The initial velocity of a car that accelerates at a constant rate of 3m/s² for 5 seconds is 12m/s.
CALCULATE INITIAL VELOCITY:
The initial velocity of the car can be calculated by using one of the equation of motion as follows:
V = u + at
Where;
- V = final velocity (m/s)
- u = initial velocity (m/s)
- a = acceleration due to gravity (m/s²)
- t = time (s)
According to this question, a car accelerates at a constant rate of 3 m/s² for 5 seconds. If it reaches a velocity of 27 m/s, its initial velocity is calculated as follows:
u = v - at
u = 27 - 3(5)
u = 27 - 15
u = 12m/s.
Therefore, the initial velocity of a car that accelerates at a constant rate of 3m/s² for 5 seconds is 12m/s.
Learn more about motion at: brainly.com/question/974124
Answer:
Explanation:
plate separation = 2.3 x 10⁻³ m
capacity C₁ = ε A / d
= ε A / 2.3 x 10⁻³
C₂ = ε A / 1.15 x 10⁻³
= 
a ) when charge remains constant
energy = 
q is charge and C is capacity
energy stored initially E₁= 
energy stored finally E₂ = 
= 
= 
= 
= 4.19 J
b )
In this case potential diff remains constant
energy of capacitor = 1/2 C V²
energy is proportional to capacity as V is constant .


= 16.76 .
<em>500 sec</em>
<em>8 min 20 sec</em>
<em>Hi there !</em>
<em />
<em>8 m ................ 1 s </em>
<em>4000 m ........ x s</em>
<em>x = 4000m×1s/8m = 500 sec = 8 min 20 sec</em>
<em />
<em>Good luck ! </em>
Answer:
Δμ = hΔf/B
Explanation:
If the photon energy , ΔE = hΔf where Δf = small frequency shift and since the potential energy change of the magnetic dipole moment μ in magnetic field B from parallel to anti-parallel state is ΔU = ΔμB. where Δμ = small shift in magnetic moment.
Since the magnetic energy change equals the photon energy,
ΔE = ΔU
hΔf = ΔμB
Δμ = hΔf/B
Solution :
Given data is :
Density of the milk in the tank, 
Length of the tank, x = 9 m
Height of the tank, z = 3 m
Acceleration of the tank, 
Therefore, the pressure difference between the two points is given by :

Since the tank is completely filled with milk, the vertical acceleration is 

Therefore substituting, we get




Therefore the maximum pressure difference in the tank is Δp = 47.87 kPa and is located at the bottom of the tank.