The velocity of the object s calculated as 22.1 m/s.
<h3>What is the speed of the object?</h3>
Given that we can write that;
v^2 = u^2 + 2gh
Now u = 0 m/s because the object was dropped from a height
v^2 = 2gh
v = √2 * 9.8 * 25
v = 22.1 m/s
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Answer: Escaped volume = 0.0612m^3
Explanation:
According to Boyle's law
P1V1 = P2V2
P1 = initial pressure in the tire = 36.0psi + 14.696psi = 50.696psi (guage + atmospheric pressure)
P2 = atmospheric pressure= 14.696psi
V1 = volume of tire =0.025m^3
V2 = escaped volume + V1 ( since air still remain in the tire)
V2 = P1V1/P2
V2 = 50.696×0.025/14.696
V2 = 0.0862m^3
Escaped volume = 0.0862 - 0.025 = 0.0612m^3
Answer:
T=0.372 s, f=2.7 Hz, w=16.9 rad/s, k=179.2 N/m, v= 8.78 m/s, F= 48.4 N
Explanation:
a.)
Period: It is already given in the question "oscillator repeats its motion every 0.372 s".
So T=0.372 s
b)
frequency= f = 1/ T
f = 1/ 0.372
f=2.7 Hz
c).
Angular frequency= w= 2πf
w= 2*π*2.7
w=16.9 rad/s
d)
Spring Constant:
As w=
⇒w²= k/m
⇒k= m*w²
⇒k= 0.628 * 16.9² N/m
⇒k=179.2 N/m
e)
The mass will have maximum speed when it passes through the mean position.
At mean position
Maximum elastic potential energy = Maximum kinetic energy
1/2 k A² = 1/2 m v² ( A is amplitude of oscillation)
⇒ v=
⇒ v=
\
⇒ v= 8.78 m/s
f)
Maximum force will be exerted on the block when it is at maximum distance.
F= k* A ( A is amplitude of oscillation)
F= 179.2 * 0.27 N
F= 48.4 N
Given :
A spring with a spring constant of 1730 N/m is compressed 0.136 m from its equilibrium position with an object having a mass of 1.72 kg.
To Find :
The embankment in the height.
Solution :
Since no external force is acting in the system, therefore total energy will be conserved.
Initial kinetic energy of the object = Energy stored in spring

Also, initial potential energy is 0.
Now,

Therefore, the embankment height is 0.64 m.
I think it would be wires!