Answer:
x = 1474.9 [m]
Explanation:
To solve this problem we must use Newton's second law, which tells us that the sum of forces must be equal to the product of mass by acceleration.
We must understand that when forces are applied on the body, they tend to slow the body down to stop it.
So as the body continues to move to the left, it is slowing down. Therefore we must calculate this deceleration value using Newton's second law. We must perform a sum of forces on the x-axis equal to the product of mass by acceleration. With leftward movement as negative and rightward forces as positive.
ΣF = m*a
![10 +12*sin(60)= - 6*a\\a = - 3.39[m/s^{2}]](https://tex.z-dn.net/?f=10%20%2B12%2Asin%2860%29%3D%20-%206%2Aa%5C%5Ca%20%3D%20-%203.39%5Bm%2Fs%5E%7B2%7D%5D)
Now using the following equation of kinematics, we can calculate the distance of the block, before stopping completely. The initial speed must be 100 [m/s].

where:
Vf = final velocity = 0 (the block stops)
Vo = initial velocity = 100 [m/s]
a = - 3.39 [m/s²]
x = displacement [m]
![0 = 100^{2}-2*3.39*x\\x=\frac{10000}{2*3.39}\\x=1474.9[m]](https://tex.z-dn.net/?f=0%20%3D%20100%5E%7B2%7D-2%2A3.39%2Ax%5C%5Cx%3D%5Cfrac%7B10000%7D%7B2%2A3.39%7D%5C%5Cx%3D1474.9%5Bm%5D)
Answer:an electrical Current can be defined as the free flow of electrons through a circuit
Explanation:
Answer:h=19.4 m
Explanation:
Given
mass of automobile 
Initial height of automobile 
Velocity at this instant 
If the car stops somewhere at a height 
Thus conserving total energy we get






Answer:
a) 1.082 × 10⁻¹⁹C ( e = 1.6 × 10⁻¹⁹C)
b) 3.466 × 10¹¹ N/C
Explanation:
a)
p(r) = -A exp ( - 2r/a₀)
Q = ₀∫^∞ ₀∫^π ₀∫^2xπ p(r)dV = -A ₀∫^∞ ₀∫^π ₀∫^2π exp ( - 2r/a₀)r² sinθdrdθd∅
Q = -4πA ₀∫^∞ exp ( - 2r/a₀)r²dr = -e
now using integration by parts;
A = e / πa₀³
p(r) = - (e / πa₀³) exp (-2r/a₀)
Now Net charge inside a sphere of radius a₀ i.e Qnet is;
= e - (e / πa₀³) ₀∫^a₀ ₀∫^π ₀∫^2π r² exp (-2r/a₀)dr
= e - e + 5e exp (-2) = 1.082 × 10⁻¹⁹C ( e = 1.6 × 10⁻¹⁹C)
b)
Using Gauss's law,
E × 4πa₀ ² = Qnet / ∈₀
E = 4πa₀ ² × Qnet × 1/a₀²
E = 3.466 × 10¹¹ N/C