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Artist 52 [7]
3 years ago
15

A stone is thrown straight downward with a speed of 20 m/s from the top of a tall building. If the stone strikes the ground 3.0

seconds later, about how tall is the building?
A) 45m
B) 60m (wrong)
C) 90m
D) 105 m
Physics
1 answer:
Lady_Fox [76]3 years ago
6 0
Do you remember this formula for the distance traveled while accelerated ?

<u>Distance = (initial speed) x (t)  plus  (1/2) x (acceleration) x (t²)</u>

I think this is exactly what we need for this problem.

initial speed = 20 m/s down
acceleration = 9.81 m/s² down
t = 3.0 seconds

Distance down = (20) x (3)  plus  (1/2) x (9.81) x (3)²

Distance = (60)  plus  (4.905) x (9)

Distance = (60)  plus  (44.145)  =  104.145 meters

Choice  <em>D)</em>  is the closest one.
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A steady current I flows through a wire of radius a. The current density in the wire varies with r as J = kr, where k is a const
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Answer:

Explanation:

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

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since current density is given

J=kr

then , current through this element will be,

di_{thru}=JdA=(kr)(2\pi\,r\,dr)=2\pi\,kr^2\,dr

integrating on both sides between the appropriate limits,

\int_0^Idi_{thru}=\int_0^a2\pi\,kr^2\,dr&#10;\\\\&#10;I=\frac{2\pi\,ka^3}{3} -------------------------------(1)

Magnetic field can be found by using Ampere's law

\oint{\vec{B}\cdot\,d\vec{l}}=\mu_0\,i_{enc}

for points inside the wire ( r<a)

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.

Hence,

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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.

applying the Ampere's law

\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}

8 0
3 years ago
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Answer:

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ΔL = (1.2×10⁻⁵ / °C) (800 ft) (-48°C)

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Therefore, the final length is approximately 800 ft − 0.46 ft = 799.54 ft.

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