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1 year ago

Three equal positive charges 'q' are at the corners of an equilateral triangle of side 'a'.

1 answer:

5 0

(a) The location of a point where the electric field is zero is at the center of the triangle which is equal to ¹/₆√3a.

(b) The magnitude and direction of the electric field at the top corner due to the two charges at the base is 1.732 kq/a².

<h3>Position where the electric field is zero</h3>

The electric field is zero at the center of the equilateral triangle whose magnitude is equal to √3a/6.

<h3>Electric field at top corner due to two charges at the base</h3>

E = E₁ + E₂

where;

E₁ is electric field at the left base corner
E₂ is electric field at the right base corner

E = kq/a²[(cos 60i + sin 60j) + (-cos 60i + sin 60j)]

E = kq/a²[2(sin 60j)] = 1.732 kq/a²

Thus, the location of a point where the electric field is zero is at the center of the triangle which is equal to ¹/₆√3a.

The magnitude and direction of the electric field at the top corner due to the two charges at the base is 1.732 kq/a².

Learn more about electric field here: brainly.com/question/14372859

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3 years ago

9. a. Determine the mass of a football which has a weight of 0.80 N on a planet where the gravitational field strength is 2 N/kg

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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)

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