m / s²

this velocity is directed to the left

What is the total kinetic energy of a 0.15 kg hockey puck sliding at 0.5 m/s and rotating about its center at 8.4 rad/s

Answers

Answer:

[tex]K=0.023J[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=0.15[/tex]

Velocity [tex]v=0.5m/s[/tex]

Angular Velocity [tex]\omega=8.4rad/s[/tex]

Generally the equation for Kinetic Energy is mathematically given by

[tex]K=\frac{1}{2}M(v^2+\frac{1}{2}R^2\omega^2)[/tex]

[tex]K=\frac{1}{2}0.15(0.5^2+\frac{1}{2}(0.038)^2.(8.4rad/s^2))[/tex]

[tex]K=0.023J[/tex]

Object A has a mass m and a speed v, object B has a mass m/2 and a speed 4v, and object C has a mass 3m and a speed v/3. Rank the objects according to the magnitude of their momentum.

Required:
Rank from smallest to largest.

Answers

Answer:

Momentum of object A = Momentum of object C < momentum of B.

Explanation:

The momentum of an object is equal to the product of mass and velocity.

Object A has a mass m and a speed v. Its momentum is :

p = mv

Object B has a mass m/2 and a speed 4v. Its momentum is :

p = (m/2)×4v = 2mv

Object C has a mass 3m and a speed v/3. Its momentum is :

p = (3m)×(v/3) = mv

So,

Momentum of object A = Momentum of object C < momentum of B.

1L of a solution of oxalic acid [(). ] contains 2.52g of acid.
Calculate the molarity of the solution.

Answers

Answer:

0.028 M.

Explanation:

NOTE: This question is a chemistry question. However, the answer to the question can be obtained as shown below:

We'll begin by calculating the number of mole in 2.52 g of oxalic acid, C₂H₂O₄. This can be obtained as follow:

Mass of C₂H₂O₄ = 2.52 g

Molar mass of C₂H₂O₄ = (2×12) + (2×1) + (4×16)

= 24 + 2 + 64

= 90 g/mol

Mole of C₂H₂O₄ =?

Mole = mass / molar mass

Mole of C₂H₂O₄ = 2.52 / 90

Mole of C₂H₂O₄ = 0.028 mole

Finally, we shall determine the molarity of the solution. This can be obtained as follow:

Mole of C₂H₂O₄ = 0.028 mole

Volume = 1 L

Molarity =?

Molarity = mole / Volume

Molarity = 0.028 / 1

Molarity = 0.028 M

Therefore, the molarity of the solution is 0.028 M.

A spring of stiffness 11000 N/m is connected to a wall. The other end is attached to a block of mass 2 kg. A second spring of stiffness 12200 N/m extends between the first block and a second block of mass 2 kg. You are holding the second block with 300 N of force and everything is motionless.

Required:
How stretched is the spring connecting the first block to the wall?

Answers

Answer:

Explanation:

Let the stretch produced in first spring and second spring be x₁ and x₂ respectively. 300 N is the restoring force created in them .

k₁x₁ = k₂x₂ = 300 N .

k₁ = 11000 N/m

11000 x₁ = 300N

x₁ = 300 N / 11000

= 0.027 m

= 2.7 cm .

Which person ha the most freedom to make his or her own lifestyle decisions?
A. Frieda is 10 years old and lives with her grandmother
B. Vladimir is 3 years old and attends preschool
C. Quincy is 16 years old and lives in a dormitory
D. Lucinda is 32 years old and has two children

Answers

Answer:

C Quincy

Explanation:

Both Frieda and Vladimir are too young to be making their own decisions. Lucinda has limited freedom due to having two children, while Quincy is just now becoming an adult and has his whole life still ahead of him. Therefore, Quincy has the most freedom to make their own lifestyle decisions.

I hope this helps!

