Stars combine Hydrogen to make Helium during nuclear fusion. Living things are made of heavier elements like Carbon, Oxygen, Iro
1 answer:
You might be interested in
Answer:
a) If the mass is doubled, then the period is increased by . Hence, the period of the system is 2.828 seconds.
b) If the mass is halved, then the period is reduced by . Hence, the period of the system is 1.414 seconds.
c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.
d) If the spring constant is doubled, then the period is reduced by . Hence, the period of the system is 1.414 seconds.
Explanation:
The statement is incomplete. We proceed to present the complete statement: <em>A block attached to a spring with unknown spring constant oscillates with a period of 2.00 s. What is the period if </em><em>a. </em><em>The mass is doubled? </em><em>b.</em><em> The mass is halved? </em><em>c.</em><em> The amplitude is doubled? </em><em>d.</em><em> The spring constant is doubled? </em>
We have a block-spring system, whose angular frequency () is defined by the following formula:
(1)
Where:
- Spring constant, measured in newtons per meter.
- Mass, measured in kilograms.
And the period (), measured in seconds, is determined by the following expression:
(2)
By applying (1) in (2), we get the following formula:
a) If the mass is doubled, then the period is increased by . Hence, the period of the system is 2.828 seconds.
b) If the mass is halved, then the period is reduced by . Hence, the period of the system is 1.414 seconds.
c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.
d) If the spring constant is doubled, then the period is reduced by . Hence, the period of the system is 1.414 seconds.
Answer:
a) x = v₀² sin 2θ / g
b) t_total = 2 v₀ sin θ / g
c) x = 16.7 m
Explanation:
This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity
sin θ = / vo
cos θ = v₀ₓ / vo
v_{oy} = v_{o} sin θ
v₀ₓ = v₀ cos θ
v_{oy} = 13.5 sin 32 = 7.15 m / s
v₀ₓ = 13.5 cos 32 = 11.45 m / s
a) In the x axis there is no acceleration so the velocity is constant
v₀ₓ = x / t
x = v₀ₓ t
the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
t = v_{o} sin θ / g
we substitute
x = v₀ cos θ (2 v_{o} sin θ / g)
x = v₀² /g 2 cos θ sin θ
x = v₀² sin 2θ / g
at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,
b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time
at the highest point the vertical speed is zero
v_{y} = v_{oy} - gt
v_{y} = 0
t = v_{oy} / g
t = v₀ sin θ / g
as the time to get on and off is the same the total time or flight time is
t_total = 2 t
t_total = 2 v₀ sin θ / g
c) we calculate
x = 13.5 2 sin (2 32) / 9.8
x = 16.7 m