A 1870 kg car traveling at 13.5 m/s collides with a 2970 kg car that is initally at rest at a stoplight. The cars stick together and move 1.93 m before friction causes them to stop. Determine the coefficient of kinetic friction between the cars and the road, assuming that the negative acceleration is constant and all wheels on both cars lock at the time of impact.

Answers

Answer 1

Answer:

The value  is [tex]  \mu  = 0.72  [/tex]

Explanation:

From the question we are told that

   The mass of the first car is  [tex]m_1  =  1870\ kg[/tex]

    the initial  speed of the car  is  [tex]u  =  13.5 \  m/s[/tex]

    The  mass of the second car is  [tex]m_2 =  2970\  kg[/tex]

    The distance move by both cars is  s =  1.93  m

Generally from the law of momentum conservation

    [tex]m_1 * u_1 + m_2 *  u_2  =  (m_1 + m_2 ) *  v_f[/tex]

Here [tex]u_2  =  0[/tex] because the second car is at rest

and  [tex]v_f[/tex] is the final  velocity of the the two car

So

    [tex]1870*  13.5+ 0=  ( 1870 + 2970 ) *  v_f[/tex]      

=> [tex]v_f  =  5.22\  m/s[/tex]

Generally from kinematic equation

    [tex]v_f^2 = u_2^2  +  2as[/tex]

here a is the deceleration

So

    [tex]5.22^2 = 0  +  2 *a  *  1.93[/tex]

=> [tex]a =  7.06 \  m/s^2 [/tex]

Generally the frictional  force is equal to the force propelling the car , this can be mathematically represented as

   [tex]F_f  =  F[/tex]

Here  F is mathematically represented as

[tex]F =  (m_1 + m_2) *  a[/tex]

[tex]F =  (1870 + 2970) *  7.06 [/tex]    

[tex]F =34170.4 \ N[/tex]  

and

[tex]F_f  =  \mu *  (m_1 + m_2 ) *  g[/tex]

[tex]F_f  =  47432 * \mu [/tex]

So

[tex]  47432 * \mu   = 34170.4  [/tex]

=> [tex]  47432 * \mu   = 34170.4  [/tex]

=> [tex]  \mu  = 0.72  [/tex]


Related Questions

Grass and plants get energy from
А
the sun.
B
eating food.
с
windmills.
D
electrons.

Answers

A is the answer for that question

Answer:

From the Sun

Explanation:

Plants can't eat any food. They don't ue or need windmills to get energy. They are plants so they don't have any electrons. The only way that they can recive energy from is the sun. Sometimes plants die when they don't get enough sun because they don't have any energy to live.

What is the change in internal energy (in J) of a system that absorbs 0.523 kJ of heat from its surroundings and has 0.366 kcal of work done on it

Answers

Answer:

The change in internal energy of the system is 2,054 J

Explanation:

The first law of thermodynamics relates the work and the transferred heat exchanged in a system through internal energy. This energy is neither created nor destroyed, it is only transformed.

Taking into account that the internal energy is the sum of all the energies of the particles that the system has, you have:

ΔU= Q + W

where U is the internal energy of the system (isolated), Q is the amount of heat contributed to the system and W is the work done by the system.

By convention, Q is positive if it goes from the environment to the system, or negative otherwise, and W is positive if it is carried out on the system and negative if it is carried out by the system.

In this case:

Q= 0.523 kJ (because the energy is absorbed, this is,it goes from the environment to the system)W= 0.366 kcal= 1.531 kJ  (because the work is done on the system, and being 1 kcal= 4.184 kJ)

Replacing:

ΔU= 0.523 kJ + 1.531 kJ

Solving:

ΔU= 2.054 kJ = 2,054 J (being 1 kJ=1,000 J)

The change in internal energy of the system is 2,054 J

In a mattress test, you drop a 7.0 kg bowling ball from a height of 1.5 m above a mattress, which as a result compresses 15 cm as the ball comes to a stop. (a) What is the kinetic energy of the ball just bef

Answers

Answer:

(a) The kinetic energy of the bowling ball just before it hits the matress is 102.974 joules.

(b) The work done by the gravitational force of Earth on bowling ball during the first part of the fall is 102.974 joules.

(c) Work done by gravitational force on bowling ball when mattress is compressed is 10.298 joules.

