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const2013 [10]
3 years ago
11

One strategy in a snowball fight is to throw a snowball at a high angle over level ground. While your opponent is watching the f

irst one, a second snowball is thrown at a low angle timed to arrive before or at the same time as the first one. Assume both snowballs are thrown with a speed of 25.0 m/s. The first one is thrown at an angle of 75.0° with respect to the horizontal. How many seconds later should the second snowball be thrown after the first in order for both to arrive at the same time?
Physics
1 answer:
marin [14]3 years ago
3 0

Answer:

3.60342 seconds

Explanation:

v = Initial velocity of snowball = 25 m/s

g = Acceleration due to gravity = 9.81 m/s²

\theta_1 = First angle = 75°

The second angle will be

\theta_2=90-\theta_1=90-75=15^{\circ}

Horizontal speed for first throw

v_{x1}=vcos\theta_1\\\Rightarrow v_{x1}=25\times cos75\\\Rightarrow v_{x1}=6.47047\ m/s

Horizontal speed for second throw

v_{x2}=vcos\theta_1\\\Rightarrow v_{x2}=25\times cos15\\\Rightarrow v_{x2}=24.14814\ m/s

Horizontal range is given by

R=\frac{v^2sin(2\theta)}{g}\\\Rightarrow R=\frac{25^2\times sin(2\times 75)}{9.81}\\\Rightarrow R=31.85\ m

Time period for first throw

t_1=\frac{R}{v_{x1}}\\\Rightarrow t_1=\frac{31.85}{6.47047}\\\Rightarrow t_1=4.92236\ s

Time period for second throw

t_2=\frac{R}{v_{x1}}\\\Rightarrow t_2=\frac{31.85}{24.14814}\\\Rightarrow t_2=1.31894\ s

The time difference is

t=4.92236-1.31894=3.60342\ s

The second ball should be thrown 3.60342 seconds later

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If a transmission line in a cold climate collects ice, the increased diameter tends to cause vortex formation in a passing wind.
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b) f_{n+1}-f_n=5.587Hz

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a) So for the <em>fundamental mode</em> (n=1) we have, substituting our values:

f_1=\frac{1}{2(347m)} \sqrt{\frac{65.4\times10^6N}{4.35kg/m}}=5.587Hz

b) The <em>frequency difference</em> between successive modes is the fundamental frequency, since:

f_{n+1}-f_n=\frac{n+1}{2L} \sqrt{\frac{T}{\mu}}-\frac{n}{2L} \sqrt{\frac{T}{\mu}}=(n+1-n)\frac{1}{2L} \sqrt{\frac{T}{\mu}}=\frac{n}{2L} \sqrt{\frac{T}{\mu}}=f_1=5.587Hz

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Answer:

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Explanation:

Given :

Current passing through both wires = 5.0 A

Separation between both wires = 8.0 cm

We have to find magnetic field at a point which is 5 cm from any of wires.

From biot savert law,

We know the magnetic field due to long parallel wires.

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Where B = magnetic field due to long wires, \mu_{0} = 4\pi \times10^{-7}, R = perpendicular distance from wire to given point

From any one wire R_{1}  = 5 cm, R_{2}  = 3 cm

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∴ B = B_{1} + B_{2}

 B = \frac{\mu_{0} i}{2\pi R_{1} } +  \frac{\mu_{0} i}{2\pi R_{2} }

 B =\frac{ 4\pi \times10^{-7} \times5}{2\pi } [\frac{1}{0.03} + \frac{1}{0.05} ]

 B = 5.33\times10^{-5}  T

Therefore, the magnitude of magnetic field at given point = 5.33\times10^{-5} T

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