Answer:
longitudinal wave
Explanation:
it is perpendicular to the direction of the wave
Answer:
H = 109.14 cm
Explanation:
given,
Assume ,
Total energy be equal to 1 unit
Balance of energy after first collision = 0.78 x 1 unit
= 0.78 unit
Balance after second collision = 0.78 ^2 unit
= 0.6084 unit
Balance after third collision = 0.78 ^3 unit
= 0.475 unit
height achieved by the third collision will be equal to energy remained
H be the height achieved after 3 collision
0.475 ( m g h) = m g H
H = 0.475 x h
H = 0.475 x 2.3 m
H = 1.0914 m
H = 109.14 cm
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.
Answer:
The distance the ball moves up the incline before reversing its direction is 3.2653 m.
The total time required for the ball to return to the child’s hand is 3.2654 s.
Explanation:
When the girl is moving up:
The final velocity (v) = 0 m/s
Initial velocity (u) = 4 m/s
a = -0.25g = -0.25*9.8 = -2.45 m/s². (Negative because it is in opposite of the velocity and also it deaccelerates while going up).
Let time be t to reach the top.
Using
v = u + a×t
0 = 4 - 2.45*t
t = 1.6327 s
Since, this is the same time the ball will come back. So,
<u>Total time to go and come back = 2* 1.6327 = 3.2654 s
</u>
To find the distance, using:
v² = u² + 2×a×s
0² = 4² + 2×(-2.45)×s
s = 3.2653 m
<u>Thus, the distance the ball moves up the incline before reversing its direction is 3.2653 m.</u>