0 mark is your answer because you want to start at 0
acceleration = change in velocity/change in time
so...
a = 20 m/s / 2 seconds
a = 10
hope that helps :)
P.S. found this from Brainly User, sometimes all you have to do is search to find the answer.
Answer:
encontré esto fuera de google, "Estos conceptos físicos que parecen necesarios en cualquier teoría física suficientemente amplia son los llamados conceptos físicos fundamentales, una lista no exhaustiva de los mismos podría ser: espacio, tiempo, energía, masa, carga eléctrica, etc."
espero que esto ayude, que tengas un gran día, Y mantente segura! :) :D :3
Answer:
Option C. 210 J.
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 0.75 Kg
Height (h) = 12 m
Velocity (v) = 18 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Total Mechanical energy (ME) =?
Next, we shall determine the potential energy of the plane. This can be obtained as follow:
Mass (m) = 0.75 Kg
Height (h) = 12 m
Acceleration due to gravity (g) = 9.8 m/s²
Potential energy (PE) =?
PE = mgh
PE = 0.75 × 9.8 × 12
PE = 88.2 J
Next, we shall determine the kinetic energy of the plane. This can be obtained as follow:
Mass (m) = 0.75 Kg
Velocity (v) = 18 m/s
Kinetic energy (KE) =?
KE = ½mv²
KE = ½ × 0.75 × 18²
KE = ½ × 0.75 × 324
KE = 121.5 J
Finally, we shall determine the total mechanical energy of the plane. This can be obtained as follow:
Potential energy (PE) = 88.2 J
Kinetic energy (KE) = 121.5 J
Total Mechanical energy (ME) =?
ME = PE + KE
ME = 88.2 + 121.5
ME = 209.7 J
ME ≈ 210 J
Therefore, the total mechanical energy of the plane is 210 J.
Answer:
0.893 rad/s in the clockwise direction
Explanation:
From the law of conservation of angular momentum,
angular momentum before impact = angular momentum after impact
L₁ = L₂
L₁ = angular momentum of bullet = + 9 kgm²/s (it is positive since the bullet tends to rotate in a clockwise direction from left to right)
L₂ = angular momentum of cylinder and angular momentum of bullet after collision.
L₂ = (I₁ + I₂)ω where I₁ = rotational inertia of cylinder = 1/2MR² where M = mass of cylinder = 5 kg and R = radius of cylinder = 2 m, I₂ = rotational inertia of bullet about axis of cylinder after collision = mR² where m = mass of bullet = 0.02 kg and R = radius of cylinder = 2m and ω = angular velocity of system after collision
So,
L₁ = L₂
L₁ = (I₁ + I₂)ω
ω = L₁/(I₁ + I₂)
ω = L₁/(1/2MR² + mR²)
ω = L₁/(1/2M + m)R²
substituting the values of the variables into the equation, we have
ω = L₁/(1/2M + m)R²
ω = + 9 kgm²/s/(1/2 × 5 kg + 0.02 kg)(2 m)²
ω = + 9 kgm²/s/(2.5 kg + 0.02 kg)(4 m²)
ω = + 9 kgm²/s/(2.52 kg)(4 m²)
ω = +9 kgm²/s/10.08 kgm²
ω = + 0.893 rad/s
The angular velocity of the cylinder bullet system is 0.893 rad/s in the clockwise direction-since it is positive.