(d) The work done by the mattress on the bowling ball is 113.272 joules.

Explanation:

The statement is incomplete. The complete question is:

In a mattress test, you drop a 7.0 kg bowling ball from a height of 1.5 m above a mattress, which as a result compresses 15 cm as the ball comes to a stop.

(a) What is the kinetic energy of the ball just before it hits the mattress?  

(b) How much work does the gravitational force of the earth do on the ball as it falls, for the first part of the fall (from the moment you drop it to just before it hits the mattress)?  

(c) How much work does the gravitational force do on the ball while it is compressing the mattress?

(d) How much work does the mattress do on the ball? (You’ll need to use the results of parts (a) and (c)

(a) Based on the Principle of Energy Conservation, we know that ball-earth system is conservative, so that kinetic energy is increased at the expense of gravitational potential energy as ball falls:

[tex]K_{1}+U_{g,1} = K_{2}+U_{g,2}[/tex] (Eq. 1)

Where:

[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Kinetic energies at top and bottom, measured in joules.

[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Gravitational potential energies at top and bottom, measured in joules.

Now we expand the expression by definition of gravitational potential energy:

[tex]U_{g,1}-U_{g,2} = K_{2}-K_{1}[/tex]

[tex]K_{2}= m\cdot g \cdot (z_{1}-z_{2})+K_{1}[/tex] (Eq. 1b)

Where:

[tex]m[/tex] - Mass of the bowling ball, measured in kilograms.

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

[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Initial and final heights of the bowling ball, measured in meters.

If we know that [tex]m = 7\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{1}= 1.5\,m[/tex], [tex]z_{2} = 0\,m[/tex] and [tex]K_{1} = 0\,J[/tex], the kinetic energy of the ball just before it hits the matress:

[tex]K_{2} = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (1.5\,m-0\,m)+0\,m[/tex]

[tex]K_{2} = 102.974\,J[/tex]

The kinetic energy of the bowling ball just before it hits the matress is 102.974 joules.

(b) The gravitational work done by the gravitational force of Earth ([tex]\Delta W[/tex]), measured in joules, is obtained by Work-Energy Theorem and definition of gravitational potential energy:

[tex]\Delta W = U_{g,1}-U_{g,2}[/tex]

[tex]\Delta W = m\cdot g\cdot (z_{1}-z_{2})[/tex] (Eq. 2)

If we know that [tex]m = 7\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{1}= 1.5\,m[/tex] and [tex]z_{2} = 0\,m[/tex], then the gravitational work done is:

[tex]\Delta W = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.5\,m-0\,m)[/tex]

[tex]\Delta W = 102.974\,J[/tex]

The work done by the gravitational force of Earth on bowling ball during the first part of the fall is 102.974 joules.

(c) The work done by the gravitational force of Earth while the bowling when mattress is compressed is determined by Work-Energy Theorem and definition of gravitational potential energy:

[tex]\Delta W = U_{g,2}-U_{g,3}[/tex]

Where [tex]U_{g,3}[/tex] is the gravitational potential energy of the bowling ball when mattress in compressed, measured in joules.

[tex]\Delta W = m\cdot g \cdot (z_{2}-z_{3})[/tex]

Where [tex]z_{3}[/tex] is the height of the ball when mattress is compressed, measured in meters.

If we know that [tex]m = 7\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{2}= 0\,m[/tex] and [tex]z_{3} = -0.15\,m[/tex], the work done is:

[tex]\Delta W = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot [0\,m-(-0.15\,m)][/tex]

[tex]\Delta W = 10.298\,J[/tex]

Work done by gravitational force on bowling ball when mattress is compressed is 10.298 joules.

(d) The work done by the mattress on the ball equals the sum of kinetic energy just before mattress compression and the work done by the gravitational force when mattress is compressed:

[tex]\Delta W' = K_{2}+\Delta W[/tex]

([tex]K_{2} = 102.974\,J[/tex], [tex]\Delta W = 10.298\,W[/tex])

[tex]\Delta W' = 113.272\,J[/tex]

The work done by the mattress on the bowling ball is 113.272 joules.

