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Evgesh-ka [11]
3 years ago
12

Which of the following statements about inductors are correct? A. When an inductor and a resistor are connected in series with

a DC battery, the current in the circuit is reduced to zero in one time constant. B. When an inductor and a resistor are connected in series with a DC battery, the current in the circuit is zero after a very long time. C. An inductor always resists any change in the current through it. D. When it is connected in a circuit, an inductor always resists having current flow through it. E. Inductors store energy by building up charge.
Physics
1 answer:
vladimir2022 [97]3 years ago
7 0

Answer: A,C,D,E are correct statements of an inductor

A. When an inductor and a resistor are connected in series with a DC battery, the current in the circuit is reduced to zero in one time constant.

C. An inductor always resists any change in the current through it.

D. When it is connected in a circuit, an inductor always resists having current flow through it.

E. Inductors store energy by building up charge.

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The speed of an arrow fired from a compound
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Answer:

A.) The arrow`s range is 624,996 m

B.) The arrow`s range is 846.887 m, when the horse is galloping

Explanation:

We have a case of oblique movement. In these cases the movement in the X axis is a Uniform Rectelinear Movement (URM), and a Uniform Accelerated Movement (UAM) in the Y axis.

By the way, the equations that we use for the X axis will be from URM, and those for the Y axis wiil be from UAM.

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X axis:

X=v_{ox}*t

v_{0x} =v_0cos(\alpha)

Y axis:

Y= Y_0 +v_{y0} t - \frac{g}{2} t^2

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To figure out t value, we use UAM, since time is determined by this movement.

Now, at the end of the movement, Y=0, then

0= Y_0 +v_{y0} t - \frac{g}{2} t^2

0=2.4m+79m/s*sin(39)t-(1/2*9.81m/s^2)t^2

Caculate the segcond degree equation to obtain the two possible values for t:

t_1= 10.18 \\t_2= -0.04046

But, in physics, time it could not be negative, so we take t_1= 10.18

Caculate now:

X=79m/s*cos(\39)*10.18s= 624.996 m

B.) Now, the narrow has an additional speed, that could be sum to the speed due to the bow.

v_0= 79m/s+13m/s= 92m/s

Using the same procedure that item A, caculate X

First, we need to know the new time

0=2.4m+92m/s*sin(39)t-(1/2*9.81m/s^2)t^2

And we obtain:

t_1=11.845s\\t_2=-0.041s

One more time, we take the positive time: t_1=11.845s

Finally:

X=92m/s *cos(39)*11.845s=846.887 m

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