Answer:
5m/s²
Explanation:
Given parameters:
Mass of wagon = 10kg
Force of pull = 70N
Frictional force = 20N
Unknown:
Acceleration of the wagon = ?
Solution:
Frictional force is a force that opposes motion.
The net force is given as:
Net force = mass x acceleration
Force of pull - Frictional force = mass x acceleration
Insert the parameters and solve;
70 - 20 = 10 x acceleration
50 = 10 x acceleration
Acceleration = 5m/s²
Answer: 217.52 N
Explanation: The applied force is 20 N, the distance covered is 12.0 m and the angle is 25° above the horizontal.
Hence the formulae that defines work done is given by
W = Force × distance
But since the force has been inclined at an angle θ above the horizontal, the horizontal component of force is neccesary to produce the required motion to make the child do work on the wagon.
Hence
Work done = (horizontal component of force) × distance
Work done = F cos θ × distance
Work done = 20 cos 25 × 12 = 217.52 N
Given:
u = 10⁵ m/s, the entrance velocity
v = 2.5 x 10⁶ m/s, the exit velocity
s = 1.6 cm = 0.016 m, distance traveled
Let a = the acceleration.
Then
u² + 2as = v²
(10⁵ m/s)² + 2*(a m/s²)*(0.016 m) = (2.5 x 10⁶ m/s)²
0.032a = 6.25 x 10¹² - 10¹⁰ = 6.24 x 10¹²
a = 1.95 x 10¹⁴ m/s²
Answer: 1.95 x 10¹⁴ m/s²
The rate constant of a reaction can be computed by the ratio of the changes in the concentration and time take taken for it to decompose. Thus, if the rate constant is given to be 14 M/s, we have
where C are the concentration values and t is the time taken for it to decompose.
Thus, it will take 0.003 s for it to decompose.
Answer: 0.003 s
