An area with excess electrons has a net _____ charge; an area with a deficit of electrons has a net _____ charge. Question 10 op
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4.) Quantitative: notice the root word 'quantity', which is a certain percentage of something.
a.) Quantitative, since it represents a certain percentage. The weight before the camp could also be written as an improper fraction.
b.) This is also quantitative because it can be written as a ratio, which can also be written as a fraction
or percentage.
c.) This is qualitative. Joan has a higher percentage than Emile.
d.) Qualitative. It judges the quality of something instead of amount. To make it quantitative requires a fraction of percentage. In this case, maybe you could add how he can run y miles every x seconds.
5.)
a.)This depends on whether or not the air is moving. If the classroom's AC is on, or the window is open, then the correct unit to use is CFM (cubic feet per minute). If not, the common unit is milliliters, but must also include temperature and air pressure as well as the volume of the room.
b.) Simply calculate the volume of the can using inches cubed, since volume is the measure of how much space an object takes up.
c.) The unit would be . If
, then simply multiply your density by your mass.
d.) I personally would use centimeters. Inches would work too, but centimeters would be much more accurate.
e.) Use the same formula for c.), which means kilograms per cubic foot.
Hope this helped.
1) Time in the air: 0.78 s
The motion of the ball is a projectile motion, which consists of two independent motions:
- A horizontal motion with constant horizontal velocity
- A vertical motion with constant downward acceleration of
(acceleration of gravity)
The initial vertical velocity is
where the negative sign means the direction is downward.
The vertical position of the ball is given by
where
h = 4.50 m is the initial heigth of the ball when it starts falling down
The ball reaches the ground when y = 0, so we have:
This is a second-order equation; solving for t, we get
t = -1.18 s
t = 0.78 s
We discard the negative solution since it has no physical meaning, so we can say that the ball spent 0.78 s in the air.
2) Horizontal distance: 1.83 m
For this second part of the problem, we just have to consider the horizontal motion of the ball.
As we said previously, the motion of the ball along the horizontal direction is a uniform motion with constant velocity, which is given by
where u = 3.05 m/s is the initial speed and the angle of projection.
For a uniform motion, we can use the following relationship between distance covered and velocity:
and substituting t = 0.78 s, we find the total distance travelled along the horizontal direction by the ball before reaching the ground: