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Rom4ik [11]
3 years ago
14

Two long, straight wires are fixed parallel to one another a distance do apart. The wires carry equal constant currents 1, in th

e same direction. The attractive magnetic force per unit length between them if f = F/L. What is the force per unit length between the wires if their separation is 2d, and each carries current 210? 10 da lo A. f/4 B. f/2 C. 3f/2 D. 2f
Physics
1 answer:
Alenkinab [10]3 years ago
3 0

Answer:

The option (B) is correct.

Explanation:

The magnetic force between the two current carrying wires is given by

F =\frac{\mu o}{4\pi}\times \frac{2 I I'}{r}

where,  I and I' be the currents in the wires and r is the distance between the two wires.

Here, we observe that the force between the wires is inversely proportional to the distance between them.

So, when the distance is doubled , let the new force is F'.

\frac{F'}{F}=\frac{r}{r'}\\\\\frac{F'}{F}=\frac{d}{2d}\\\\F'=\frac{F}{2}

So, option (B) is correct.

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What are the factors of works?
olga2289 [7]

Answer:

Explanation:

The seven factors are work load, family life, transportation, compensation policy and benefits, colleagues and supervisor, working environment and working condition and career growth.

4 0
3 years ago
A particle of mass m collides with a second particle of mass m. Before the collision, the first particle is moving in the x-dire
oee [108]

Answer:

a) v, v

b) 2mv^2

c) Elastic collion

Explanation:

(a) The velocity of the second particle after the collision is (v2x,v2y)=(v,−v).  From momentum conservation in x-direction

Here x, y represent direction.They are not variable. 1 and 2 represent before and after.

2vm=v1xm+v2xm, we find v1x=v.

From momentum conservation in y-direction

0 =v1ym+v2ym, we findv1y=v.

(b) By energy conservation principle

Before: K=1/2m(2v)^2=2mv^2.

After: K=1/2m(v^2(1x)+v^2(1y))+12m(v22x+v22y)=2mv^2

(c) The collision is elastic

6 0
3 years ago
Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is one kilocalorie, defined as 1 kc
Sergeeva-Olga [200]

Answer:

a) The student must run flight of stairs to lose 1.00 kg of fat 709.5 times.

b) Average power

P(w)= 1062.07 [w]

P(hp)=1.42 [hp]

c) This activity is highly unpractical, because the high amount of repetitions he has to due in order to lose, just 1 Kg of fat.

Explanation:

First, lets consider the required amount of work to move the mass of the student. (considering running stairs just as a vertical movement)

Work:

W= F*d= m*g*d

Where m is the mass of the student, g is gravity (9.8 m/s) and d is the total distance going up the stairs (0.15m *85steps= 12.75m )

W= F*d= m*g*d=85* 9.8*12.75=10620.75 [J]

Converting from Joules to Kcals:

\frac{10620.75}{4186} =2.537 Kcal

Now lets take into account the efficiency of the human body (20%)

2.537 ---> 20%

 x       ---> 100%

x=\frac{2.537*100}{20} =12.685

So the student is consuming 12.685 KCals each time he runs up the stairs.

Now,

1 g --> 9 Kcals

1000 g --> 9000KCals

Burning 1 g of fat, requieres 9 KCals, 1000g burns 9000KCals. So in order to burn a 1Kg of fat:

\frac{9000Kcals}{12.685Kcals} =709.5 times

He must run up the stairs 709.5 times, to burn 1 Kg of fat.

********************

For b) just converting units, taking into account the time lapse. (53103.75 is the 100% of the energy in joules, from converting 12.685Kcals to joules)

Power=\frac{Joules}{Seconds} =\frac{53103.75}{50} =1062.075 [W]\\

P(hp)=\frac{P(w)}{745.7} =\frac{1062.075}{745.7} =1.42[hp]

*****

4 0
3 years ago
QUICK WILL MARK BRAINLIEST
Tju [1.3M]

Answer:

37.1°

Explanation:

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3 years ago
Which option correctly matches the chemical formula of a compound with its name?
77julia77 [94]

Answer:it’s A

Explanation:

4 0
3 years ago
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