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4
5
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10
Dressing appropriately for exercise includes
A. wearing the same clothes for all exercises
B. choosing dark colored clothing when exercising at night
C. wearing sunscreen when exercising outside
D. making sure you wear the best brand-name clothes
Please select the best answer from the choices provided.
A
B.
C.
D.

The answer is C.

Answers

Answer:

the answer is C

Explanation:

you said its c

Which factor affects the amount of runoff that occurs in an area?

land use
the water table
the saturation zone
amount of nutrients in soil

Answers

Answer:

A. land use

Explanation:

Answer:

a

Explanation:

if you are driving 110 km/h along a straight road and you look to the side for 2.0 s , how far do you travel during this inattentive period ? explain. ​

Answers

Explanation:

hope this helps, have a good one :D

Answer:

60.12m

Explanation:

Distance = Velocity x Time

To use this formula we must first convert 110km/h to m/s, which we can do by dividing the value by 3.6:

110/3.6 = 30.56m/s (2dp)

Velocity = 30.56m/s

Time = 2s

Distance = 30.56x2

Distance = 61.12m

You travel 60.12m during this inattentive period.

Hope this helped!

3. A wye-connected load has a voltage of 480 V applied to it. What is the voltage dropped across each phase?

Answers

Answer:

[tex]E_s = 277.13V[/tex]

Explanation:

Given

[tex]Load\ Voltage = 480V[/tex]

Required

Determine the voltage dropped in each stage.

The relation between the load voltage and the voltage dropped in each stage is

[tex]E_l = E_s * \sqrt3[/tex]

Where

[tex]E_l = 480[/tex]

So, we have:

[tex]480 = E_s * \sqrt3[/tex]

Solve for [tex]E_s[/tex]

[tex]E_s = \frac{480}{\sqrt3}[/tex]

[tex]E_s = \frac{480}{1.73205080757}[/tex]

[tex]E_s = 277.128129211[/tex]

[tex]E_s = 277.13V[/tex]

Hence;

The voltage dropped at each phase is approximately 277.13V

The bending of rocks due to the compression of tectonic plates is called
Ofaulting
O folding
subduction
plyometrics

Answers

Answer:

Folding

Explanation:

On what part of the eye are rods and cones found?

Answers

They are located in the retina.

Answer:

retina.

Explanation:

Which of these statements best describes the impact of ocean thermal power and current power on the environment?

A. Current power may decrease the fish population.
B. Current power may decrease the gravitational pull of the moon.
C. Ocean thermal power may increase the fish population.
C. Ocean thermal power may increase the gravitational pull of the moon.

Answers

Answer:

A. Current power may decrease the fish population.

Explanation:

The statement that best describes the impact of ocean thermal power and current power on the environment is that current power may decrease the fish population.

The environment is made up of living and non-living components that co-exist and interact with one another.

Harnessing current power from ocean movement will seriously affect the fish population. Most fishes are not sedentary. They move and glide through the water. When current power causes a change in the environment of the fish. This will definitely affect the normal condition prevalent in the body of water.

Answer:

A

Explanation:

I took the test good luck :D

g A child bounces a 50 g super ball on the sidewalk. The velocity change of the super bowl is from 27 m/s downward to 17 m/s upward. If the contact time with the sidewalk is 1 800 s, what is the magnitude of the average force exerted on the superball by the sidewalk

Answers

Answer:

The average force exerted on the superball by the sidewalk is 0.00122 N.

Explanation:

Given;

mass of the super ball, m = 50 g = 0.05 kg

initial velocity of the super bowl, u = -27 m/s (assuming downward motion to be negative)

final velocity of the super bowl, u = 17 m/s (assuming upward motion to be positive)

time of motion, t = 1800 s

The average force exerted on the superball by the sidewalk is given by;

[tex]F = ma\\\\F = \frac{m(v-u)}{t} \\\\F = \frac{0.05(17-(-27))}{1800}\\\\ F = \frac{0.05(44)}{1800}\\\\F = 0.00122 \ N[/tex]

Therefore, the average force exerted on the superball by the sidewalk is 0.00122 N.

What must be true if a wave's wavelength is short?
A
Its frequency is low.
B
It does not carry energy.
с
Its frequency is high.
D
The waves are visible.

