The value of normal force as the slider passes point B is
The value of h when the normal force is zero
<h3>How to solve for the normal force</h3>
The normal force is calculated using the work energy principle which is applied as below
K₁ + U₁ = K₂
k represents kinetic energy
U represents potential energy
the subscripts 1,2 , and 3 = a, b, and c
for 1 to 2
K₁ + W₁ = K₂
0 + mg(h + R) = 0.5mv²₂
g(h + R) = 0.5v²₂
v²₂ = 2g(1.5R + R)
v²₂ = 2g(2.5R)
v²₂ = 5gR
Using summation of forces at B
Normal force, N = ma + mg
N = m(a + g)
N = m(v²₂/R + g)
N = m(5gR/R + g)
N = 6mg
for 1 to 3
K₁ + W₁ = K₃ + W₃
0 + mgh = 0.5mv²₃ + mgR
gh = 0.5v²₃ + gR
0.5v²₃ = gh - gR
v²₃ = 2g(h - R)
at C
for normal force to be zero
ma = mg
v²₃/R = g
v²₃ = gR
and v²₃ = 2g(h - R)
gR = 2gh - 2gR
gR + 2gR = 2gh
3gR = 2gh
3R/2 = h
Learn more about normal force at:
brainly.com/question/20432136
#SPJ1
Answer:
Time of flight A is greatest
Explanation:
Let u₁ , u₂, u₃ be their initial velocity and θ₁ , θ₂ and θ₃ be their angle of projection. They all achieve a common highest height of H.
So
H = u₁² sin²θ₁ /2g
H = u₂² sin²θ₂ /2g
H = u₃² sin²θ₃ /2g
On the basis of these equation we can write
u₁ sinθ₁ =u₂ sinθ₂=u₃ sinθ₃
For maximum range we can write
D = u₁² sin2θ₁ /g
1.5 D = u₂² sin2θ₂ / g
2 D =u₃² sin2θ₃ / g
1.5 D / D = u₂² sin2θ₂ /u₁² sin2θ₁
1.5 = u₂ cosθ₂ /u₁ cosθ₁ ( since , u₁ sinθ₁ =u₂ sinθ₂ )
u₂ cosθ₂ >u₁ cosθ₁
u₂ sinθ₂ < u₁ sinθ₁
2u₂ sinθ₂ / g < 2u₁ sinθ₁ /g
Time of flight B < Time of flight A
Similarly we can prove
Time of flight C < Time of flight B
Hence Time of flight A is greatest .
Answer:
Microlensing.
Explanation:
This techniques is called Microlensing.
Microlensing is a method of gravitational lensing where light from a backdrop point of origin is curved to develop distorted, numerous and/or lightened images by the gravity field of a foreground lens.
This method is very effective in discovering planets that are far-far from earth.It is actually an astronomical effect that was predicted by Albert Einstein's general theory of relativity.
Answer:
Part 1 
Part 2 
Part 3 
Explanation:
Given
Number of protons 
Radius of nucleus 
Distance of the electrons 
Part 1
Electric field produced by just outside its surface

Part 2
Electric field produced by just outside its surface

Part 3
The net electric field inside a uniform shell of negative charge is zero because the electric flux lines cancel out each other
hence, the solution is
Part 1 
Part 2 
Part 3 