Answer:
F = 2.26 × 10⁻³ N
Explanation:
given,
length of rod = 11 cm
charge = 19 nC
linear charge density = 3.9 x 10⁻⁷ C/m
electric force at 2 cm away.

F = E q

integrating from 0.02 to 0.02 + L
![F= \dfrac{2K\lambda\ q}{L}[ln(0.02+L)-ln(0.002)]](https://tex.z-dn.net/?f=F%3D%20%5Cdfrac%7B2K%5Clambda%5C%20q%7D%7BL%7D%5Bln%280.02%2BL%29-ln%280.002%29%5D)
![F= \dfrac{2\times 9 \times 10^9\times 3.9\times 10^{-7}\times 19 \times 10^{-9}}{0.11}[ln(0.02+0.11)-ln(0.002)]](https://tex.z-dn.net/?f=F%3D%20%5Cdfrac%7B2%5Ctimes%209%20%5Ctimes%2010%5E9%5Ctimes%203.9%5Ctimes%2010%5E%7B-7%7D%5Ctimes%2019%20%5Ctimes%2010%5E%7B-9%7D%7D%7B0.11%7D%5Bln%280.02%2B0.11%29-ln%280.002%29%5D)
F = 2.26 × 10⁻³ N
Answer:
the answer is 104568948 degress
Explanation:
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Answer:
a) F = 1.26 10⁵ N, b) F = 2.255 10³ N, c) F_ {soil} = 3078 N
Explanation:
For this exercise we will use the relationship between momentum and moment
I = Δp
F t = p_f -p₀
a) with stiff legs, final speed is zero, initial velocity is down
Ft = 0-p₀
F = m v / t
let's calculate
F = 84.0 6.82 / 4.56 10⁻³
F = 1.26 10⁵ N
b) bending the legs
let's calculate
F = 84.0 6.82 / 0.254
F = 2.255 10³ N
c) It is requested to calculate the force of the ground on the man
∑ F = F_soil -W
F_soil = F + W
F_ {soil} = 2.255 103 + 84 9.8
F_ {soil} = 3078 N
Wind-driven currents are found NEAR THE SURFACE of the oceans
Explanation:
Surface currents are at the interphase between the hydrosphere and atmosphere. Therefore the feel the greatest effect of drag by wind currents, especially prevailing winds (that blow predominantly in one direction like westerlies and easterlies) within the lower atmosphere. The deep currents, on the other hand, are more influenced by Coriolis effect of the earth’s rotation. It is these differences in influences of surface and deeper currents that cause Ekman transport in oceans.
Learn More:
For more on surface and deep ocean currents check out;
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