Answers

C because it’s frequently is high

Answer:

im sure its A

Explanation:

A certain superconducting magnet in the form of a solenoid of length 0.300 m can generate a magnetic field of 8.90 T in its core when its coils carry a current of 95 A. Find the number of turns in the solenoid.

Answers

Answer:

The number of turns in the solenoid is 22366.

Explanation:

The number of turns in the solenoid can be found using the following equation:

[tex] B = \mu_{0} I\frac{N}{L} [/tex]

Where:

B: is the magnetic field = 8.90 T

L: is the solenoid's length = 0.300 m

N: is the number of turns =?

I: is the current = 95 A

μ₀: is the magnetic constant = 4π×10⁻⁷ H/m

By solving equation (1) for N we have:

[tex] N = \frac{BL}{\mu_{0} I} = \frac{8.90 T*0.300 m}{4\pi \cdot 10^{-7} H/m*95 A} = 22366 turns [/tex]

Therefore, the number of turns in the solenoid is 22366.

I hope it helps you!

A bird is flying in a room with a velocity field of . Calculate the temperature change that the bird feels after 9 seconds of flight, as it flies through x

Answers

Complete Question

The complete question is shown on the first uploaded image

Answer:

The temperature change is [tex]\frac{dT}{dt} = 1.016 ^oC/m[/tex]

Explanation:

From the question we are told that

   The velocity field with which the bird is flying is  [tex]\vec V =  (u, v, w)= 0.6x + 0.2t - 1.4 \ m/s[/tex]

   The temperature of the room is  [tex]T(x, y, u) =  400 -0.4y -0.6z-0.2(5 - x)^2 \  ^o C[/tex]

    The time considered is  t =  10 \  seconds

    The  distance that the bird flew is  x  =  1 m

 Given that the bird is inside the room then the temperature of the room is equal to the temperature of the bird

Generally the change in the bird temperature with time is mathematically represented as

      [tex]\frac{dT}{dt} = -0.4 \frac{dy}{dt} -0.6\frac{dz}{dt} -0.2[2 *  (5-x)] [-\frac{dx}{dt} ][/tex]

Here the negative sign in [tex]\frac{dx}{dt}[/tex] is because of the negative sign that is attached to x in the equation

 So

       [tex]\frac{dT}{dt} = -0.4v_y  -0.6v_z -0.2[2 *  (5-x)][ -v_x][/tex]

From the given equation of velocity field

    [tex]v_x  =  0.6x[/tex]

    [tex]v_y  =  0.2t[/tex]

     [tex]v_z  =  -1.4 [/tex]

So

[tex]\frac{dT}{dt} = -0.4[0.2t]  -0.6[-1.4] -0.2[2 *  (5-x)][ -[0.6x]][/tex]    

substituting the given values of x and t

[tex]\frac{dT}{dt} = -0.4[0.2(10)]  -0.6[-1.4] -0.2[2 *  (5-1)][ -[0.61]][/tex]      

[tex]\frac{dT}{dt} = -0.8 +0.84 + 0.976[/tex]  

[tex]\frac{dT}{dt} = 1.016 ^oC/m[/tex]  

A force of 15 newtons is used to push a box along the floor a distance of 3 meters. How much work was done?

Answers

Answer:

The answer is 45 J

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question

distance = 3 meters

force = 15 newtons

We have

workdone = 15 × 3

We have the final answer as

45 J

Hope this helps you

How is the voltage V across the resistor related to the current I and the resistance R of the resistor? (Use I for current and R for resistance.)

Answers

Answer:

This relationship is explained by Ohm's law

Explanation:

Ohm's law states that the current flowing through a circuit or a resistor is directly proportional to the voltage across the resistor and inversely proportional to the resistance. Where current is i, voltage is v and resistance is r, Ohm's law can be represented mathematically as

V= IR

Calculate the ratio of the mechanical energy at B and mechanical energy at a (eb,ea) and (ec,ea). What do these ratios tell you about the conservation of energy?

A) is the mechanical conserved between a and b? explain

B) is the mechanical energy conserved between b and c ?explain

Answers

Answer:

Yes at A the mechanical energy is conserved.

Yes at B the part of mechanical energy is conserved potential energy and  kinetic energy and some is lost as frictional force.

