Answer:
8.3m/s
Explanation:
Given parameters:
mass of clay ball = 5kg
Speed of clay ball = 25m/s
mass of clay ball at rest = 10kg
speed of clay ball at rest = 0m/s
Unknown:
Velocity after collision = ?
Solution:
Since the balls stick together, this is an inelastic collision:
m1v1 + m2v2 = v(m1 + m2)
5(25) + 10(0) = v (5 + 10)
125 = 15v
v = 8.3m/s
Answer:
car is moving away its direction is negative.
the speed must also be negative.
speed this distance decreases the acceleration is negative and if the speed is increasing the acceleration is positive.
Explanation:
In the exercise they indicate that the direction to the motion sensor is positive, as they indicate that the car is moving away its direction is negative.
The speed of the car is
v = (x₂-x₁) / t
As the positions are negative, and the car moves away the speed must also be negative.
The analysis for acceleration must be very careful if the speed this distance decreases the acceleration is negative and if the speed is increasing the acceleration is positive.
Answer: through digested food
Explanation: cells get their sugar molecules from the food that is taken in by an organism. The food taken in is digested by the stomach into base materials like sugar, fatty acids and amino acids. These base materials are circulated through the blood till they get to the cells where they are needed.
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.
