The speed of the 0.8 kg ball immediately after collision is 0.625 m/s in opposite direction to the stationary ball.
The given parameters;
- mass of the ball, m₁ = 0.8 kg
- speed of the ball, u₁ = 2.5 m/s
- mass of the object at rest, m₂ = 2.5 kg
- final velocity of the object at rest, v₂ = 1 m/s
Let the final velocity of the 0.8 kg ball immediately after collision = v₁
Apply the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(0.8 x 2.5) + (2.5 x 0) = (0.8)v₁ + 2.5(1)
2 = 2.5 + (0.8)v₁
-0.5 = (0.8)v₁
Thus, the speed of the 0.8 kg ball immediately after collision is 0.625 m/s in opposite direction to the stationary ball.
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Answer:
The temperature is 2584.5 K
Explanation:
Given:
Activation energy 
Preexponential 
Diffusion flux 
Thickness of plate 
m
Concentration of carbon at two faces 
From the formula of temperature in terms of diffusion flux,
Where 
8.314 
( gas constant )
Put the values and find the temperature,
K
Therefore, the temperature is 2584.5 K
Answer:
Explanation:
Let the potential difference between the middle point and one of the plate be ΔV .
electric potential energy will be lost and it will be converted into kinetic energy .
Electrical potential energy lost = Vq , where q is charge on charge particle .
For proton
ΔV× q = 1/2 M V² ( kinetic energy of proton )
where M is mass and V be final velocity of proton .
For electron
ΔV× q = 1/2 m v² ( kinetic energy of electron )
where m is mass and v be final velocity of electron . Charges on proton and electron are same in magnitude .
As LHS of both the equation are same , RHS will also be same . That means the kinetic energy of both proton and electron will be same
1/2 M V² = 1/2 m v²
(V / v )² = ( m / M )
(V / v ) = √ ( m / M )
In other words , their velocities are inversely proportional to square root of their masses .
Answer:
θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²
Explanation:
This is an angular kinematic exercise the equation for the angular position
the particle A
θ = θ₀ + ω₀ t + ½ α t²
They say for the particle B
w₀B = ½ w₀
αB = 2 α
In addition, the particle begins at a time t_1 after particle A, in order to use the same timer, we must subtract this time from the initial
t´ = t - t_1
l
et's write the equation of particle B
θ = θ₀ + w₀B t´ + ½ αB t´2
replace
θ = θ₀ + ½ w₀ (t -t_1) + ½ 2α (t -t_1)²
θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²
The answer is B, Law of Kinetic Energy
