The speed of A and B immediately after collision is 5.28m/s
<u>Explanation:</u>
Mass of A is 6275kg
Speed of A is 6.5m/s
Mass of B is 5155kg
Speed of B is 3.8m/s
Track is frictionless.
A and B stick together.
speed of attached A and B = ?
mₐsₐ + mᵇsᵇ = (mₐ + mb) s
Therefore, The speed of A and B immediately after collision is 5.28m/s
Answer:
A. 420 J
Explanation:
Given the following data;
Mass = 30.45 g
Specific heat capacity = 4.18 J/g °C.
Temperature = 3.3°C
To find the quantity of heat;
Heat capacity is given by the formula;
Where;
Q represents the heat capacity or quantity of heat.
m represents the mass of an object.
c represents the specific heat capacity of water.
t represents the temperature.
Substituting into the equation, we have;
Q = 420.03 ≈ 420 Joules
Answer:
Acceleration of Sea Lion is 4.41 g
This is 49% of maximum jet acceleration given as a = 9g
Explanation:
As we know that the radius of the circular loop is given as
R = 0.37 m
The speed of the fish is given as
Now the centripetal acceleration of the sea lion is given as
as we know that
so we have
Now Percentage of this acceleration wrt maximum jet acceleration is given as
%
Answer:
The force on q₁ due to q₂ is (0.00973i + 0.02798j) N
Explanation:
F₂₁ =
Where;
F₂₁ is the vector force on q₁ due to q₂
K is the coulomb's constant = 8.99 X 10⁹ Nm²/C²
r₂₁ is the unit vector
|r₂₁| is the magnitude of the unit vector
|q₁| is the absolute charge on point charge one
|q₂| is the absolute charge on point charge two
r₂₁ = [(9-5)i +(7.4-(-4))j] = (4i + 11.5j)
|r₂₁| =
(|r₂₁|)² = 148.25
= 0.050938(0.19107i + 0.54933j) N
= (0.00973i + 0.02798j) N
Therefore, the force on q₁ due to q₂ is (0.00973i + 0.02798j) N