For any object thrown upwards where only the force of gravity is acting upon it, uses the following formula for the maximum height attained.
H= v²/2g, where g = 9.81 m/s²
There are two information of velocities are given. However, we use the 20 m/s information because this is the launch velocity. Hence, the solution is as follows:
H = (20 m/s)²/2(9.81 m/s²)
<em>H = 20.4 m</em>
Answer:
(A) No
(B) Speed decreases
Explanation:
(A) since there is nothing propelling the boat and the friction between the ice and the boat and also air resistance is negligible the net force of the system in the horizontal direction is zero and hence there is no change in the horizontal momentum of the boat.
(B) Since the person had not velocity in the horizontal direction before landing on the boat but now has one after landing on the boat, the speed of the boat will decrease because the momentum has to be conserved (remember there is no change in it).
Answer:
a)At the mean position
b)At the extremes positions
Explanation:
Given that mass is having oscillation motion.
We know that
1. At the mean position -The velocity of the mass is maximum and the acceleration of the mass is minimum.The net force on the mass will be zero.
2. At the extreme position-The velocity of the mass is minimum and the acceleration of the mass is maximum.The net force on the mass will not be zero.
Therefore
a)At the mean position
b)At the extremes positions
Answer:
(a) 23.946 kV
(b) -0.077 J
Explanation:
(a) The electric potential is given by the following formula:
(1)
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
q1 = q2 = 1.60*10^{-6}C
r1 and r2 are the distance from the charges to the point in which electric potential is evaluated.
Firs, you calculate the distance r1 and r2 by taking into account the position of the charges
Next, you replace the values of the parameters to calculate V:
(b) The potential electric energy is given by:
![U_T=U_{1,2}+U_{1,3}+U_{2,3}\\\\U_T=k\frac{q_1q_2}{r_{1,2}}+k\frac{q_1q_3}{r_{1,3}}+k\frac{q_2q_3}{r_{2,3}}\\\\r_{1,2}=2.00m\\\\r_{1,3}=1.20m\\\\r_{2,3}=1.20m\\\\U_T=(8.98*10^9)[\frac{(1.6*10^{-6})^2}{2.00m}+\frac{(1.6*10^{-6})(-3.70*10^{-6})}{1.20}+\frac{(1.6*10^{-6})(-3.70*10^{-6})}{1.20}]J\\\\U_T=-0.077J](https://tex.z-dn.net/?f=U_T%3DU_%7B1%2C2%7D%2BU_%7B1%2C3%7D%2BU_%7B2%2C3%7D%5C%5C%5C%5CU_T%3Dk%5Cfrac%7Bq_1q_2%7D%7Br_%7B1%2C2%7D%7D%2Bk%5Cfrac%7Bq_1q_3%7D%7Br_%7B1%2C3%7D%7D%2Bk%5Cfrac%7Bq_2q_3%7D%7Br_%7B2%2C3%7D%7D%5C%5C%5C%5Cr_%7B1%2C2%7D%3D2.00m%5C%5C%5C%5Cr_%7B1%2C3%7D%3D1.20m%5C%5C%5C%5Cr_%7B2%2C3%7D%3D1.20m%5C%5C%5C%5CU_T%3D%288.98%2A10%5E9%29%5B%5Cfrac%7B%281.6%2A10%5E%7B-6%7D%29%5E2%7D%7B2.00m%7D%2B%5Cfrac%7B%281.6%2A10%5E%7B-6%7D%29%28-3.70%2A10%5E%7B-6%7D%29%7D%7B1.20%7D%2B%5Cfrac%7B%281.6%2A10%5E%7B-6%7D%29%28-3.70%2A10%5E%7B-6%7D%29%7D%7B1.20%7D%5DJ%5C%5C%5C%5CU_T%3D-0.077J)
