Answer:
D) - 0.72 secs
Explanation:
Parameters given:
Height of bridge = 40ft = 12.19 m
Initial velocity of Bill's stone = 0m/s
Initial velocity of Ted's stone = 10m/s
We find the time it take Bill's stone to bit the river and the time it takes Ted's stone to hit the river. Then we find the time difference.
Using one of the equations of motion:
For Bill:
S = ut + ½gt²
Where g = 9.8 m/s
12.19 = 0 + ½*9.8*t²
t² = 12.19/4.9 = 2.49
t = 1.58 secs
For Ted:
S = uT + ½gT²
12.19 = 10*T + ½*9.8*T²
=> 4.9T² + 10T - 12.19 = 0
Using quadratic formula and retaining only the positive value, we get that:
T = 0.86 secs
Time difference between Bill's throw and Ted's throw is:
0.86 - 1.58 = - 0.72 secs
In reality, this means that Ted must throw his stone 0.72 secs before Bill throws his for both stones to land the same time.
The particles in diagram 3 move fast enough to break away from each other.
Answer:
(a) Magnitude of static friction force is 109 N
(b) Minimum possible value of static friction is 0.356
Solution:
As per the question;
Horizontal force exerted by the girl, F = 109 N
Mass of the crate, m = 31.2 kg
Now,
(a) To calculate the magnitude of static friction force:
Since, the crate is at rest, the forces on the crate are balanced and thus the horizontal force is equal to the frictional force, f:
F = f = 109 N
(b) The maximum possible force of friction between the floor and the crate is given by:
where
N = Normal reaction = mg
Thus
For the crate to remain at rest, The force exerted on the crate must be less than or equal to the maximum force of friction.
1. that's 2 joules 1*2 = 2 j
2.that will be 20*3.5 = 70 j
3.that will be = 500*2.2= 1100 j