Answer:3.56 nanosecond
In this case, you are asked the time and given the light distance(3.5ft)
To answer this question you would need to know the velocity of light. Speed of light is <span>299792458m/s. Then the calculation would be:
time= distance/speed
time= 3.5 ft / (</span>299792458m/s) x 0.3048 meter/ 1 ft= 3.56

second or 3.56 nanosecond
Answer:
the ship's energy is greater than this and the crew member does not meet the requirement
Explanation:
In this exercise to calculate kinetic energy or final ship speed in the supply hangar let's use the relationship
W =∫ F dx = ΔK
Let's replace
∫ (α x³ + β) dx = ΔK
α x⁴ / 4 + β x = ΔK
Let's look for the maximum distance for which the variation of the energy percent is 10¹⁰ J
x (α x³ + β) =
- K₀
= K₀ + x (α x³ + β)
Assuming that the low limit is x = 0, measured from the cargo hangar
Let's calculate
= 2.7 10¹¹ + 7.5 10⁴ (6.1 10⁻⁹ (7.5 10⁴) 3 -4.1 10⁶)
Kf = 2.7 10¹¹ + 7.5 10⁴ (2.57 10⁶ - 4.1 10⁶)
Kf = 2.7 10¹¹ - 1.1475 10¹¹
Kf = 1.55 10¹¹ J
In the problem it indicates that the maximum energy must be 10¹⁰ J, so the ship's energy is greater than this and the crew member does not meet the requirement
We evaluate the kinetic energy if the System is well calibrated
W = x F₀ =
–K₀
= K₀ + x F₀
We calculate
= 2.7 10¹¹ -7.5 10⁴ 3.5 10⁶
= (2.7 -2.625) 10¹¹
= 7.5 10⁹ J
The angular speed of the device is 1.03 rad/s.
<h3>What is the conservation of angular momentum?</h3>
A spinning system's ability to conserve angular momentum ensures that its spin will not change until it is subjected to an external torque; to put it another way, the rotation's speed will not change as long as the net torque is zero.
Using the conservation of angular momentum

Here, = the system's angular momentum before the collision
= 0 + mv
= (0.005)(450)(0.752)
= 1.692 kgm²/s
The moment of inertia of the system is given by
I = 2(M₁R₁² + M₂R₂²)+ mR₁²
= 2[(1.2)(0.8)² +(0.5)(0.3)²]+0.005(0.8)²
= 1.6292 kgm²
Here, = Iω
So,
1.692 = 1.6292(ω)
ω = 1.03 rad/s
To know more about the conservation of angular momentum, visit:
brainly.com/question/1597483
#SPJ1
Based on Hooke's law, the spring constant of the the body's muscle mechanism is the ratio of force to extension, the effective mass is m/3 and the potential energy that can be stored is ke^2 / 2.
<h3>What is the spring constant?</h3>
The spring constant or stiffness constant of an elastic spring is constant which describes the extent a bit forceapplied to an elastic spring will extend it.
- Spring constant, K = force/extension
Assuming, a body's muscle mechanism is a spring obeying Hooke's law, the effective mass of the spring with mass m is 1/3 of the mass of the spring = m/3
The potential energy that can be stored = ke^2 / 2
where K is spring constant and e is the extension produced.
Therefore, the spring constant of the the body's muscle mechanism is the ratio of force to extension, the effective mass is m/3 and the potential energy that can be stored is ke^2 / 2.
Learn more about Hooke's law at: brainly.com/question/12253978