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27 Nov 2019

You move from location i at < 5, 3, 5 > m to location f at< 7, 5, 11 > m. All along this path there is a nearly uniformelectric field = < 1100, 170, -520 > N/C. Calculate ?V = Vf -Vi, including sign and units.


_______ V

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Hubert Koch
Hubert KochLv2
18 May 2019

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Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

Related questions

<p>A Simple Atwood Machine consists of two masses m1 and m1 that are connected by a string wound over a pulley, as seen in the figure below. m2 is larger than m1.<br />On a piece of paper, draw two free body diagrams; one for each of the masses, showing all forces acting on each mass. Then answer the following questions.<br /><br /><br />(a) Based on your free body diagram, identify the forces acting on the mass m1. Select all that apply.<br />(m1+m2)a<br />m1g<br />FN<br />m1a<br />(m1+m2)g<br />m2a<br />f<br />m2g<br />T<br /><br /><br /><br />(b) Using the direction rosette indicate the direction for each of the forces acting on m1 that you identified in part (a). If any of the forces listed below is not applicable, enter NA.<br />m1a= <br />m2a= <br />m1g= <br />m2g= <br />(m1+m2)a= <br />(m1+m2)g= <br />f= <br />T= <br />FN= <br /><br />(c) Based on your free body diagram, identify the forces acting on the mass m2. Select all that apply.<br />T<br />(m1+m2)g<br />m2g<br />m1g<br />(m1+m2)a<br />m1a<br />m2a<br />f<br />FN<br /><br /><br /><br />(d) Using the direction rosette indicate the direction for each of the forces acting on m2 that you identified in part (c). If any of the forces listed below is not applicable, enter NA.<br />m1a= <br />m2a= <br />m1g= <br />m2g= <br />(m1+m2)a= <br />(m1+m2)g= <br />f= <br />T= <br />FN= <br /><br />(e) Write Newton's Second Law equation (in the form Fnet = ma) for m1 in the y-direction.<br />Use the first answer box for the left hand side (LHS) of the equation and the second answer box for the right hand side (RHS) of the equation.<br />It is important to use the correct signs for each of the forces. Motion in the upward direction is positive and motion in the downward direction is negative. (Use m1, g, a, and T as appropriate.)<br />LHS = <br />RHS = <br /><br />(f) Write Newton's Second Law equation (in the form Fnet = ma) for m2 in the y-direction.<br />Use the first answer box for the left hand side (LHS) of the equation and the second answer box for the right hand side (RHS) of the equation.<br />It is important to use the correct signs for each of the forces. Motion in the upward direction is positive and motion in the downward direction is negative.(Use m2, g, a, and T as appropriate.)<br />LHS = <br />RHS = <br /><br />(g) Use your answers from parts (e) and (f) to get an expression for the magnitude of the tension T. (Use m1, m2, and g as appropriate.)<br />magnitude of T = <br /><br />(h) Use your answers from parts (e), (f), and (g) to find the magnitude of a, the acceleration of the system. (Use m1, m2, and g as appropriate.)<br />magnitude of a = <br /><br />(i) In what direction will each mass accelerate.<br />For mass m1: <br />For mass m2: <br /><br />(j) Suppose m1 = 7 kg and m2 = 11 kg. What is the tension in the string? <br />T = <br /><br />(k) Suppose m1 = 7 kg and m2 = 11 kg. What is the value of the acceleration?<br />a = <br /><br />(l) Suppose that m2 starts from rest at a height of 5 m. Use the Kinematic Equations to determine how long it takes for m2 to hit the ground.</p>

Assume the Earth is spherical (radius R) and has uniform
density (mass M). A tunnel is now dug along any secant. A
box of mass m (very well wrapped!) is held just above one
opening and dropped from rest. Assume:
· You can ignore friction and all effects due to the various
motions of the Earth
· The tunnel is so thin it has no measurable effect on the
gravitational field of the Earth.
a) Use Gauss’ law to find an expression for F(r), themagnitude
of the gravitational force felt by mass at distance r fromthe
center of the Earth, where r < R. Now, find FT(r), the(tangential) component of this force along
the direction of the tunnel.
b) Show that m will move through the Earth in simple harmonicmotion. Find T, the period of that
motion. Use: M Earth = 5.98x1024 kg, R Earth = 6.37x106 m, and G =6.67x10-11 N-m2/kg2 to get a
numerical answer.

A boy in a wheelchair (total mass 50.5 kg) has a speed of 1.40 m/s at the crest of a slope 2.40 m high and 12.4 m long. At the bottom of the slope, his speed is 6.50 m/s. Assume air resistance and rolling resistance can be modeled as a constant friction force of 41.0 N. Find the work he did in pushing forward on his wheels during the downhill ride.

