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NemiM [27]
3 years ago
12

The vector product of vectors A⃗ and B⃗ has magnitude 12.0 m2 and is in the +z-direction.Vector A⃗ has magnitude 4.0 m and is in

the −x-direction. Vector B⃗ has no x-component.Part A: What is the magnitude of vector B⃗ ? (I solved this question the answer is 3 and it's correct)Part B: What is the direction angle θ of vector B⃗ measured from the +y-direction to the +z-direction? (This is the part that I didn't get it correct)
Physics
1 answer:
g100num [7]3 years ago
8 0

Answer:

θ=180°

Explanation:

The problem says that the vector product of A and B is in the +z-direction, and that the vector A is in the -x-direction. Since vector B has no x-component, and is perpendicular to the z-axis (as A and B are both perpendicular to their vector product), vector B has to be in the y-axis.

Using the right hand rule for vector product, we can test the two possible cases:

  • If vector B is in the +y-axis, the product AxB should be in the -z-axis. Since it is in the +z-axis, this is not correct.

  • If vector B is in the -y-axis, the product AxB should be in the +z-axis. This is the correct option.

Now, the problem says that the angle θ is measured from the +y-direction to the +z-direction. This means that the -y-direction has an angle of 180° (half turn).

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T_2 = T_1\frac{P_2V_2}{P_1V_1} = 298\frac{0.77*1.8}{1.2*1} = 298*1.155 = 344.19 K = 344.19 - 273 = 71.19 C

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At an accident scene on a level road, investigators measure a car's skid mark to be 84 m long. It was a rainy day and the coeffi
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The given data is incomplete. The complete question is as follows.

At an accident scene on a level road, investigators measure a car's skid mark to be 84 m long. It was a rainy day and the coefficient of friction was estimated to be 0.36.  Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes. (why does the car's mass not matter?)

Explanation:

Let us assume that v is the final velocity and u is the initial velocity of the car. Let s be the skid marks and \mu be the friction coefficient and m be the mass of car.

Hence, the given data is as follows.

                v = 0,     s = 84 m,     \mu = 0.36

According to Newton's law of second motion the expression for acceleration is as follows.

                      F = ma

                 -\mu N = ma

                 -\mu mg = ma

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Also,    

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              (0)^{2} = u^{2} + 2(-\mu g)s

                  u^{2} = 2(\mu g)s

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Thus, we can conclude that the speed of the car when the driver slammed on (and locked) the brakes is 24.36 m/s.

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