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3 years ago

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A coil with a wire that is wound around a 2.0 m1355_files/i0130000.jpg hollow tube 35 times. A uniform magnetic field is applied

perpendicular to the plane of the coil. If the field changes uniformly from 0.00 T to 0.55 T in 0.85 s, what is the induced emf in the coil? a. –45 V c. 33 V b. –33 V d. 45 V

1 answer:

Blizzard [7]3 years ago

5 0

The <span>induced emf in the coil is -45V.</span>

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Two transverse waves travel along the same taut string. Wave 1 is described by y1(x, t) = A sin(kx - ωt), while wave 2 is descri

Vadim26 [7]

Answer:

6) Wave 1 travels in the positive x-direction, while wave 2 travels in the negative x-direction.

Explanation:

What matters is the part $kx \pm \omega t$ , the other parts of the equation don't affect time and space variations. We know that when the sign is - the wave propagates to the positive direction while when the sign is + the wave propagates to the negative direction, but <em>here is an explanation</em> of this:

For both cases, + and -, after a certain time $\delta t$ ( $\delta t >0$ ), the displacement <em>y</em> of the wave will be determined by the $kx\pm\omega (t+\delta t)$ term. For simplicity, if we imagine we are looking at the origin (x=0), this will be simply $\pm \omega (t+\delta t)$ .

To know which side, right or left of the origin, would go through the origin after a time $\delta t$ (and thus know the direction of propagation) we have to see how we can achieve that same displacement <em>y</em> not by a time variation but by a space variation $\delta x$ (we would be looking where in space is what we would have in the future in time). The term would be then $k(x+\delta x)\pm\omega t$ , which at the origin is $k \delta x \pm \omega t$ . This would mean that, when the original equation has $kx+\omega t$ , we must have that $\delta x>0$ for $k\delta x+\omega t$ to be equal to $kx+\omega\delta t$ , and when the original equation has $kx-\omega t$ , we must have that $\delta x$ for $k\delta x-\omega t$ to be equal to $kx-\omega \delta t$

<em>Note that their values don't matter, although they are a very small variation (we have to be careful since all this is inside a sin function), what matters is if they are positive or negative and as such what is possible or not .</em>

<em />

In conclusion, when $kx+\omega t$ , the part of the wave on the positive side ( $\delta x>0$ ) is the one that will go through the origin, so the wave is going in the negative direction, and viceversa.

4 0

3 years ago

A 5,000 kg satellite is orbiting the Earth in a circular path. The height of the satellite above the surface of the Earth is 800

Arada [10]

Explanation:

The given data is as follows.

m = 5000 kg, h = 800 km = $0.8 \times 10^{6} m$

$R_{e} = 6.37 \times 10^{6} m$ , r = R + h = $7.17 \times 10^{6} m$

$M_{e} = 5.98 \times 10^{24}$ kg, G = $6.67 \times 10^{-11} Nm^{2}/kg^{2}$

As we know that,

$\frac{mv^{2}}{r} = \frac{GmM_{e}}{r^{2}}$

v = $\sqrt{\frac{GM_{e}}{r^{2}}}$

And, it is known that formula to calculate angular velocity is as follows.

$\omega = \frac{v}{r} = \sqrt{\frac{GM_{e}}{r^{3}}}$

v = $\sqrt{\frac{GM_{e}}{r^{3}}}$

= $\sqrt{\frac{6.67 \times 10^{-11} Nm^{2}/kg^{2} \times 5.98 \times 10^{-24} kg^{-2}}{(7.17 \times 10^{6} m)^{3}}}$

= $1.0402 \times 10^{-3} rad/s$

Thus, we can conclude that speed of the satellite is $1.0402 \times 10^{-3} rad/s$ .

6 0

3 years ago

Its not b...................

vladimir1956 [14]

sir what's the question you have all you wrote is it's not b

4 0

3 years ago

I need the science answer step by step

sukhopar [10]

What is the question

7 0

3 years ago

A sky diver with a mass of 70kg jumps from an aircraft. The aerodynamic drag force acting on the sky diver is known to be Fd=kV^

xeze [42]

Answer:

$v_{max}=52.38\frac{m}{s}$

$v_{100}=33.81$

Explanation:

the maximum speed is reached when the drag force and the weight are at equilibrium, therefore:

$\sum{F}=0=F_d-W$

$F_d=W$

$kv_{max}^2=m*g$

$v_{max}=\sqrt{\frac{m*g}{k}} =\sqrt{\frac{70*9.8}{0.25}}=52.38\frac{m}{s}$

To calculate the velocity after 100 meters, we can no longer assume equilibrium, therefore:

$\sum{F}=ma=W-F_d$

$ma=W-F_d$

$ma=mg-kv_{100}^2$

$a=g-\frac{kv_{100}^2}{m}$ (1)

consider the next equation of motion:

$a = \frac{(v_{x}-v_0)^2}{2x}$

If assuming initial velocity=0:

$a = \frac{v_{100}^2}{2x}$ (2)

joining (1) and (2):

$\frac{v_{100}^2}{2x}=g-\frac{kv_{100}^2}{m}$

$\frac{v_{100}^2}{2x}+\frac{kv_{100}^2}{m}=g$

$v_{100}^2(\frac{1}{2x}+\frac{k}{m})=g$

$v_{100}^2=\frac{g}{(\frac{1}{2x}+\frac{k}{m})}$

$v_{100}=\sqrt{\frac{g}{(\frac{1}{2x}+\frac{k}{m})}}$ (3)

$v_{100}=\sqrt{\frac{9.8}{(\frac{1}{2*100}+\frac{0.25}{70})}}$

$v_{100}=\sqrt{\frac{9.8}{(\frac{1}{200}+\frac{1}{280})}}$

$v_{100}=\sqrt{\frac{9.8}{(\frac{3}{350})}}$

$v_{100}=\sqrt{1,143.3}$

$v_{100}=33.81$

To plot velocity as a function of distance, just plot equation (3).

To plot velocity as a function of time, you have to consider the next equation of motion:

$v = v_0 +at$

as stated before, the initial velocity is 0:

$v =at$ (4)

joining (1) and (4) and reducing you will get:

$\frac{kt}{m}v^2+v-gt=0$

solving for v:

$v=\frac{ \sqrt{1+\frac{4gk}{m}t^2}-1}{\frac{2kt}{m} }$

Plots:

5 0

3 years ago