 

 
 
 

2 Answers

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  • electron1
    Lv 7
    4 years ago

    There are two forces that cause the wheelchair to accelerate. These two forces are the component of the wheelchair’s weight that is parallel to the hill and the force the boy exerts. The net force on the wheelchair is equal to the sum of these two forces minus the friction force. Let’s use the following equation to determine the acceleration.

     

    vf2 = vi^2 + 2 * a * d

    6.5^2 = 1.4^2 + 2 * a * 12.4

    24.8 * a = 40.29

    a = 40.29 ÷ 24.8

    This is approximately 1.62 m/s^2.

     

    Net force = 50.5 * 40.29 ÷ 24.8 = 2034.645 ÷ 24.8

    This is approximately 82 N.

     

    Force parallel = 50.5 * 9.8 * 2.40 ÷ 12.4 = 1187.76 ÷ 12.4

     

    This is approximately 95.8 N. This force and the force that the boy exerts are the forces that cause the wheelchair to accelerate.

     

    Net force = F + (1187.76 ÷ 12.4) – 41

    Set the two net forces equal to each other.

     

    F + (1187.76 ÷ 12.4) – 41 = 2034.645 ÷ 24.8

    F = 41 + (2034.645 ÷ 24.8) – (1187.76 ÷ 12.4)

    F is approximately 27.3 N

     

    Work = [41 + (2034.645 ÷ 24.8) – (1187.76 ÷ 12.4)] * 12.4 = 337.9625 J

     
  • JupiterSky
    Lv 6
    4 years ago

    His change in kinetic energy is due to gravitational- and his own work,

    while the friction does negative work.

     

    ∆K = ∆U +∆W + ∆Wf

    ½m(Vf²−Vi²) = mgh + ∆W + Ff*d

    ∆W = ½m(Vf²−Vi²) − mgh − Ff*d

    ∆W = ½*50.5(6.5²−1.4²) − (50.5*9.8*2.4) − (−41*12.4)

    ∆W = 508 J

     
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Next Question
---
asked in Science & MathematicsPhysics · 8 years ago

What is the spring constant? please help me!!!!?

A block of mass 487 g is pushed against the

spring (located on the left-hand side of the

track) and compresses the spring a distance

4.7 cm from its equilibrium position (as shown

in the figure below). The block starts from

rest, is accelerated by the compressed spring,

and slides across a frictionless horizontal track

(as shown in the figure below). It leaves the

track horizontally, flies through the air, and

subsequently strikes the ground.

acceleration = 9.81

 

A) What is the spring constant?

B)What is the speed v of the block when it leaves

the track?

C) What is the total speed of the block when it

hits the ground?

...Show more
Update:

Height Above ground = 1.9 m

Lands = 4.59m away

 
What is the spring constant? please help me!!!!?

3 Answers

Relevance
  • Jánošík
    Lv 7
    8 years ago
    Favorite Answer

    I'm sure there is some crucial information in "the figure below" !!!

     

    Edit:

    OK... now we got something to work with ... ;-)

     

    Time needed to fall a vertical distance of 1.9 m is:

    t = √( 2h / g )

    t = √[ 2×(1.9 m) / (9.81 m/s²) ]

    t = 0.6224 s

     

    In that time, the block must travel a horizontal distance of 4.59 m, so the horizontal velocity is:

    r = v×t

    (4.59 m) = v×(0.6224 s)

    v = 7.37 m/s < - - - - - - - - - - - - - - - - - answer B

     

    That is also the speed of the block as it leaves the spring.

    Its kinetic energy at that time is:

    KE = m×v²/2

    KE = (0.487 kg)×(7.3749 m/s)²/2

    KE = 13.244 J

     

    That is also the elastic potential energy stored in the spring at maximum compression. So the spring constant is:

    KE = PEe = k×d²/2

    (13.244 J) = k×(0.047 m)²/2

    k = 12 N/m < - - - - - - - - - - - - - - - - - - answer A

     

    The gravitational potential energy of the block on the table is:

    PEg = m×g×h

    PEg = (0.487 kg)×(9.81 m/s²)×(1.9)

    PEg = 9.077 J

     

    The total energy of the block on the table is:

    TE = PEg + PEe

    TE = (9.077 J) + (13.244 J)

    TE = 22.32 J

     

    When hitting the ground, all that energy is converted to kinetic energy, so

    TE = KE = m×v²/2

    (22.32 J) = (0.487 kg)×v²/2

    v = 9.57 m/s < - - - - - - - - - - - - - - - - - - answer C

    ...Show more
     
  • Technobuff
    Lv 7
    8 years ago

    There must be a height above ground and a distance from the end of the track on the "as shown in the figure below".??

     
  • STALKER
    Lv 4
    8 years ago

    use f=kl

     

    where k is the spring constant and l is extension from the natural length.

