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3 years ago

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If wave A has twice the amplitude and three times the frequency as wave B, then the energy carried by wave A must be ____ times

that of wave B.

1 answer:

soldier1979 [14.2K]3 years ago

4 0

Answer:

4 times

Explanation:

As we know that the energy of a wave is directly proportional to the square of the amplitude of the wave,

Here, the amplitude of the wave A is twice as compared to B.

So, the energy of wave A is 4 times the energy of wave B.

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Consider the vector field. f(x, y, z) = xy2z2i x2yz2j x2y2zk (a) find the curl of the vector field?

Marat540 [252]

Observe that the given vector field is a gradient field:

Let $f(x,y,z)=\nabla g(x,y,z)$ , so that

$\dfrac{\partial g}{\partial x} = x y^2 z^2$

$\dfrac{\partial g}{\partial y} = x^2 y z^2$

$\dfrac{\partial g}{\partial z} = x^2 y^2 z$

Integrating the first equation with respect to $x$ , we get

$g(x,y,z) = \dfrac12 x^2 y^2 z^2 + h(y,z)$

Differentiating this with respect to $y$ gives

$\dfrac{\partial g}{\partial y} = x^2 y z^2 + \dfrac{\partial h}{\partial y} = x^2 y z^2 \\\\ \implies \dfrac{\partial h}{\partial y} = 0 \implies h(y,z) = i(z)$

Now differentiating $g$ with respect to $z$ gives

$\dfrac{\partial g}{\partial z} = x^2 y^2 z + \dfrac{di}{dz} = x^2 y^2 z \\\\ \implies \dfrac{di}{dz} = 0 \implies i(z) = C$

Putting everything together, we find a scalar potential function whose gradient is $f$ ,

$f(x,y,z) = \nabla \left(\dfrac12 x^2 y^2 z^2 + C\right)$

It follows that the curl of $f$ is 0 (i.e. the zero vector).

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1 year ago

Which of these forces are acting within a nucleus? Check all that apply.

zavuch27 [327]

Answer:

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Explanation:

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2 years ago

A factory has 1200 workers of which 720 are male and the rast are female what percent of workers are female

umka21 [38]

Answer:

Percent of Female Workers = 40%

Explanation:

The percentage of the female workers in the given group of workers can be easily found by the following formula:

$Percent\ of\ Female\ Workers = \frac{No.\ of\ Female\ Workers}{Total\ No.\ of\ Workers}\ x\ 100\%$

where,

Total No. of Workers = 1200

No. of Female Workers = Total Workers - No. of Male Workers

No. of Female Workers = 1200 - 720 = 480

Therefore,

$Percent\ of\ Female\ Workers = \frac{480}{1200}\ x\ 100\%$

<u>Percent of Female Workers = 40%</u>

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2 years ago

What are the different isotopes

Norma-Jean [14]

Answer:

Isotopes can both be the same element but have a different number of electrons

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2 years ago

A 95kg fullback (football player for those not into sports) moving south with a speed of 5.0 m/s has a perfectly inelastic colli

Lunna [17]

Answer:

a. $v=3.11mls$ , $29.4^{0}$

b. $K.E =-697.8J$

Explanation:

To calculate the values in the question, a deep understanding of perfect inelastic collision is important.

When two bodies undergo inelastic collision, two important parameters must be well understood i.e

Momentum: the momentum is always conserved in perfectly inelastic collision. i.e the total momentum after collision is the sum of the individual momentum before collision

Kinetic energy: Kinetic energy is not conserved due to dissipative force.

a.To calculate the velocity, we first find the total momentum before collision

Momentum of player 1 $p_{1} =mv=95kg*5m/s\\p_{1} =475kgm/s\\$

Momentum of player 2 $p_{2} =mv=90kg*3m/s\\p_{1} =270kgm/s\\$

Hence the total momentum $p_{12}=p_{1}+p_{2}\\$

Note, since the direction of movement before collision is due south and due north respectively we have to represent the velocity using the rectangular coordinate

Hence $p_{12}=(m_{1}+m_{2})v=p_{1}i+p_{2}j\\$

$(95+90)v=475i+270j\\$

$v=2.57i+1.45j\\$

solving for the resultant velocity, we have

$v=\sqrt{2.75^{2} +1.45^{2}}\\ v=3.11mls$

To calculate the direction of movement, we have

$\alpha =tan^{-1}=\frac{v_{j} }{v_{i}}\\ \alpha =tan^{-1}=\frac{1.45}{2.57}\\\alpha =29.4^{0}$

b. to calculate the decrease in total kinetic energy, before collision, the total kinetic was

$K.E_{initial} =\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}.\\K.E_{initial} =((1/2)*95*5^{2})+((1/2)*90*3^{2})\\K.E_{initial} =1187.5+405\\K.E_{initial} =1592.5J\\$

And the final kinetic energy after collision is

$K.E_{final} =\frac{1}{2}(m_{1}+m_{2} )v^{2}\\ K.E_{final} =\frac{1}{2}(95+90)* 3.11^{2}\\ K.E_{final} =894.7J$

The decrease in Kinetic energy is

$K.E =K.E_{final}- K.E_{initial}=894.7-1592.5$

$K.E =-697.8J$

The negative sign indicate a decrease in Kinetic energy

4 0

2 years ago