Answer:
The point at which the electrical potential is zero is x = +0.33 m.
Explanation:
By definition the electrical potential is:
Where:
K: is Coulomb's constant = 9x10⁹ N*m²/C²
q: is the charge
r: is the distance
The point at which the electrical potential is zero can be calculated as follows:
(1)
q₁ is the first charge = +3 mC
r₁ is the distance from the point to the first charge
q₂ is the first charge = -6 mC
r₂ is the distance from the point to the second charge
By replacing r₁ = 1 - r₂ into equation (1) we have:
(2)
By solving equation (2) for r₂:
Therefore, the point at which the electrical potential is zero is x = +0.33 m.
I hope it helps you!
Answer:
Explanation:
Given
velocity of mug with which it leaves the bar is 1.7 m/s
Also
height of bar=1 m
Considering motion in vertical direction
here
t=0.451 s
so horizontal distance traveled is
here a=0
speed of mug will be combination of horizontal and vertical velocity
thus
for direction
is with x axis in clockwise sense
Refer to the diagram shown below.
The basket is represented by a weightless rigid beam of length 0.78 m.
The x-coordinate is measured from the left end of the basket.
The mass at x=0 is 2*0.55 = 1.1 kg.
The weight acting at x = 0 is W₁ = 1.1*9.8 = 10.78 N
The mass near the right end is 1.8 kg.
Its weight is W₂ = 1.8*9.8 = 17.64 N
The fulcrum is in the middle of the basket, therefore its location is
x = 0.78/2 = 0.39 m.
For equilibrium, the sum of moments about the fulcrum is zero.
Therefore
(10.78 N)*(0.39 m) - (17.64 N)*(x-0.39 m) = 0
4.2042 - 17.64x + 6.8796 = 0
-17.64x = -11.0838
x = 0.6283 m
Answer: 0.63 m from the left end.
Answer: 317.0 m/s
Explanation:
The motion of the bullet is a projectile motion, with:
- a uniform motion with constant speed v along the horizontal direction
- an accelerated motion with constant acceleration toward the ground
We know that the starting height of the bullet is h=1.4 m. If we consider the vertical motion only, the initial velocity is zero, so we can write:
The bullet reach the ground when y(t)=0, so the time taken is
During this time, the bullet travels d=168 m horizontally, so its horizontal speed (which is equal to the initial speed of the bullet) is given by