Answer:
E = 3544.44 N/C
Explanation:
Given:
- charge Q = 2.2 *10^-6 C
- Length L = 1.3 m
Find:
The Electric Field strength E @ a = 1.8 m
Solution:
- The differential electric field dE due to infinitesimal charge dq can be considered as a point charge at a distance of r is given by:
dE = k*dq / r^2
- The charge Q is spread over entire length L, hence:
dq = (Q / L ) * dx
-The resulting dE:
dE = (k*Q/L)*(dx / r^2)
- point P lies on the x- axis with distance (x+a) from differential charge from:
dE = (k*Q/L)*(dx / (x+a)^2)
- Integrate dE over length 0 to L
E = (-k*Q/L)*( 1 / (x+a) )
E = (-k*Q/L)* (1 / a - 1 / (L+a))
E = (-k*Q/L)* (L / a(L+a))
E = (k*Q / a(L+a))
- Evaluate E @ a = 1.8 m
E =(8.99*10^9 * 2.2*10^-6 / 1.8*(1.3+1.8))
E = 3544.44 N/C
The acceleration of gravity is
9.8 m/s^2 down.
When an object falls out of a hand, its speed after 1.8s is
(9.8)x(1.8) = 17.6 m/s down.
It doesn't matter what it is, how much it weighs, or how high it was dropped from.
If it's more than 17.6 m/s, then this happened on a different, bigger planet.
If it's less than 17.6 m/s, then it must have hit something on the way down, like some air or something.
Answer:
The correct answer is c. More total rainfall from a slower moving storm.
When the the forward speed of the hurricanes and tropical storms slows down they tend to increase the rainfall. Because of the slow movement the storm can be for few days over a given region and produce rainfall without stopping, thus create major flooding, pilling up of the coastal water, and produce persistent strong winds even though they have decreased in their forward speed.
Explanation:
Answer:
g_x = 3.0 m / s^2
Explanation:
Given:
- Change in length of spring [email protected] = 22.6 cm
- Time taken for 11 oscillations t = 19.0 s
Find:
- The value of gravitational free fall g_x at plant X:
Solution:
- We will assume a simple harmonic motion of the mass for which Time is:
T = 2*pi*sqrt(k / m ) ...... 1
- Sum of forces in vertical direction @equilibrium is zero:
F_net = k*x - m*g_x = 0
(k / m) = (g_x / x) .... 2
- substitute Eq 2 into Eq 1:
2*pi / T = sqrt ( g_x / x )
g_x = (2*pi / T )^2 * x
- Evaluate g_x:
g_x = (2*pi / (19 / 11) )^2 * 0.226
g_x = 3.0 m / s^2
