In what way are liquids different from solids?.
1 answer:
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Answer:
Velocity of truck is 21.875 m/s.
Explanation:
Given:
Mass of the car (m) = 2500 kg
Initial speed of the car (u) = 35 m/s
Mass of the truck (M) = 4000 kg
Initial speed of the truck (U) = 0 m/s (Rest)
As per question, the total momentum of the car gets transferred to the truck.
Therefore, the final momentum of the car is 0.
Momentum is given as the product of mass and velocity.
So, if momentum becomes zero means the velocity becomes 0.
Therefore, final velocity of the car (v) = 0 m/s
Let the final velocity of truck be 'V' m/s.
Now, during collision, the total momentum remains conserved.
Initial momentum = Final momentum.
Initial momentum of car + Initial momentum of truck = Final momentum of car + Final momentum of truck.
⇒
Therefore, the velocity of the truck is 21.875 m/s.
Answer:
E = Q / 4 π ε₀ (r-R_int) / (R_put³ -R_int³)
Explanation:
For this exercise we can use Gauss's law
Ф = ∫ E. dA = / ε₀
Where q is the charge inside the surface.
In this case the surface must be a sphere, the electric field lines and the radius of the sphere are parallel, so the scalar product is reduced to the algebraic product
∫ E dA = q_{int}/ ε₀
The area of a sphere is
A = 4π r²
- The electric field for a distance r < R_int
The charge inside is zero, so the electric field
E = 0 r <R_in t
- The field for a radio inside the shell
Let's use the concept of density
ρ = Q / V
q = ρ (4/3 π r³)
dq = ρ 4π r² dr
We substitute in the Gaussian equation
E ∫ dA = ρ 4π r² dr / ε₀
E 4π r² = ρ 4π/ε₀ r³ / 3
E = ρ / 3ε₀ r
We evaluate between the lower limit r = R_int, E = 0 and the upper limit r = r, E = E
E- 0 = ρ / 3ε₀ (r –R_int)
Density is
ρ = q / 4/3 π (R_out³ - R_int³)
Where R <r
E = Q / 4 π ε₀ (r-R_int) / (R_put³ -R_int³)