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3 years ago

4. Interference is an example of which aspect of electromagnetic radiation?

Physics

2 answers:

Lisa [10]3 years ago

6 0

Answer:

D is the answer wave behavior

soldi70 [24.7K]3 years ago

4 0

Interference, refraction, diffraction, and dispersion are all aspects of <em>wave behavior. (D)</em>. That is, particles don't do these things.

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Three people pull simultaneously on a stubborn donkey. Jack pulls directly ahead of the donkey with a force of 64.7 N64.7 N , Ji

WARRIOR [948]

Answer:

(a) Magnitude of force is 262.51 N

(b) Angle with East direction is $-14.75^{o}$

Explanation:

Force by Jack in vector form

$\overrightarrow F _1} = 64.7{\rm{ N}}\left( {\hat i} \right)$

Force by Jill in Vector form is given by

$\begin{array}{c}\\{\overrightarrow F _2} = 86.5{\rm{ N }}\cos {\rm{4}}{{\rm{5}}^{\rm{o}}}\left( {\hat i} \right) + 86.5{\rm{ N }}\sin {\rm{4}}{{\rm{5}}^{\rm{o}}}\left( {\widehat j} \right)\\\\ = 61.16{\rm{ N}}\left( {\hat i} \right) + 61.16{\rm{ N}}\left( {\widehat j} \right)\\\end{array}$

Force by Jane is

$\begin{array}{c}\\{\overrightarrow F _3} = 181{\rm{ N }}\cos {\rm{4}}{{\rm{5}}^{\rm{o}}}\left( {\hat i} \right) + 181{\rm{ N }}\sin {\rm{4}}{{\rm{5}}^{\rm{o}}}\left( { - \widehat j} \right)\\\\ = 128{\rm{ N}}\left( {\hat i} \right) + 128{\rm{ N}}\left( { - \widehat j} \right)\\\end{array}$

Net force is:

$\overrightarrow F = {\overrightarrow F _1} + {\overrightarrow F _2} + {\overrightarrow F _3}$

Hence

$\begin{array}{c}\\\overrightarrow F = 64.7{\rm{ N}}\left( {\hat i} \right) + 61.16{\rm{ N}}\left( {\hat i} \right) + 61.16{\rm{ N}}\left( {\widehat j} \right) + 128{\rm{ N}}\left( {\hat i} \right) + 128{\rm{ N}}\left( { - \widehat j} \right)\\\\ = 253.86{\rm{ N}}\left( {\hat i} \right) - 66.84{\rm{ N}}\left( {\widehat j} \right)\\\end{array}$

The net force will be given by

$F = \sqrt {{{\left( {{F_x}} \right)}^2} + {{\left( {{F_y}} \right)}^2}$

Since $F_{x}=253.86N$ and $F_{y}=-66.84N$

$\begin{array}{c}\\F = \sqrt {{{\left( {253.86{\rm{ N}}} \right)}^2} + {{\left( { - 66.84{\rm{ N}}} \right)}^2}} \\\\ = {\bf{262.51 N}}\\\end{array}$

The direction of net force is:

$\theta = {\tan ^{ - 1}}\left {\frac{{{F_y}}}{{{F_x}}}}$

Since $F_{x}=253.86N$ and $F_{y}=-66.84N$

$\begin{array}{c}\\\theta = {\tan ^{ - 1}}\left( {\frac{{ - 66.84{\rm{ N}}}}{{253.86{\rm{ N}}}}} \right)\\\\ = {\tan ^{ - 1}}\left( { - 0.2633} \right)\\\\ = {\bf{ - 14}}{\bf{.7}}{{\bf{5}}^{\bf{o}}}\\\end{array}$

The angle with East direction is $-14.75^{o}$

Net force exerted on the donkey is in the south-east direction. So, the angle of net force from the east direction is $-14.75^{o}$ and it is $14.75^{o}$ from the south.

5 0

3 years ago

Define velocity??????

ira [324]

Answer:

the speed of something in a given direction.

Explanation:

3 0

3 years ago

Read 2 more answers

Answer fast enough and ill brainiest u

denis-greek [22]

Answer:

There is no question to answer friend

Well HI anyway

3 0

4 years ago

Read 2 more answers

Why just indirect evidence be used to study the structure of atoms?

ArbitrLikvidat [17]

It has to be used because you can not visually see them

6 0

3 years ago

A water trough is 12 feet long, and its cross section is an equilateral triangle with sides 2 feet long. Water is pumped into th

MAXImum [283]

Answer:

The water level rises at 11.76 $ft^{2} /s$

$h=\sqrt{3/2} s$ let be h: height and s: side of an equilateral triangle

Explanation:

The picture shows a diagram of the situation, first we have to determine the height of the trough as follows:

With the Pytagorean Theorem we can find out that:

$h^{2} =s^{2} -s^{2} /4 =\sqrt{3} /2 s$

Then, the area of an equilateral triangle, as any triangle, is a half of its base times its height:

$A=\frac{1}{2} hs=\frac{1}{2} \frac{\sqrt{3} }{2} s^{2} =\frac{\sqrt{3} }{4} s^{2}$

Replacing values, we have:

$A=1.73ft^{2}$

That is the total area of the trough, but the problem specifies that it has been filled until 0.5 ft. Therefore, we have to find the cross section area of the water flow by substracting the total area minus the unfilled area of the trough:

$h_{u}: height of the unfilled triangle\\s_{u} : side of the unfilled triangle\\A_{u}: area of the unfilled triangle\\h_{u}=\sqrt{3} -0.5 =1.23\\s_{u}= \frac{2}{\sqrt{3} } 1.23=1.42\\A_{u}=\frac{\sqrt{3} }{4} 1.42^{2} =0.87ft^{2} A_{u}$

Then, the cross section area of water flow is

$A_{s} =1.73ft^{2} -0.87ft^{2} =0.85ft^{2}$

Finally, to determine the speed of water flow at this point we solve for v, the flow formula:

$Q: water flow\\v: speed\\Q=vA\\v=Q/A =\frac{10ft^{3}/s }{0.85ft^{2} } =11.76 \frac{ft^{2}}{s}$

6 0

3 years ago