Explanation:

Ratio = Eb/ Ea=  1058.3 J/2940 J= 0.3599

Ratio = Ec/ Eb= 0J/ 1058.3 J= 0

At point A the skater is at rest  or it is the starting point and the whole energy is due to the position of the  skater i.e= mgh = 50 *9.8*6=  2940 J

Since there's no movement there is no Kinetic energy = 0 J

Yes at A the mechanical energy is conserved.

At point B the skater has traveled for some of the distance . It has potential energy and  kinetic energy.

Yes at B the part of mechanical energy is conserved as potential energy and  kinetic energy.

The total Mechanical energy = 1058.3 J

At point B Total Mechanical energy = PE+ KE

1058.3J = 980 J + 78.3 J

1058.3 J = mgh + 1/2mv²

      = 50*2*9.8 + 1/2 *50*(8.85)²

       = 980 J + 78.3 J

As the total energy of the system must remain the same some of the mechanical energy is lost as frictional force at point B .

2940 J-1058.3 J= 1881.7

At Point C the skater has arrived at the end point and the height , speed, PE, KE and ME  all are zero.

(a) The ratio of the mechanical energy at B and mechanical energy at A is 0.36.

(b) The ratio of the mechanical energy at C and mechanical energy at A is 0.

(c) mechanical energy is conserved between a and b.

(d) mechanical energy is not conserved between b and c.

The given parameters;

mechanical energy at A, [tex]E_a = 2,940 \ J[/tex]mechanical energy at B, [tex]E_b =1,058.3 \ J[/tex]mechanical energy at C, [tex]E_c = 0[/tex]

The ratio of the mechanical energy at B and mechanical energy at A;

[tex]ratio = \frac{E_b}{E_a} = \frac{1058.3}{2940} = 0.36[/tex]

The ratio of the mechanical energy at C and mechanical energy at A;

[tex]ratio = \frac{E_c}{E_a} = \frac{0}{2940} = 0[/tex]

The change mechanical energy between A and B from the given position;

[tex]\Delta E = mg(h_b - h_a) - \frac{1}{2}m(v_b^2 - v_a^2)\\\\ \Delta E = 50\times 9.8(2-6) \ - \ \frac{1}{2} \times 50(8.85^2 - 0)\\\\\Delta E =- 1960 + 1960\\\\\Delta E = 0 \ J[/tex]

Thus, we can conclude that mechanical energy is conserved between a and b.

The change mechanical energy between A and B from the given position;

[tex]\Delta E = mg(h_c - h_b) - \frac{1}{2}m(v_c^2 - v_b^2)\\\\ \Delta E = 50\times 9.8(0-2) \ - \ \frac{1}{2} \times 50(0^2 - 8.85^2)\\\\\Delta E = -980 + 1960 \\\\\Delta E = 980 \ J[/tex]

Thus, we can conclude that mechanical energy is not conserved between b and c.

Learn more here:https://brainly.com/question/19969393

12, The error in measurement may occur due to
S
a, inexperience of a person
b, The faulty apparatus
c, Inappropriate method
d, Due to all reasons in a, b and c​

Answers

Answer:

d, Due to all reasons in a, b and c​

Explanation:

All of the reasons from the choices can lead to errors in measurement. An error is an uncertainty introduced to a scientific process.

Errors can be due to the following reasons;

It can be to a poor technician; an inexperienced scientist can introduce serious into their set up. Even when recording their observation, a dearth of experience can lead to error. Faulty apparatus can also lead to errors in measurement. An instrument that is poorly calibrated can also result in measurement errors. An inappropriate scientific method or measuring guidelines can also lead to errors in measurements.

Luck walked to a store that is 250m away and it took him 50secs while Layne walked to the mall that is 1000m away and took her 200s to do. What do they have in common?

A. Average speed
B. Acceleration
C. Displacement
D.mass

Answers

Answer:

Average speed

Explanation:

250/50=5

1000/20=5

If Mary runs 5 miles in 50 minutes, what is her speed with the correct
label?

Answers

1 mile = 10 mins or 1 mile takes 10 mins to run

Pls help pls pls pls pls

Answers

1.cool down
2.activity log
3.specific warm up
4.activities of daily living
5.planned exercise
6.general warm up

g An angry rhino with a mass of 2700 kg charges directly toward you with a speed of 3.70 m/s. Before you start running, as a distraction, you throw a 0.180 kg rubber ball directly at the rhino with a speed of 9.05 m/s. Determine the speed of the ball (in m/s) after it bounces back elastically toward you.

