ANSWER:
The answer will be OT
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.
Coffee mug is cylindrical shape. Therefore,
Volume of mug = volume of cylinder = πr²h
r = inner radius of mug. h = height of coffee = 6 cm.
Density of coffee = Density of water = 1 g/ml
Therefore, volume of coffee = mass/density = 400/1 = 400 ml = 400 cm³
Volume of coffee = πr²h
400= πr²(6)
r² = 21.23
r = inner radius of mug = 4.607 cm
Answer:
1.6 m
Explanation:
Given that the launch velocity of a toy car launcher is determined to be 5 m/s. If the car is to be launched from a height of 0.5 m.
The time for landing should be calculated by using the second equation of motion formula
h = Ut + 1/2gt^2
Let U = 0
0.5 = 1/2 × 9.8 × t^2
0.5 = 4.9t^2
t^2 = 0.5 / 4.9
t^2 = 0.102
t = 0.32 s
The target should be placed so that the toy car lands on it at:
Distance = 5 × 0.32
distance = 1.597 m
Distance = 1.6 m
Therefore, the target should be placed so that the toy car lands on it 1.6 metres away.