Answer:

The correct answer is C. Quincy is 16 years old and lives in a dormitory

Explanation:

Quincy being 16 may be able to get a job and hold a membership at a gym even if not at a gym he still has the most freedom period with Lucinda having children and Frieda and Vladimir being to young to make the choices of exercising on their own.

Please tell me if I'm wrong so I may give you the correct answer!

Happy Holidays!!  

TRUE or FALSE: The acceleration of projectile is 0 m/s/s at the peak of the trajectory. Identify the evidence which supports your answer.

Answers

FALSE The vertical velocity of a projectile is 0 m/s at the peak of its trajectory; but the horizontal component of the velocity at the peak is whatever the value was when first launched

The vertical acceleration of the projectile is at 0 m/s while the horizontal acceleration of the projectile at the peak of the trajectory = initial acceleration  ( i.e. statement in the question is False )

Projectile motion follows a parabolic path with x and y components of its velocity and acceleration. also the acceleration of a projectile is subject only to the acceleration due to gravity unlike other kinds of motions.

In a parabolic motion an object ( projectile ) is thrown into the air and left to move through a parabolic path under the effect of acceleration due to gravity.

Hence we can conclude that the statement is false, because horizontal acceleration of the projectile at the peak of the trajectory = initial acceleration

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If the accuracy in measuring the position of a particle increases, the accuracy in measuring its velocity will Group of answer choices remain the same. decrease. increase. It is impossible to say since the two measurements are independent and do not affect each other.

Answers

Answer:

The correct answer is Option A (decrease).

Explanation:

According to Heisenberg's presumption of unpredictability, it's impossible to ascertain a quantum state viewpoint as well as momentum throughout tandem.Also, unless we have accurate estimations throughout the situation, we will have a decreased consistency throughout the velocity as well as vice versa though too.

Other given choices are not connected to the given query. Thus the above is the right answer.

I need the help
Please. I’m terrible at physics

Answers

Answer:

Explanation:

so opposite and equal , right?    forces are. soooo..

528+52= 580 N    is the force that is being exerted up on the scale

A small object with mass 0.20 kg swings as a pendulum on the end of a long light rope. For small amplitude of swing, the period of the motion is 3.0 s. If the object is replaced by one with mass 0.400 kg, what is the period for small amplitude of swing? (a) 1.5 s (b) 3.05 (c) 6.0 s (d) 12.0 s (e) none of the above answers

Answers

Answer:

The correct option is (e) "none of the above".

Explanation:

Given that,

A small object with mass 0.20 kg swings as a pendulum on the end of a long light rope. For small amplitude of swing, the period of the motion is 3.0 s.

If the object is replaced by one with mass 0.400 kg, then we have to find the period for small amplitude of the swing.

We know that the time period can be calculated as :

[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]

Where

l is the length

g is acceleration due to gravity

It means the time period is independent of the mass. So, if the mass is replaced by one with mass 0.400 kg, there is no effect on the time period.

please help me .finish this paper​

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Solution-1:-

[tex]\boxed{\sf \dfrac{10\times 1000}{60\times 60}}[/tex]

Solution:-2

[tex]\boxed{\sf Sodium\:and\:Potassium}[/tex]

Solution:-3

[tex]\boxed{\sf 320m}[/tex]

Solution:-4

[tex]\boxed{\sf Rough\:tiles\:are\:used\:in\:bathroom}[/tex]

Solution:-5

[tex]\boxed{\sf Mg_3N_2}[/tex]

Solution:-6

[tex]\boxed{\sf Grapes\:and\:Rambutan}[/tex]

Solution:-7

[tex]\boxed{\sf {}^{}_{}N}[/tex]

Solution:-8

[tex]\boxed{\sf Galactuse}[/tex]

Solution:-9

[tex]\boxed{\sf Y-X}[/tex]

Solution:-10

[tex]\boxed{\sf Cell\:wall}[/tex]

A force of 15 N toward the WEST is applied to a 4.0 kg box. Another force of 42 N toward the EAST is also applied to the 4.0 kg box. The net force on the 4.0 kg box
is

Answers

Answer :

[tex]\implies F_1 < F_2[/tex]

[tex] \implies F_{net} = F_2 - F1[/tex]

[tex]\implies F_{net} = 42 -15[/tex]

[tex]\implies \underline{ \boxed{ F_{net} = 27 \: N}}[/tex]

The net force on the 4.0 kg box is 27 N towards EAST.

