Two equal masses are attached to separate identical springs next to one another. One mass is pulled so its spring stretches 20 c
2 answers:
Answer:
Both reaches equilibrium simultaneously.
Explanation:
Given that,
First stretch spring= 20 cm
Second stretch spring = 10 cm
Two equal masses are suspended from identical spring.
Their forces constants are equal time period for oscillation
Using formula of time period
Where, T = time period
m = mass
k = spring constant
Time period independent of amplitude.
Hence, Both reaches equilibrium simultaneously.
Answer:
both reaches at the same time.
Explanation:
The time period of the mass is given by
where, K be the spring constant and m be the mass attached.
As both the masses are same and springs are identical so K is same for both the springs.
Thus, both the masses will reach at the equilibrium position at the same time.
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Answer:
"Apparent weight during the "plan's turn" is 519.4 N
Explanation:
The "plane’s altitude" is not so important, but the fact that it is constant tells us that the plane moves in a "horizontal plane" and its "normal acceleration" is
Given that,
v = 420 m/s
R = 11000 m
Substitute the values in the above equation,
It has a horizontal direction. Furthermore, constant speed implies zero tangential acceleration, hence vector a = vector a N. The "apparent weight" of the pilot adds his "true weight" "m" "vector" "g" and the "inertial force""-m" vector a due to plane’s acceleration, vector
In magnitude,
Because vector “a” is horizontal while vector g is vertical. Consequently, the pilot’s apparent weight is vector
Which is quite heavier than his/her true weigh of 519.4 N