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Semenov [28]
3 years ago
15

A car heading north collides at an intersection with a truck heading east. if they lock together and travel 28 m.s at 46 degrees

north of east just after the collision, how fast was the car initially traveling? assume that the two vehicles have the same mass.
A) 30 m/s
B) 80 m/s
C) 20 m/s
D) 40 m/s
Physics
1 answer:
Dominik [7]3 years ago
4 0

Answer:

C) 20 m/s

Explanation:

The collision of the car and truck will give a single resultant velocity due to equal masses.

We'll treat this question as a vector question.

The velocity profile will form a right angle triangle.

To calculate the velocity of the car

Let the velocity be x

Sina = x/28

Sin46= x/28

0.7193*28 = x

20 = x

X = 20 m/s

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What is the minimum frequency with which a 200-turn, flat coil of cross sectional area 300 cm2 can be rotated in a uniform 30-mT
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A long, hollow, cylindrical conductor (inner radius 3.4 mm, outer radius 7.3 mm) carries a current of 36 A distributed uniformly
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Answer:

a. B= 9.45 \times10^{-3} T

b. B= 0.820 T

c. B= 0.0584 T

Explanation:

First, look at the picture to understand the problem before to solve it.

a. d1 = 1.1 mm

Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:

To solve the equations we have to convert all units to those of the international system. (mm→m)

B=\frac{u_{0}I_{w}}{2\pi d_{1}} =\frac{52 \times4\pi \times10^{-7} }{2\pi 1.1 \times 10^{-3}} =9.45 \times10^{-3} T\\

μ0 is the constant of proportionality

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b. d2=3.6 mm

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J= \frac {-I_{c}}{\pi(c^{2}-b^{2}  ) }}

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B=\frac{u_{0}(I_{w}-JA_{s})}{2\pi d_{2} } \\A_{s}=\pi (d_{2}^{2}-b^2)=4.40\times10^{-6} m^2\\

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B=\frac{u_{0}(I_w-I_c)}{2\pi d_3 } =\frac{2.011\times10^-5}{3.441\times10^{-4}} =0.0584 T

As we see, the magnitud of the magnetic field is greater inside the conductor, because of the density of current and the material's nature.

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