Which of the following careers would physicists be least qualified to perform,
based on their training?
A. Artist
B. Mechanical engineer
C. Laser designer
D. Power plant engineer

Answers

D

Source? Trust Me Bro

A container is filled with water and the pressure at the container bottom is P. If the container is instead filled with a liquid having specific gravity 1.05, what new bottom pressure will be measured

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eat ahah is when s susuehs

what is the prefix notation of 0.0000738?​

Answers

Answer:

The scientific notation of 738 is 7.38 x 1002.

I agree with the answer above me…

Convert the unit of 0.0063 milliseconds into microseconds. (Answer in scientific notation)

Answers

Answer:

microsecond = 1 × 10-6 seconds

1 second = 1 × 100 seconds

1 microsecond = (1 / 1) × 10-6 × 10-0 seconds

1 microsecond = (1) × 10-6-0 seconds

1 microsecond = (1) × 10-6 seconds

1 microsecond = 1 × 1.0E-6 seconds

1 microsecond = 1.0E-6 seconds,just do like it

Answer:

6.3 x [tex]10^{-3}[/tex]

Explanation:

1 millisecond = 1000 microseconds

0.0063 millisecond = 6.3 microseconds

A baseball pitcher throws a fastball by spinning his arm at 27.7m/s. The ball has a mass of 0.700kg and experiences a net centripetal force of 625N. How long is the pitchers arm (the radius of the curve)?

Answers

In the historical sense, postmodern society is simply a society that occurs after the modern society. ... Many of the elements of a society like this are reactions to what the modern society stood for: industrialism, rapid urban expansion, and rejection of many past principles.

A 2 nC point charge is embedded at the center of a nonconducting sphere of radius = 4.8 cm which has a charge of 1 nC distributed uniformly throughout its volume. What is the magnitude of the electric field at a point that is 2.4 cm from the center of the sphere? Write your answer in terms of kN/C.

Answers

Answer: [tex]3.32\times 10^4\ kN/C[/tex]

Explanation:

Given

Charge at the center of the sphere is [tex]q=2\ nC[/tex]

Charge distributed over the entire sphere [tex]Q=1\ nC[/tex]

Radius of sphere [tex]r=4.8\ cm[/tex]

Using Guass law

[tex]\Rightarrow \oint \vec{E}\cdot \vec{dA}=\dfrac{q_{enc}}{\epsilon_o}\\\\\Rightarrow E\cdot 4\pi r^2=\dfrac{1}{\epsilon}(q+\dfrac{Q}{4\pi R^2}\times 4\pi r^2)\\\\\Rightarrow E\cdot 4\pi r^2=\dfrac{1}{\epsilon}(q+Q\dfrac{r^2}{R^2})\\\\\Rightarrow E\cdot (2.4\times 10^{-2})^2=9\times 10^9(2+1\cdot \dfrac{1}{4})\times 10^{-9}\\\\\Rightarrow E=3.32\times 10^4\ kN/C[/tex]

Set the initial bead height to 3.00 m. Click Play. Notice that the ball makes an entire loop. What is the minimum height required for the ball to make an entire loop

Answers

Answer:

h> 2R

Explanation:

For this exercise let's use the conservation of energy relations

starting point. Before releasing the ball

       Em₀ = U = m g h

Final point. In the highest part of the loop

       Em_f = K + U = ½ m v² + ½ I w² + m g (2R)

where R is the radius of the curl, we are considering the ball as a point body.

