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Westkost [7]
3 years ago
15

Which velocity-time graph matches the position-time graph?

Physics
2 answers:
skelet666 [1.2K]3 years ago
7 0

The answer is Graph C. To explain, this is because as we look at the position vs time graph, we see that after the first second, it was 30 meters from the start. That would mean that it took 1 second to get to 30 meters. That is shown in Graph c

Leona [35]3 years ago
6 0

Explanation:

In this question, a position time graph is shown. We need to find the corresponding velocity time graph.

In position time graph, till 1 second the object is in uniform motion and reaches a distance of 30 meters. Then the object is at till 3 seconds. After 3 seconds, the object again reaches a distance of 40 meters uniformly.

Corresponding velocity-time graph is shown in figure (3). The object is moving initially with a speed of 30 m/s. After 1 second, it comes to rest till 3 seconds then it again moves. So, the correct option is (c). Hence, this is the required solution.

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3 0
2 years ago
A block of wood mass 0.60kg is balanced on top of a vertical port 2.0m high. A 10gm bullet is fired horizontally into the block
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Answer:

Mass of bullet is m=0.01kg

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Coefficient=0.25,distance=20m

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7 0
3 years ago
If the moon were twice as far from years as it is now the following would be true
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7 0
3 years ago
g An electron enters a region of space containing a uniform 1.63 × 10 − 5 T magnetic field. Its speed is 121 m/s and it enters p
kolbaska11 [484]

Answer:

i. The radius 'r' of the electron's path is 4.23 × 10^{-5} m.

ii. The frequency 'f' of the motion is 455.44 KHz.

Explanation:

The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.

                 r = \frac{mv}{qB}

Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.

From the question, B = 1.63 × 10^{-5}T, v = 121 m/s, Θ = 90^{0} (since it enters perpendicularly to the field), q = e  = 1.6 × 10^{-19}C and m = 9.11 × 10^{-31}Kg.

Thus,

         r = \frac{mv}{qB} ÷ sinΘ

But,  sinΘ =  sin 90^{0} = 1.

So that;

          r = \frac{mv}{qB}

            = (9.11 × 10^{-31} × 121) ÷ (1.6 × 10^{-19}  × 1.63 × 10^{-5})

            = 1.10231 × 10^{-28}   ÷ 2.608 × 10^{-24}

            = 4.2266 × 10^{-5}

            = 4.23 × 10^{-5} m

The radius 'r' of the electron's path is 4.23 × 10^{-5} m.

B. The frequency 'f' of the motion is called cyclotron frequency;

           f = \frac{qB}{2\pi m}

             =  (1.6 × 10^{-19}  × 1.63 × 10^{-5}) ÷ (2 ×\frac{22}{7} × 9.11 × 10^{-31})

             =  2.608 × 10^{-24} ÷  5.7263 × 10^{-30}

             = 455442.4323

          f  = 455.44 KHz

The frequency 'f' of the motion is 455.44 KHz.

3 0
3 years ago
Read 2 more answers
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