Answer:
The frequency of sound heard by the boy is 1181 Hz.
Explanation:
Given that,
Frequency of sound from alarm
Speed = -8.25 m/s
Negative sign show the boy riding away from the car
Speed of sound = 343
We need to calculate the heard frequency
Using formula of frequency
Where, = frequency of source
= speed of observer
= speed of source
= speed of sound
Put the value into the formula
here, source is at rest
Hence, The frequency of sound heard by the boy is 1181 Hz.
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
Answer:
v = 6.79 m/s
Explanation:
It is given that,
Mass of a train car, m₁ = 11000 kg
Speed of train car, u₁ = 21 m/s
Mass of other train car, m₂ = 23000 kg
Initially, the other train car is at rest, u₂ = 0
It is a case based on inelastic collision as both car couples each other after the collision. The law of conservation of momentum satisied here. So,
V is the common velocity after the collisions
So, the two car train will move with a common velocity of 6.79 m/s.