How is fitness walking beneficial?
2 answers:
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Answer:
Time period of the osculation will be 0.0671 sec
Explanation:
It is given a vertical spring is stretched by 4 cm
So change in length of the spring x = 4 cm = 0.04 m
Mass which is hung from it m = 12 gram = 0.012 kg
Sprig force will be equal to weight of the mass
So
k = 244.7 N/m
Now new mass is m = 28 gram = 0.028 kg
So time period with new mass will be
(a) 30.9 m
Let's analyze the motion of the parachutist. Its vertical position above the ground is given by
where
h = 45 m is the initial height
u = 10 m/s is the initial velocity (upward)
t is the time
g = -9.8 m/s^2 is the acceleration of gravity (downward)
Substituting t=3 s , we find the height of the parachutist when it opens the parachute:
(b) 44.1 m
Here we have to find first the height of the balloon 3 seconds after the parachutist has jumped off from it. The vertical position of the balloon is given by
y = h + ut
where
h = 45 m is the initial height
u = 10 m/s is the initial velocity (upward)
t is the time
Substituting t = 3 s, we find
y = 45 m + (10 m/s)(3 s) = 75 m
So the distance between the balloon and the parachutist after 3 s is
d = 75 m - 30.9 m = 44.1 m
(c) 8.2 m/s downward
The velocity of the parachutist at the moment he opens the parachute is:
v = u +gt
where
u = 10 m/s is the initial velocity (upward)
t is the time
g = -9.8 m/s^2 is the acceleration of gravity (downward)
Substituting t = 3 s,
v = 10 m/s + (-9.8 m/s^2)(3 s)= -19.4 m/s
where the negative sign means it is downward
After t=3 s, the parachutist open the parachute and it starts moving with a deceleration of
a =+5 m/s^2
where we put a positive sign since this time the acceleration is upward.
The total distance he still has to cover till the ground is
d = 30.9 m
So we can find the final velocity by using
where this time we have u = 19.4 m/s as initial velocity. Taking the downward direction as positive, the deceleration must be considered as negative:
a = -5 m/s^2
Solving for v,
(d) 5.24 s
We can find the duration of the second part of the motion of the parachutist (after he has opened the parachute) by using
where
a = -5 m/s^2 is the deceleration
v = 8.2 m/s is the final velocity
u = 19.4 m/s is the initial velocity
t is the time
Solving for t, we find
And added to the 3 seconds between the instant of the jump and the moment he opens the parachute, the total time is
t = 3 s + 2.24 s = 5.24 s
The glassware with the highest degree of accuracy is 250 mL beaker.
<h3>What is accuracy?</h3>When in an experiment, a value is measured 5 times, then if the values measured are same for most of the time or like three times out of five, it said to be accurate. The phenomenon is accuracy.
Accuracy compares the experimental value to the theoretical value.
Accepted density of water = 0.99 g/mL
1. 10 mL cylinder
Mass of water = 6.76 g
Volume of water = 6.8 mL
Density = mass /Volume
density = 6.76/6.8 = 0.99412 g/mL
Percentage error = [Observed value - Accepted value/ Accepted value ] x 100
% error = [0.99412 - 0.99 / 0.99 ] x 100
% error = 0.416 %
2. 50 mL cylinder
Mass of water = 24 g
Volume of water = 24.2 mL
density = 24/24.2 = 0.9917 g/mL
Percentage error = [Observed value - Accepted value/ Accepted value ] x 100
% error = [0.9917 - 0.99 / 0.99 ] x 100
% error = 0.172 %
3. 25 mL cylinder
Mass of water = 17 g
Volume of water = 17.1 mL
density = 17/17.1 = 0.99415 g/mL
Percentage error = [Observed value - Accepted value/ Accepted value ] x 100
% error = [0.99415 - 0.99 / 0.99 ] x 100
% error = 0.419 %
4. 250 mL beaker
Mass of water = 35 g
Volume of water = 35.3 mL
density = 35/35.3 = 0.9915 g/mL
Percentage error = [Observed value - Accepted value/ Accepted value ] x 100
% error = [0.9915 - 0.99 / 0.99 ] x 100
% error = 0.152 %
Lesser the percentage error, higher is the accuracy.
Thus, 250 mL beaker has the highest degree of accuracy.
Learn more about accuracy.
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