     

    k=f/l

     

    since you have mass in grams. use w=mg

     

    w=0.487 x 9.8 = 4.77N

     

    Now use the equation,

     

    k=4.77/4.7

     

    k=1.012

     

    Note: my answer could be wrong since I do not know the springs original length from the diagram shown. But, the method of working out spring constant is same.

    ...Show more
     
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Next Question
ひいらぎ
Lv 5
asked in Science & MathematicsPhysics · 1 decade ago

Physics Momentum, Rolling?

Pat builds a track for his model car out of wood. The track is 10.00 cm wide and 3.00 m long, and is solid. The runway is cut such that it forms a parabola by the equation y=(x - 3)^2/9. Locate the horizontal position of the center of gravity of this track.

 

A washing machine load in the "spin" cycle, (much like a centrifuge used to separate fluids), can be modeled as a cylinder which has zero density toward the center, and starting from some inner radius where the clothes (or fluids) begin, increases in density as you move radially outward. The washing machine has density of clothes (in SI units) of p=14r+5, a height of 0.51 m, and only has clothes present from inner radius 0.2m out to 0.7m from the center.

-What is the moment of Inertia of the wash load?

-What is the power of motor that drives the clothes from rest up to 4 rev/sec, in 14 seconds if the basin has moment of Inertia of 4 kg*.m2

...Show more
 
Physics Momentum, Rolling?

1 Answer

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  •  
    Anonymous
    1 decade ago
    Favorite Answer
     

    I can't visualize the first problem.

     

    The second: Moment of inertia is I = m r^2. In a shell of thickness dr there is a mass of p * 2 * pi * r * h * dr. The moment of inertia of the shell is 2 pi h p r^3 dr. Now integrate this from r = 0.2m to r = 0.7m.

     

    Kinetic energy is 1/2 m v^2. In rotations the moment of inertia is like the mass and the angular velocity is like the velocity, so rotational energy is 1/2 I omega^2. Omega is in rad/s. The total inertia of the system is the sum of the inertia of the clothes and the basin. Divide the energy by the time needed to get the power.

    ...Show more
     
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Next Question
AZ
asked in Science & MathematicsPhysics · 1 decade ago

Physics:You and your friends push a 75-kg...Help!?

You and your friends push a 75-kg greased pig up an aluminum slide at the county fair, starting from the low end of the slide. The coefficient of kinetic friction between the pig and the slide is 0.070.

 

(a) All of you pushing together (parallel to the incline) manage to accelerate the pig from rest at the constant rate of 4.9 m/s2 over a distance of 1.8 m, at which point you release the pig. The pig continues up the slide, reaching a maximum vertical height above its release point of 50 cm. What is the angle of inclination of the slide? __degree

 

(b) At the maximum height the pig turns around and begins to slip down the slide, how fast is it moving when it arrives at the low end of the slide? __m/s

...Show more
 
Physics:You and your friends push a 75-kg...Help!?

1 Answer

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  •  
    Anonymous
    1 decade ago
    Favorite Answer
     

    ♣the distance the pig was pushed up is

    s=0.5*a*t^2, where s=1.8m, a=4.9m/s^2, t is time of push-up;

    speed of the pig gained after s=1.8m is v=a*t; thus v^2=2as;

    ♠ kinetic energy of the pig at the moment it’s released after push-up is

    E=0.5*m*v^2, which is enough for the pig to move by itself L distance more along the ramp, where L=h/sin(b), h=50cm=0.5m, b is angle in question;

    ♦pig’s potential energy at vertex (when stopped) is E1=mgh;

    the energy E2=E-E1 is gone on friction, that is E2=f*L is work of friction, where

    f=μ*w1, w1=mg*cos(b) is component of pig’s weight normal to the aluminum ramp; or;

    f= μ*mg*cos(b);

    ♦Thus E2=E-E1; or; f*L = 0.5*m*v^2 –mgh; or;

    (μ*mg*cos(b)) * (h/sin(b)) =0.5*m* 2as –mgh; or;

    μ*gh*cos(b)/sin(b) =as –gh, hence cot(b) =(as –gh)/(μ*gh), hence

    (a); b=atan(μ*gh/(as-gh))= atan(0.07*9.8*0.5/(4.9*1.8 -9.8*0.5)) =5°;

     

    ♥ pig’s potential energy above the low end of the ramp is

    W= mg*(s*sin(b) +h); pig’s ramp position is p=s +L = s + h/sin(b);

    Pig’s final kinetic energy is W1=W-W2, where W2=f*p is work of friction;

    and W1=0.5*m*v^2; thus

    0.5*m*v^2 = mg*(s*sin(b) +h) – μ*mg*cos(b)*(s + h/sin(b));

    v^2 =2g*(s*sin(b) +h)*(1 –μ*cot(b)) =2.575;

    (b); v =1.605 m/s;

    ...Show more
     
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