Answers

Answer:

9.05m/s

Explanation:

given data

m1= 2700kg

v1=3.7m/s

m2=0.18kg

v2=9.05m/s

v3=?

We know that the velocity of the rhino will remains unchanged after impact as the mass of the rubber ball is negligible

m1v1+m2v2=m1v1+m2v3

2700*3.7+0.18*9.05=2700*3.7+0.18*v3

9990+1.629=9990+0.18v3

9991.629-9990=0.18v3

1.629=0.18v3

v3=1.629/0.18

v3=9.05m/s

am I right? be honest

Answers

Answer:

I chose c because it is the greater slope at point c

A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 555 N. As the drawing shows, she is closer to the left cliff than to the right cliff, with the result that the tensions in the left and right sides of the rope are not the same. Find the tension in the rope to the left of the mountain climber.

Answers

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The tension in the rope on the left of the mountain climber is [tex] T_a = 1106 \ N [/tex]

Explanation:

From the question we are told that

The weight of the mountain climber is m = 555 N

Generally from the diagram , the total amount of force acting on the rope along the vertical axis at equilibrium is mathematically represented as  

       [tex]T_a*  cos 65 -555 + T_b * cos(85) =  0[/tex]

Here  [tex]T_a, T_b[/tex] are the tension of the rope on the left and on the right hand side

 So

    [tex]0.423T_a   + 0.0871T_b  =  555[/tex]

=>   [tex] 0.0871T_b  =  555 - 0.423T_a[/tex]

=>   [tex] T_b  =  \frac{555 - 0.423T_a}{0.0871}[/tex]

Generally from the diagram , the total amount of force acting on the rope along the horizontal  axis at equilibrium is mathematically represented as

      [tex]T_a*  sin 65 - T_b * sin(85) =  0[/tex]

=>     [tex] 0.9063T_a - 0.9962T_b =  0[/tex]

=>     [tex] 0.9063T_a =   0.9962T_b [/tex]

=>     [tex] 0.9063T_a =   0.9962[\frac{555 - 0.423T_a}{0.0871}] [/tex]

=>     [tex] 0.9063T_a =   [\frac{552.891 - 0.421T_a}{0.0871}] [/tex]

=>    [tex] 0.0789T_a =   [552.891 - 0.421T_a[/tex]

=>    [tex] 0.4999T_a =   552.891 [/tex]

=>      [tex] T_a = 1106 \ N [/tex]

A
6. All other changeable factors that must
be kept the same to ensure a fair test
(what you keep the same).​

Answers

Answer:

a constant variable?

Explanation:

A constant variable is any aspect of an experiment that a researcher intentionally keeps unchanged throughout an experiment.

Experiments are always testing for measurable change, which is the dependent variable. You can also think of a dependent variable as the result obtained from an experiment. It is dependent on the change that occurs

Why wouldn't carbon dating work to determine the age of the earth?
A) Carbon dating works best on other planets
B) The half life of carbon is too short
C) The age of the earth cannot be determined
D) The half life of carbon is too long.

Answers

Answer:

The half-life of carbon is too short.

Explanation:

The answer is B.

Marisa’s car accelerates at an average rate of 2.6m/s^2. Calculate how long it takes her car to accelerate from 24.6m/s to 26.8m/s? Show your work.

Answers

given info is... Acceleration(a)=2.6m/s^2

                       final velocity(v)=26.8m/s

                       initial velocity(u)=24.6m/s

need to find.... time(t)=?

[tex]a=\frac{v-u}{t} \\2.6=\frac{26.8-24.6}{t} \\\\[/tex]

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

[tex]t=\frac{26.8-24.6}{2.6}[/tex]

[tex]t=0.846s[/tex]

Explanation:

It takes 0.84 second her car to accelerate from 24.6m/s to 26.8m/s.

What is acceleration?

Acceleration is the rate at which speed and direction of velocity vary over time. A point or object going straight ahead is accelerated when it accelerates or decelerates. Even if the speed is constant, motion on a circle accelerates because the direction is always shifting.

Given parameters:

Initial speed of the car: u = 24.6 m/s

Final speed of the car: v = 26.8 m/s.