      I = m R²

      v = w R

we substitute

       Em_f = ½ m v² + ½ m R² (v/R) ² + 2 m g R

       em_f = m v² + 2 m g R

Energy is conserved

       Emo = Em_f

       mgh = m v² + 2m g R

       h = v² / g + 2R

 

The lowest velocity that the ball can have at the top of the loop is v> 0

      h> 2R

Physics question plz help

Answers

Answer:

B. [tex]30\,s\,< t \le 40\,s[/tex].

Explanation:

The vehicle is travelling eastwards when its velocity is positive and westwards when velocity is negative. According to the graph, [tex]v > 0[/tex] for [tex]30\,s\,< t \le 40\,s[/tex]. Hence, correct answer is B.

1/012=1/0.05+1/d' hiiiiiiiiii

Answers

Correct question is;

1/0.12 = (1/0.05) + (1/d')

Answer:

d' = -1/700

Explanation:

1/0.12 = (1/0.05) + (1/d')

Let's rearrange to get;

(1/d') = (1/0.12) - (1/0.05)

(1/d') = (1/(12/100)) - (1/(5/100))

(1/d') = 100/12 - 100/5

Let's multiply through by 60 to get rid of the denominators on the right side;

> (1/d') = 500 - 1200

> (1/d') = -700

> d' = -1/700

list the factors that affecting frictional force ?​

Answers

Answer:

The frictional force between two bodies depends mainly on three factors: (I) the adhesion between body surfaces (ii) roughness of the surface (iii) deformation of bodies

A 650-kg elevator starts from rest. It moves upward for 3.00 s with constant acceleration until it reaches its cruis- ing speed of 1.75 m/s. (a) What is the average power of the elevator motor during this period

Answers

Answer: [tex]P=5573.43\ W[/tex]

Explanation:

Given

Mass of the elevator is [tex]M=650\ kg\\\[/tex]

Time period of ascension [tex]t=3\ s[/tex]

cruising speed [tex]v=1.75\ m/s[/tex]

Distance moved by elevator during this time

Suppose Elevator starts from rest

[tex]\Rightarrow v=u+at\\\Rightarrow 1.75=0+a(3)\\\Rightarrow a=0.583\ s[/tex]

Distance moved

[tex]\Rightarrow h=ut+0.5at^2\\\Rightarrow h=0+0.5\times 0.5833\times (3)^2\\\Rightarrow h=2.62\ m[/tex]

Gain in Potential Energy is

[tex]\Rightarrow E=mgh\\\Rightarrow E=650\times 9.8\times 2.62\\\Rightarrow E=16,720.3\ N[/tex]

Average power during this period is

[tex]\Rightarrow P=\dfrac{E}{t}\\\\\Rightarrow P=\dfrac{16,720.3}{3}\\\\\Rightarrow P=5573.43\ W[/tex]

Answer:

The power is 331.7 W.

Explanation:

mass, m = 650 kg

time, t= 3 s

initial  velocity,  u = 0 m/s

final velocity, v = 1.75 m/s

(a) The power is defined as the rate of doing work.

Work is given by the change in kinetic energy.

W = 0.5 m (V^2 - u^2)

W = 0.5 x 650 x 1.75 x 1.75 =  995.3 J

The power is given by

P = W/t = 995.3/3 = 331.7 W

12. A car travels in a straight line with an average velocity of 80 km/h for 2.5h and then an average velocity of 40 km/h from 1.5 h (a) what is the total displacement for the 4 h trip? (b) What is the average velocity for the total trip?

Answers

Answer:

a. Total displacement = 140 km/h

b. Average velocity = 35 km/h

Explanation:

Given the following data;

Average velocity A = 80 km/h

Time A = 2.5 hours

Average velocity B = 40 km/h

Time B = 1.5 hours

a. To find the total displacement for the 4 h trip;

Total time = Time A + Time B

Total time = 2.5 + 1.5

Total time = 4 hours

Next, we would determine the displacement at each velocity.

