I believe c is correct......
It would be option A (a decrease in mass with an increase in volume)
Answer:
Explanation:
Given that
The window height is 2m
And the window is 7.5m from the ground
Then the total height of the window from the ground is 7.5+2=9.5m
It takes the ball 0.32sec travelled pass the window.
When the ball get to the window, it has an initial velocity (u') and when it gets to the top of the window it has a final velocity ( v')
Now using the equation of free fall during this window travels
S=ut-½gt² against motion.
S=2, g=9.81, t=0.32sec
Then,
S=u't-½gt²
2=u'×0.32-½×9.81×0.32²
2=0.32u'-0.5023
2+0.5032=0.32u'
Then, 0.32u'=2.5032
u'=2.5032/0.32
u'=7.82m/s
This is the initial velocity as the ball got the the window
Now, let analyse from the window bottom to the ground which is a distance of 7.5m
Using the equation of free fall again
v²=u²-2gH
In this case the final velocity (v) is the velocity when the ball reach the bottom of the window i.e u'=7.82m/s,
While u is the original initial velocity from the throw of the ball
Then,
u'²=u²-2gH
7.82²=u²-2×9.81×7.5
61.146=u²-147.15
61.146+147.15=u²
Then, u²=208.296
So, u=√208.296
u=14.43m/s
The initial velocity of the ball form the throw is 14.43m/s
Answer:
a) k = 891.82 N/m
b) e = 0.0143 m = 1.43 cm
c) W = 5.02 J
Explanation:
Step 1: Data given
Mass = 2.60 kg
the spring stretches 2.86 cm = 0.0286
Step 2: What is the force constant of the spring?
Force constant, k = force applied / extension produced
k = (2.60kg * 9.81N/kg) / 0.0326 m
k = 891.82 N/m
b) If the 2.60-kg object is removed, how far will the spring stretch if a 1.30-kg block is hung on it
Extension = F/k = (1.30 kg * 9.81) / 891.82 = 0.0143 m = 1.43 cm
Half the mass means half the extension
c) How much work must an external agent do to stretch the same spring 7.50 cm from its unstretched position?
W = average force used * distance
W = 1/2 * k*e * e = 1/2 k*e²
W = 1/2 * 891.82 * (0.075)² = W = 5.02 J