Acceleration of the car: a = 2.6 m/s²

Time interval: t = ?

change is speed = final speed - initial speed

= 26.8 m/s - 24.6 m/s

= 2.2 m/s

From the definition of acceleration,

acceleration = change is speed / time interval

So, time interval  =  change is speed / acceleration

= 2.2 m/s/2.6 m/s²

= 0.84 second.

Hence, it takes 0.84 second her car to accelerate from 24.6m/s to 26.8m/s.

Learn more about acceleration here:

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An object moving 20 m/s
experiences an acceleration of 4 m/s' for 8
seconds. How far did it move in that time?
Variables:
Equation and Solve:

Answers

Answer:

We are given:

initial velocity (u) = 20m/s

acceleration (a) = 4 m/s²

time (t) = 8 seconds

displacement (s) = s m

Solving for Displacement:

From the seconds equation of motion:

s = ut + 1/2 * at²

replacing the variables

s = 20(8) + 1/2 * (4)*(8)*(8)

s = 160 + 128

s = 288 m

Determine the electrical force of attraction between two balloons
that are charged with the opposite type of charge but the same
quantity of charge. The charge on the balloons is 6.0 x 10-7 C and they
are separated by a distance of 0.50 m.

Answers

Answer:

F=1.3x10^-2N

Explanation:

Fe= k(6x10^-7C)^2/(0.5)^2

Electrical force of attraction between the balloons is F=1.3x10^-2N

The electric force of attraction between two balloons should be F=1.3x10^-2N.

Calculation of the electric force;

Since The charge on the balloons is 6.0 x 10-7 C and they are separated by a distance of 0.50 m.

So, here the electric force is

Fe= k(6x10^-7C)^2/(0.5)^2

F=1.3x10^-2N

hence, The electric force of attraction between two balloons should be F=1.3x10^-2N.

Learn more about force here: https://brainly.com/question/19848845

Find the magnitude of the net force exerted on the loop by the magnetic field created by the long wire. Answer in units of N.

Answers

The question is incomplete. Here is the complete question.

A current in the long, straight wire, which lies in the plane of rectangular loop, that also carries a current, as shown in the figure.

Find the magnitude of the net force exerted on the loop by the magnetic field created by the long wire. Answer in units of N.

Answer: Net Force = [tex]50.215.10^{-7}[/tex]N

Explanation: Force and Magnetic field are related through the following formula:

F = I.L.B.sinθ

Magnetic field (B) in a straight long wire is given by

[tex]B=\frac{\mu_{0}.I}{2.\pi.r}[/tex]

in which

[tex]\mu_{0}[/tex] is permeability of free space and is [tex]4.\pi.10^{-7}[/tex]T.m/A

I is current in the wire;

r is distance to the wire;

Examining the square loop and using the right hand rule, the top, which we will name it F₂, and the bottom, named F₄, have angle θ = 0, giving sin(0) = 0 and therefore, F₁ = F₃ = 0.

So, for the net force, the relevant forces will be on the sides parallel to the wire.

For the other forces, angle is 90°, sin(90°) = 1, then:

F = I.L.B

Replacing magnetic field:

F = [tex]\frac{\mu_{0}.I_{w}.L.I_{l}}{2.\pi.r}[/tex]

Note: The side closest to the wire is F₁, while the farthest is F₃.

Note2: As the constant unit is in meters, distance and length of side of the square loop are also in meters.

Calculating forces:

F₁ = [tex]\frac{4*\pi*10^{-7}*4.3*0.19*14}{2.\pi.0.082}[/tex]

F₁ = [tex]278.975*10^{-7}[/tex]N

Current in F₃ is flowing thoruhg the negative side of the referential, so:

F₃ = [tex]-\frac{4*\pi*10^{-7}*4.3*0.19*14}{2.\pi.0.1}[/tex]

F₃ = [tex]-228.76*10^{-7}[/tex]N

Net force is total force:

[tex]F_{net} = F_{1}+F_{3}[/tex]

[tex]F_{net}=(278.975-228.76).10^{-7}[/tex]

[tex]F_{net}=50.22.10^{-7}[/tex]

The total force acting on the square loop is [tex]F_{net}=50.22.10^{-7}[/tex]N.

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