Answer:
v = 29.4 m / s
Explanation:
For this exercise we can use the conservation of mechanical energy
Lowest starting point.
Em₀ = K = ½ m v²
final point. Higher
= U = m g h
Let's use trigonometry to lock her up
cos 60 = y / L
y = L cos 60
Height is the initial length minus the length at the maximum angle
h = L - L cos 60
h = L (1- cos 60)
energy is conserved
Em₀ = Em_{f}
½ m v² = mgL (1 - cos 60)
v = 2g L (1- cos 60)
let's calculate
v² = 2 9.8 3.0 (1- cos 60)
v = 29.4 m / s
Answer:
F = 6666.7 N
Explanation:
Given that,
Mass of a chip, m = 0.1 mg
Initial speed, u = 0
Final speed,
Time of collision,
We know that,
Force, F = ma
Put all the values,
So, the required force is 6666.7 N.
Complete Question
Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the surface. Find the electric field at the following locations, with radially outward defined as the positive direction and radially inward defined as the negative direction. The permittivity of free space ????0 is 8.85×10−12 C/(V⋅m). What is the electric field
E⃗ 1 inside the cell at a distance of 3.05 μm from the center?
E⃗ 2 Just inside the surface of the cell
E⃗ 3 Just outside the surface of the cell
E⃗ 4 At a point outside the cell 3.05 μm from the surface
Answer:
E⃗ 1
0 V/m
E⃗ 2
0 V/m
E⃗ 3
E⃗ 4
Explanation:
From the question we are told that
The diameter is
The charge is
The permittivity of free space is
The distance considered is
Generally the electric field inside the cell at a distance of 3.05 μm from the center is
0 V/m
This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge
Generally the electric field just inside the surface of the cell is 0 V/m
This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge
Generally the electric field just outside the cell is mathematically represented as
Here is the coulomb constant with value
r is the radius of the sphere which is mathematically as
Generally the electric field at a point outside the cell 3.05 μm from the surface is mathematically represented as
Here R is mathematically represented as
=>
So
Hooke's law establishes that the elongation of a spring is directly proportional to the modulus of the force applied to it, <u>as long as the spring is not permanently deformed</u>:
(1)
Where:
is the elastic constant of the spring. The higher its value, the more work it will cost to stretch the spring.
is the length of the spring without applying force.
is the length of the spring with the force applied.
According to this, we have a spring where only the force due gravity is applied.
In other words, the force applied is the weigth of the block:
(2)
Where is the mass of the block and
is the gravity acceleration.
(3)
(4)
Knowing the force applied and
and
, we can substitute the values in equation (1) and find
:
(5)
(6)
<u>Finally:</u>