Answer:
at d the charge will be 3q and at 3d it will be 9q
Explanation:
for V=Vp-V2d
V=KQ/d=K*6q/2d=3kq/d for potential to 2d at 6q be zero the Vp will equal 3kq/d; hence at d, Q=3q and at 3d, Q=9q
Answer:
Find the answer in the explanation
Explanation:
Given that a ball is dropped from rest from the top of a building of height h. At the same instant, a second ball is projected vertically upward from ground level, such that it has zero speed when it reaches the top of the building.
a. When do the two balls pass each other?
The the balls will pass each other before the second ball reach half of the height of the building. Because the velocity of the second ball decreases as it goes upwards while the velocity of the first ball increases as it moves downwards
b. Which ball has greater speed when they are passing?
The first ball of course because the velocity of the second ball decreases as it goes upwards while the velocity of the first ball increases as it moves downwards
c. What is the height of the two balls when they are passing?
Taking the height from the bottom of the building upward
Since the two balls meet at a certain point, they will both have the same height.
Answer:
true ............................................
Answer:
807.88N/m
Explanation:
<em>The question has some missing details in it, nevertheless, based on the given data we want to find the spring constant K</em>
Step one
given data
Unstretched length = 33.5 cm
Final length of the spring = 42.0 cm
Δx= 42-33.5
Δx=8.5cm to m= 0.085m
mass m= 7kg
The force on the spring
F=mg
F= 7*9.81
F=68.67N
Step two:
From Hooke's law, we can make k subject of formula and find the spring constant k, we have
F=kΔx---------1
make k subject of the formula
k=F/Δx
k= 68.67/ 0.085
k=807.88N/m
Gamma rays ..........................
