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3 years ago

11

*An inductor is capable of dissipating 50W of heat energy when a current 0.8A flows through it at a certain frequency. Calculate

its impedance when the reactance of the inductor is 50Ω*.
A. 62.5 Ω.
B. 40.5 Ω.
C. 92.8 Ω.
D. 1.0 Ω.

1 answer:

ale4655 [162]3 years ago

8 0

I think that your answer would be D

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The question will be on one of the comments on the answers you give just now take the points

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3 years ago

Matter is made of small particles to small to be seen. Which of these best describe evidence of this statement? 1. Tara’s crayon

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I think the answer is statement no 3.

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3 years ago

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If you weigh 690 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun an

lana [24]

Answer:

Explanation:

Given that,

Mass of star M(star) = 1.99×10^30kg

Gravitational constant G

G = 6.67×10^−11 N⋅m²/kg²

Diameter d = 25km

d = 25,000m

R = d/2 = 25,000/2

R = 12,500m

Weight w = 690N

Then, the person mass which is constant can be determined using

W =mg

m = W/g

m = 690/9.81

m = 70.34kg

The acceleration due to gravity on the surface of the neutron star is can be determined using

g(star) = GM(star)/R²

g(star) = 6.67×10^-11 × 1.99×10^30 / 12500²

g (star) = 8.49 × 10¹¹ m/s²

Then, the person weight on neutron star is

W = mg

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W = 70.34 × 8.49 × 10¹¹

W = 5.98 × 10¹³ N

The weight of the person on neutron star is 5.98 × 10¹³ N

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3 years ago

Choose the appropriate words to complete the thought. A solar powered car converts _______________ energy into _________________

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The best and most correct answer among the choices provided by your question is the second choice or letter C. A solar-powered car converts light energy to mechanical energy.

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3 years ago

Suppose you are pushing a 3 kg box with a force of 25 N (directed parallel to the ground) over a distance of 15 m. Afterward, th

alex41 [277]

Answer: 321 J

Explanation:

Given

Mass of the box $m=3\ kg$

Force applied is $F=25\ N$

Displacement of the box is $s=15\ m$

Velocity acquired by the box is $v=6\ m/s$

acceleration associated with it is $a=\dfrac{F}{m}$

$\Rightarrow a=\dfrac{25}{3}\ m/s^2$

Work done by force is $W=F\cdot s$

$W=25\times 15\\W=375\ J$

change in kinetic energy is $\Delta K$

$\Rightarrow \Delta K=\dfrac{1}{2}m(v^2-0)\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 6^2\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 36\\\\\Rightarrow \Delta K=54\ J$

According to work-energy theorem, work done by all the forces is equal to the change in the kinetic energy

$\Rightarrow W+W_f=\Delta K\quad [W_f=\text{Work done by friction}]\\\\\Rightarrow 375+W_f=54\\\Rightarrow W_f=-321\ J$

Therefore, the magnitude of work done by friction is $321\ J$

3 0

3 years ago