Which statement describes the relationship between bond strength and the melting and boiling points of a substance? A. As the fo
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The acceleration exerted by the object of mass 10 kg is
Answer: Option A
<u>Explanation:</u>
According to Newton’s second law of motion, any external force acting on a body will be directly proportional to the mass of the body as well as acceleration exerted by the body. So, the net external force acting on any object will be equal to the product of mass of the object with acceleration exerted by the object. Thus,
So,
As the force acting on the object is stated as 10 N and the mass of the object is given as 10 kg, then the acceleration will be
So, the acceleration exerted by the object of mass 10 kg is
Answer:
a) t = 746 s
b) t = 666 s
Explanation:
a)
- Total time will be the sum of the partial times between stations plus the time stopped at the stations.
- Due to the distance between stations is the same, and the time between stations must be the same (Because the train starts from rest in each station) we can find total time, finding the time for any of the distance between two stations, and then multiply it times the number of distances.
- At any station, the train starts from rest, and then accelerates at 1.1m/s2 till it reaches to a speed of 95 km/h.
- In order to simplify things, let's first to convert this speed from km/h to m/s, as follows:
- Applying the definition of acceleration, we can find the time traveled by the train before reaching to this speed, as follows:
- Next, we can find the distance traveled during this time, assuming that the acceleration is constant, using the following kinematic equation:
- In the same way, we can find the time needed to reach to a complete stop at the next station, applying the definition of acceleration, as follows:
- We can find the distance traveled while the train was decelerating as follows:
- Finally, we need to know the time traveled at constant speed.
- So, we need to find first the distance traveled at the constant speed of 26.4m/s.
- This distance is just the total distance between stations (3.0 km) minus the distance used for acceleration (x₁) and the distance for deceleration (x₃), as follows:
- x₂ = L - (x₁+x₃) = 3000 m - (316.8 m + 158.4 m) = 2525 m (6)
- The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:
- Total time between two stations is simply the sum of the three times we have just found:
- t = t₁ +t₂+t₃ = 24 s + 95.6 s + 12 s = 131.6 s (8)
- Due to we have six stations (including those at the ends) the total time traveled while the train was moving, is just t times 5, as follows:
- tm = t*5 = 131.6 * 5 = 658.2 s (9)
- Since we know that the train was stopped at each intermediate station for 22s, and we have 4 intermediate stops, we need to add to total time 22s * 4 = 88 s, as follows:
- Ttotal = tm + 88 s = 658.2 s + 88 s = 746 s (10)
b)
- Using all the same premises that for a) we know that the only difference, in order to find the time between stations, will be due to the time traveled at constant speed, because the distance traveled at a constant speed will be different.
- Since t₁ and t₃ will be the same, x₁ and x₃, will be the same too.
- We can find the distance traveled at constant speed, rewriting (6) as follows:
- x₂ = L - (x₁+x₃) = 5000 m - (316.8 m + 158.4 m) = 4525 m (11)
- The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:
- Total time between two stations is simply the sum of the three times we have just found:
- t = t₁ +t₂+t₃ = 24 s + 171.4 s + 12 s = 207.4 s (13)
- Due to we have four stations (including those at the ends) the total time traveled while the train was moving, is just t times 3, as follows:
- tm = t*3 = 207.4 * 3 = 622.2 s (14)
- Since we know that the train was stopped at each intermediate station for 22s, and we have 2 intermediate stops, we need to add to total time 22s * 2 = 44 s, as follows:
- Ttotal = tm + 44 s = 622.2 s + 44 s = 666 s (15)
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