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Nataliya [291]
3 years ago
14

You have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 1515 lb and was trav

eling eastward. Car B weighs 1125 lb and was traveling westward at 43.0 mph. The cars locked bumpers and slid eastward with their wheels locked for 17.5 ft before stopping. You have measured the coefficient of kinetic friction between the tires and the pavement to be 0.750. How fast (in miles per hour) was car A traveling just before the collision? (This problem uses English units because they would be used in a U.S. legal proceeding.)
Physics
1 answer:
serious [3.7K]3 years ago
6 0

what was the speed limit, and it was 40 mph


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An object in the shape of a thin ring has radius a and mass M. A uniform sphere with mass m and radius R is placed with its cent
madreJ [45]

Answer:

F = GMmx/[√(a² + x²)]³

Explanation:

The force dF on the mass element dm of the ring due to the sphere of mass, m at a distance L from the mass element is

dF = GmdM/L²

Since the ring is symmetrical, the vertical components of this force cancel out leaving the horizontal components to add.

So, the horizontal components add from two symmetrically opposite mass elements dM,

Thus, the horizontal component of the force is

dF' = dFcosФ where Ф is the angle between L and the x axis

dF' = GmdMcosФ/L²

L² = a² + x² where a = radius of ring and x = distance of axis of ring from sphere.

L = √(a² + x²)

cosФ = x/L

dF' = GmdMcosФ/L²

dF' = GmdMx/L³

dF' = GmdMx/[√(a² + x²)]³

Integrating both sides we have

∫dF' = ∫GmdMx/[√(a² + x²)]³

∫dF' = Gm∫dMx/[√(a² + x²)]³    ∫dM = M

F = GmMx/[√(a² + x²)]³  

F = GMmx/[√(a² + x²)]³

So, the force due to the sphere of mass m is

F = GMmx/[√(a² + x²)]³

3 0
3 years ago
A man does 4,475 J of work in the process of pushing his 2.50 103 kg truck from rest to a speed of v, over a distance of 26.0 m.
Tcecarenko [31]

Answer:

a) 1.89 m/s  b) 172.1 N

Explanation:

a)

  • Applying the work-energy theorem, if we can neglect the friction between truck and road, the total change in kinetic energy must be equal to the work done by the external forces.
  • This work, is just 4,475 J.
  • So we can write the following equation:

        \Delta K = \frac{1}{2} * m*v^{2} = 4,475 J

  • where m= mass of the truck = 2.5*10³ kg.
  • So, we can find the speed v, as follows:

        v =\sqrt{\frac{2*W}{m}} =\sqrt{\frac{2*4,475J}{2.5e3kg} }  = 1.89 m/s

b)

  • The work done by the man, is just the horizontal force applied, times the displacement produced by the force horizontally:

        W = F*d

  • We can solve for F, as follows:

        F = \frac{W}{d} = \frac{4,475 J}{26.0m} =  172.1 N

4 0
2 years ago
Holding onto a tow rope moving parallel to a frictionless ski slope, a 68.7 kg skier is pulled up the slope, which is at an angl
Furkat [3]

Answer:

a) F = 78.606\,N, b) F = 88.911\,N

Explanation:

a) Let consider two equations of equilibrium, the first parallel to ski slope and the second perpendicular to that. The equations are, respectively:

\Sigma F_{x'} = F - m\cdot g \cdot \sin \theta = 0\\\Sigma F_{y'} = N - m\cdot g \cdot \cos \theta = 0

The force on the skier is:

F = m \cdot g \cdot \sin \theta

F = (68.7\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \sin 6.7^{\textdegree}

F = 78.606\,N

b) The equations of equilibrium are the following:

\Sigma F_{x'} = F - m\cdot g \cdot \sin \theta = m\cdot a\\\Sigma F_{y'} = N - m\cdot g \cdot \cos \theta = 0

The force on the skier is:

F = m\cdot (a + g \cdot \sin \theta)

F = (68.7\,kg)\cdot (0.150\,\frac{m}{s^{2}}+9.807\,\frac{m}{s^{2}}\cdot \sin 6.7^{\textdegree})

F = 88.911\,N

3 0
3 years ago
John is hiking and notices a small stream of water flowing down the side of the mountain. What part of the water cycle is John o
Alexus [3.1K]

Answer: Runofff

Explanation: because it said that it is going down the side of the mountain

3 0
2 years ago
A charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a th
Mila [183]

Answer:

x = a + \frac{a\sqrt2}{\sqrt3 - \sqrt2}

Explanation:

Position of charge 3q is x = 0

position of charge -2q is x = a

so here we know that

when two charges are of opposite nature then the electric field will be zero on the line joining the two charges at the position near to smaller magnitude charge

So here the electric field will be zero if the field due to 3q is counterbalanced by field due to -2q

so here we can say

\frac{k(3q)}{(r+a)^2} + \frac{k(-2q)}{r^2} = 0

\frac{3}{(r + a)^2} = \frac{2}{r^2}

\frac{r+a}{r} = \sqrt{\frac{3}{2}}

1 + \frac{a}{r} =\sqrt{\frac{3}{2}}

\frac{a}{r} = \frac{\sqrt3 - \sqrt2}{\sqrt2}

so we will have

r = \frac{a\sqrt2}{\sqrt3 - \sqrt2}

so the x coordinate of this position is given as

x = a + \frac{a\sqrt2}{\sqrt3 - \sqrt2}

6 0
3 years ago
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