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3 years ago

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Is metallic wire carrying current charged or electrically neutral? Justify. ...?

1 answer:

Vladimir79 [104]3 years ago

7 0

A metallic wire carrying current is not charged - it is electrically neutral.
This is because a wire, or any type of conductor, cannot store charge - the fact that it is carrying current means that charge carriers that already existed in the wire are in motion. There are as many electrons as there are protons in the wire, which makes it neutral - there is no buildup of charge.

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A 3.0 kilogram object is thrown upward with an initial speed of 10 meters per second. what maximum height will the object reach?

tino4ka555 [31]

Time= v*sintheta/g = 10*1/9.8= 1.02s therefore h=1/2*9.8*(1.02)^2= 5.10m

3 0

3 years ago

Nicholas sends 20.0 minutes ironing shirts with his 1800-watt irn . How many joules of energy were usedby the iron

Nuetrik [128]

1 watt = 1 joule/sec
1800 watts = 1800 joules/sec

   (1800 J/sec) x (20 min) x (60 sec/min)

   = (1800 x 20 x 60) joules = 2,160,000 joules .

   ( 2.16 megajoules )

6 0

4 years ago

Two blocks A and B have a weight of 11 lb and 5 lb , respectively. They are resting on the incline for which the coefficients of

Alchen [17]

Answer:

$\theta=10.20^{\circ}$

$\Delta l=0.10 ft$

Explanation:

First of all, we analyze the system of blocks before starting to move.

$\Sum F_{x}=P_{A}sin(\theta)+P_{B}sin(\theta)-F_{fA}-F_{fB}=0$

$\Sum F_{x}=11sin(\theta)+5sin(\theta)-0.16N_{A}-0.23N_{B}=0$

$11sin(\theta)+5sin(\theta)-0.16P_{A}cos(\theta)-0.23P_{B}cos(\theta)=0$

$11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0$

$11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0$

$16sin(\theta)-2.91cos(\theta)=0$

$tan(\theta)=0.18$

$\theta=arctan(0.18)$

$\theta=10.20^{\circ}$

Hence, the incline angle θ for which both blocks begin to slide is 10.20°.

Now, if we do a free body diagram of block A we have that after the block moves, the spring force must be taken into account.

$P_{A}sin(\theta)-F_{fA}-F_{spring}=0$

Where:

$F_{spring} = k\Delta l=2.1\Delta l$

$P_{A}sin(\theta)-0.16*11cos(\theta)-2.1\Delta l=0$

$\Delta l=\frac{11sin(\theta)-0.16*11cos(\theta)}{2.1}$

$\Delta l=0.10 ft$

Therefore, the required stretch or compression in the connecting spring is 0.10 ft.

I hope it helps you!

4 0

3 years ago

An auto race takes place on a circular track. A car completes one lap in a time of 25.0 s, with an average tangential speed of 5

ludmilkaskok [199]

Answer:

(a) Angular speed will be 0.2512 rad/sec

(b) Radius will be 213.375 m

Explanation:

We have given time to complete 1 lap = 25 sec

We know that 1 lap = $2\pi$ radian

(a) So angular speed $=\frac{2\pi }{25}=\frac{2\times 3.14}{25}=0.2512rad/sec$

(B) Tangential velocity is given as v = 53.6 m/sec

We know that tangential velocity is given by

$v=\omega r$

So radius $r=\frac{v}{\omega }=\frac{53.6}{0.2512}=213.375m$

4 0

3 years ago

I DO NOT WANT LINKES OR …………..

il63 [147K]

1st B is correct option

2nd D is correct option

4 0

3 years ago

Read 2 more answers