The speed of the pin after the elastic collision is 9 m/s east.
<h3>
Final speed of the pin</h3>
The final speed of the pin is calculated by applying the principle of conservation of linear momentum as follows;
m1u1 + mu2 = m1v1 + m2v2
where;
- m is the mass of the objects
- u is the initial speed of the objects
- v is the final speed of the objects
4(1.4) + 0.4(0) = 4(0.5) + 0.4v2
5.6 = 2 + 0.4v2
5.6 - 2 = 0.4v2
3.6 = 0.4v2
v2 = 3.6/0.4
v2 = 9 m/s
Thus, The speed of the pin after the elastic collision is 9 m/s east.
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The spring constant is 4 N/m
Explanation:
When a spring is stretched/compressed by the application of a force, the relationship between the magnitude of the force applied and the elongation of the spring is given by Hooke's law:
where
F is the magnitude of the spring applied
k is the spring constant
x is the elongation of the spring, relative to its equilibrium position
For the spring in this problem, we have:
F = 0.12 N (force applied)
x = 3 cm = 0.03 m (elongation of the spring)
Therefore, we can solve the formula for k to find the spring constant:
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To do this we may use things that are good conductors - are painted dull black -
Have a air flow around them Maximised.
Answer:
Explanation:
a ) from San Antonio to Houston let distance be d km .
Average speed = total distance / total time
time = distance / speed
Total time = (d / 2 x 75 ) +( d / 2 x 106 )
= .0067 d + .0047 d
= .0114 d
Average speed = d / .0114 d = 87.72 km /h
b ) from Houston back to San Antonio
Total time = (d / 2 x 106 ) +( d / 2 x 75 )
= .0047 d + .0067 d
= .0114 d
Average speed = d / .0114 d = 87.72 km /h
c )
For entire trip :
total distance = 2d
total time = 2 x .0114 d
Average speed = 2 d / 2 x .0114 d
= 87.72 km /h .
