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ra1l [238]
3 years ago
14

A roller coaster is moving at 25m/s at the bottom of a hill. Three seconds later it reaches the top of the hill moving at 10m/s.

What was the acceleration of the coaster
Physics
1 answer:
marishachu [46]3 years ago
8 0

Answer:

5 m/s^{2}

Explanation:

acceleration is defined as the rate of change of velocity per unit time. Therefore, acceleration, a=\frac {\triangle v}{\triangle t}

where \triangle v is change in velocity and t is time

Substituting initial and final velocities with 25 m/s and 10 m/s then using time as given in the question of 3 s then

a=\frac {25-10}{3}=\frac {15}{3}=5 m/s^{2}

Therefore, the acceleration is 5 m/s^{2}

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Answer:

Explanation:

we can consider an element of radius r < a and thickness dr.  and Area of this element is

dA=2\pi r dr

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now, consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius r.

by applying the Ampere's law, we can write

\oint{\vec{B}_{in}\cdot\,d\vec{l}}=\mu_0\,i_{enc}&#10;

by symmetry \vec{B} will be of uniform magnitude on this loop and it's direction will be tangential to the loop.

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B_{in}\times2\pi\,l=\mu_0\int_0^r(kr)(2\pi\,r\,dr)=&#10;\\\\2\pi\,B_{in} l=2\pi\mu_0k \frac{r^3}{3}&#10;\\\\B_{in}=\frac{\mu_0kl^2}{3}&#10;

now using equation 1, putting the value of k,

B_{in} = \frac{\mu_{0} l^2 }{3 } \,\,\, \frac{3I}{2 \pi a^3}&#10;\\\\B_{in} = \frac{ \mu_{0} I l^2}{2 \pi a^3}&#10;

B)

now, for points outside the wire ( r>a)

consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius l.

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\oint{\vec{B}_{out}\cdot\,d\vec{l}}=\mu_0\,i_{enc}&#10;

by symmetry \vec{B} will be of uniform magnitude on this loop and it's direction will be tangential to the loop. Hence

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again using,equaiton 1,

B_{out}= \mu_0 \frac{a^3}{3r} \times \frac{3 I}{2 \pi a^3}&#10;\\\\B_{out} = \frac{ \mu_{0} I}{2 \pi r}

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