True or false questions for work and power.
1 answer:
You might be interested in
Answer:
It will still hover until the spaceship "hits" or exerts a force on it.
Explanation:
Remember, if there is no net force, there is no acceleration or movement.
In this case, our ball is hovering in the spaceship, and in space, we can assume there is no , and we can assume there is no
, nor no forces acting against it.
So, the ball would not move.
However, once the spaceship starts accelerating, the ball would still hover until the spaceship exerts a force on it.
This is because of the same thing as explained above, no forces acting on it, therefore, no acceleration.
Think about it this way.
Imagine you jumped up, then someone threw a ball at you. Now let's imagine you can't move until you hit the floor, meaning that in an ideal situation only is acting on you. Now again, let's imagine time slows really down for you, but not the ball. Before the ball comes and hits you, you are "hovering" like a ball. But after the ball hits you, you move a little because the ball exerted a force on you.
If you did not understand what I meant above, just forget about it, and think about the fact that if there is a Net force (all the force values added up), then there is acceleration and movement.
The cost of excavating a space of 84 ft long, 42 ft wide, and 9 ft deep is $45864
Information about the problem:
- Space long= 84 ft
- Space wide= 42 ft
- Space deep= 9 ft
- Cost by yard3 = $39/yd3
- Total cost= ?
To solve this problem, we have to state the equation using the information of the problem:
Calculating the volume of the total space:
space volume = space long * space wide * space deep
space volume = 84 ft * 42 ft * 9 ft
space volume = 31752 ft3
By converting the volume from ft3 to yd3, we have:
31752 ft3 * (0,037037 yd3 / 1 ft3) = 1176 yd3
Calculating the cost of excavating the volume space:
Total cost = space volume * cost by yard3
Total cost = 1176 yd3 * $39/yd3
Total cost = $45864
<h3>What is volume?</h3>It is the space occupied by a body, it is calculated by multiplying its dimensions, for example: length, height and width.
Learn more about volume at: brainly.com/question/12628341
#SPJ4
The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.
<h3>How to determine the friction factor</h3>Using the formula
μ = viscosity = 0. 06 Pas
d = diameter = 120mm = 0. 12m
V = velocity = 1m/s and 3m/s
ρ = density = 0.9
a. Velocity = 1m/s
friction factor = 0. 52 ×
friction factor = 0. 52 ×
friction factor = 0. 52 × 0. 55
friction factor
b. When V = 3mls
Friction factor = 0. 52 ×
Friction factor = 0. 52 ×
Friction factor = 0. 52 × 0. 185
Friction factor
Loss When V = 1m/s
Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter
Head loss = 0. 289 × ×
×
Head loss = 1. 80 × 10^8
Head loss When V = 3m/s
Head loss = ×
×
×
Head loss = 5. 3× 10^8
Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.
Learn more about friction here:
#SPJ1