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
Answer:
v = 7.67 m/s
Explanation:
Given data:
horizontal distance 11.98 m
Acceleration due to gravity 9.8 m/s^2
Assuming initial velocity is zero
we know that

solving for t
we have

substituing all value for time t

t = 1.56 s
we know that speed is given as


v = 7.67 m/s
When an object moves its length contracts in the direction of motion. The faster it moves the shorter it gets in the direction of motion.
The object in this question moves and then stops moving. So it's length first contracts and then expands to its original length when the motion stops.
The speed doesn't have to be anywhere near the speed of light. When the object moves its length contracts no matter how fast or slow it's moving.
Answer: The electric field is: a) r<a , E0=; b) a<r<b E=ρ (r-a)/εo;
c) r>b E=ρ b (b-a)/r*εo
Explanation: In order to solve this problem we have to use the Gaussian law in diffrengios regions.
As we know,
∫E.dr= Qinside/εo
For r<a --->Qinside=0 then E=0
for a<r<b er have
E*2π*r*L= Q inside/εo in this case Qinside= ρ.Vol=ρ*2*π*r*(r-a)*L
E*2π*r*L =ρ*2*π*r* (r-a)*L/εo
E=ρ*(r-a)/εo
Finally for r>b
E*2π*r*L =ρ*2*π*b* (b-a)*L/εo
E=ρ*b* (b-a)*/r*εo