Answer:
8.46E+1
Explanation:
From the question given above, the following data were obtained:
Charge 1 (q₁) = 39 C
Charge 2 (q₂) = –53 C
Force (F) of attraction = 26×10⁸ N
Electrical constant K) = 9×10⁹ Nm²/C²
Distance apart (r) =?
The distance between the two charges can be obtained as follow:
F = Kq₁q₂ / r²
26×10⁸ = 9×10⁹ × 39 × 53 / r²
26×10⁸ = 1.8603×10¹³ / r²
Cross multiply
26×10⁸ × r² = 1.8603×10¹³
Divide both side by 26×10⁸
r² = 1.8603×10¹³ / 26×10⁸
r² = 7155
Take the square root of both side
r = √7155
r = 84.6 m
r = 8.46E+1 m
<span>If there isn't any force then the normal contact force will be
N=m*g=7.5*9.81=73.58N
which is 73.58-23=50.58N less
so, there the person must pull at 23 degree upward
break down the tension in two components, vertical and horizontal.
vertical tension= 50.58=T*sin23
T=50.58/sin23=129.45N</span>
The answer to this question is going to be False
Answer:
1 / f = 1 / i + 1 / o thin lens equation
1 / i = 1 / f - 1 / o = (o - f) / (o * f)
i = o * f / (o - f)
i = 54.2 * 12.7 / (54.2 - 12.7) = 16.6 cm image distance
Image is real and inverted and 16.6 / 54.2 * 6 = 1.94 cm tall
Answer:
1201 lbs
Explanation:
Given that in mammals, the weight of the heart is approximately 0.5% of the total body weight.
Let the weight of the heart of a mammal be H
And the weight of the total body be B
The linear model that can gives the heart weight in terms of the total body weight will be:
H = 0.005B
B.) To find the weight of the heart of a whale whose weight is 2.402 × 105 lbs, substitute the whole weight in the formula.
H = 0.005 × 2.402 × 10^5
H = 1201 lbs
Therefore, the weight of the heart of the whale is 1201 lbs
