3. A 0.145 kg ball moving horizontally at 20 m/s is struck by a bat that causes the ball to move in the
1 answer:
Answer:
F = 1015 N
Explanation:
Given that,
Mass of a ball, m = 0.145 kg
Initial speed, u = 20 m/s
Final speed, v = -15 m/s (as it moves in opposite direction)
The ball and bat were in contact for 5 ms
We need to find the force the bat applies to the bat. We know that,
F = ma
So, the bat will apply 1015 N force to the ball.
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Answer: 0.333 h
Explanation:
This problem can be solved using the <u>Radioactive Half Life Formula</u>:
(1)
Where:
is the final amount of the material
is the initial amount of the material
is the time elapsed
is the half life of the material (the quantity we are asked to find)
Knowing this, let's substitute the values and find from (1):
(2)
(3)
Applying natural logarithm in both sides:
(4)
(5)
Clearing :
(6)
Finally:
This is the half-life of the Bismuth-218 isotope
Answer:
a)W=8.333lbf.ft
b)W=0.0107 Btu.
Explanation:
<u>Complete question</u>
The force F required to compress a spring a distance x is given by F– F0 = kx where k is the spring constant and F0 is the preload. Determine the work required to compress a spring whose spring constant is k= 200 lbf/in a distance of one inch starting from its free length where F0 = 0 lbf. Express your answer in both lbf-ft and Btu.
Solution
Preload = F₀=0 lbf
Spring constant k= 200 lbf/in
Initial length of spring x₁=0
Final length of spring x₂= 1 in
At any point, the force during deflection of a spring is given by;
F= F₀× kx where F₀ initial force, k is spring constant and x is the deflection from original point of the spring.
Change to lbf.ft by dividing the value by 12 because 1ft=12 in
100/12 = 8.333 lbf.ft
work required to compress the spring, W=8.333lbf.ft
The work required to compress the spring in Btu will be;
1 Btu= 778 lbf.ft
?= 8.333 lbf.ft----------------cross multiply
(8.333*1)/ 778 =0.0107 Btu.