Mathematically, displacement is given by the formula;

Displacement = velocity * time

Substituting into the formula, we have;

Displacement A = 80 * 2.5

Displacement A = 200 km/h

Displacement B = 40 * 1.5

Displacement B = 60 km/h

Total displacement = Displacement A - Displacement B

Total displacement = 200 - 60

Total displacement = 140 km/h

b. To find the average velocity for the total trip;

Mathematically, the average velocity of an object is given by the formula;

[tex] Average \; velocity = \frac {total \; displacement}{total \; time} [/tex]

Substituting into the formula, we have;

[tex] Average \; velocity = \frac {140}{4} [/tex]

Average velocity = 35 km/h

A barge is hauled along a straight-line section of canal by two horses harnessed to tow ropes and walking along the tow paths on
either side of the canal. Each horse pulls with a force of 839 N at an angle of 15° with the centerline of the canal. Find the sum of these
two forces on the barge.
answer in ___kN

Answers

Answer:

1.621 kN

Explanation:

Since each horse pulls with a force of 839 N at an angle of 15° with the centerline of the canal, the horizontal component of the force due to the first horse along the canal is F= 839cos15° N and its vertical component is F' = 839sin15° N(it is positive since it is perpendicular to the centerline of the canal and points upwards).

The horizontal component of the force due to the second horse along the canal is f = 839cos15° N and its vertical component is f' = -839sin15° N (it is negative since it is perpendicular to the centerline of the canal and points downwards).

So, the resultant horizontal component of force R = F + f = 839cos15° N + 839cos15° N = 2(839cos15°) N = 2(839 × 0.9659) = 2 × 810.412 = 1620.82 N

So, the resultant vertical component of force R' = F' + f' = 839sin15° N + (-839sin15° N) = 839sin15° N - 839sin15° N = 0 N

The magnitude of the resultant force which is the sum of the two forces is R" = √(R² + R'²)

= √(R² + 0²)  (since R' = 0)

= √R²

= R  

= 1620.82 N

= 1.62082 kN

≅ 1.621 kN

So, the sum of these  two forces on the barge is 1.621 kN

A compact disk with a 12 cm diameter is rotating at 5.24 rad/s.

a. What is the linear speed _______m/s
b. What is the centripetal acceleration of a point on its outer rim _______
c. Consider a point on the CD that is halfway between its center and its outer rim. Without repeating all of the calculations required for parts (a) and (b), determine the linear speed of this point. _______m/s
d. Determine the centripetal acceleration of this point. _______

Answers

Answer:

(a) 31.44 m/s (b) 164.74 m/s²

Explanation:

Given that,

The diameter of a disk, d = 12 cm

Radius, r = 6 cm

Angular speed = 5.24 rad/s

(a) Linear speed,

[tex]v=r\omega\\\\v=6\times 5.24\\\\v=31.44\ m/s[/tex]

(b) Centripetal acceleration,

[tex]a=\dfrac{v^2}{r}\\\\a=\dfrac{31.44^2}{6}\\\\a=164.74\ m/s^2[/tex]

A thin stream of water flows vertically downward. The stream bends toward a positively charged object when it is placed near it. The positively charged object is then removed. What will happen to the same stream of water when a negatively charged object is placed nearit

Answers

Answer:

the water jet and the negative object attract

Explanation:

Let's analyze the situation. Water is a good conductor of electricity, so when an object with a positive charge is brought closer, a charge of the opposite sign is created, which is why the two objects attract each other. The charge created comes from the ground

When claiming the object, the charge is reduced to zero or the charge goes to earth since water is a good conductor of electricity, therefore when approaching an object with a negative charge, more charges go to earth and the ring is left with a positive charge and attracts it to the object.

In short, the water jet and the negative object attract

Which wave has the smallest amplitude?

I will give brainliest to the right answer.

Answers

Wave-D (the red one) has the smallest amplitude of the 4 waves on this graph.

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