Explanation:
i think you have to find the velocity by using its wight times the gravity of the earth times the distance it fell?
look in google for a formula the use these sign, w and g ,d
Answer:
Within a horizontal flow fluid, points of higher fluid speed will have less pressure than points of slower fluid speed.
Explanation:
Answer:
0
Explanation:
Displacement is a vector from initial to final point. Because initial and final point are the same, so displacement is 0.
Answer:
F_net = 26.512 N
Explanation:
Given:
Q_a = 3.06 * 10^(-4 ) C
Q_b = -5.7 * 10^(-4 ) C
Q_c = 1.08 * 10^(-4 ) C
R_ac = 3 m
R_bc = sqrt (3^2 + 4^2) = 5m
k = 8.99 * 10^9
Coulomb's Law:
F_i = k * Q_i * Q_j / R_ij^2
Compute F_ac and F_bc :
F_ac = k * Q_a * Q_c / R^2_ac
F_ac = 8.99 * 10^9* ( 3.06 * 10^(-4 ))* (1.08 * 10^(-4 )) / 3^2
F_ac = 33.01128 N
F_bc = k * Q_b * Q_c / R^2_bc
F_bc = 8.99 * 10^9* ( 5.7 * 10^(-4 ))* (1.08 * 10^(-4 )) / 5^2
F_bc = - 22.137 N
Angle a is subtended between F_bc and y axis @ C
cos(a) = 3 / 5
sin (a) = 4 / 5
Compute F_net:
F_net = sqrt (F_x ^2 + F_y ^2)
F_x = sum of forces in x direction:
F_x = F_bc*sin(a) = 22.137*(4/5) = 17.71 N
F_y = sum of forces in y direction:
F_y = - F_bc*cos(a) + F_ac = - 22.137*(3/5) + 33.01128 = 19.72908 N
F_net = sqrt (17.71 ^2 + 19.72908 ^2) = 26.5119 N
Answer: F_net = 26.512 N
Answer:
1. 77.31 N/m
2. 26.2 m/s
3. increase
Explanation:
1. According to the law of energy conservation, when she jumps from the bridge to the point of maximum stretch, her potential energy would be converted to elastics energy. Her kinetic energy at both of those points are 0 as speed at those points are 0.
Let g = 9.8 m/s2. And the point where the bungee ropes are stretched to maximum be ground 0 for potential energy. We have the following energy conservation equation


where m = 75 kg is the mass of the jumper, h = 72 m is the vertical height from the jumping point to the lowest point, k (N/m) is the spring constant and x = 72 - 35 = 37 m is the length that the cord is stretched


2. At 35 m below the platform, the cord isn't stretched, so there isn't any elastics energy, only potential energy converted to kinetics energy. This time let's use the 35m point as ground 0 for potential energy

where H = 35m this time due to the height difference between the jumping point and the point 35m below the platform


3. If she jumps from her platform with a velocity, then her starting kinetic energy is no longer 0. The energy conservation equation would then be

So the elastics energy would increase, which would lengthen the maximum displacement of the cord