Mass is the physical quantity
Answer:
2.23 × 10^6 g of F- must be added to the cylindrical reservoir in order to obtain a drinking water with a concentration of 0.8ppm of F-
Explanation:
Here are the steps of how to arrive at the answer:
The volume of a cylinder = ((pi)D²/4) × H
Where D = diameter of the cylindrical reservoir = 2.02 × 10^2m
H = Height of the reservoir = 87.32m
Therefore volume of cylindrical reservoir = (3.142×202²/4)m² × 87.32m = 2798740.647m³
1ppm = 1g/m³
0.8ppm = 0.8 × 1g/m³
= 0.8g/m³
Therefore to obtain drinking water of concentration 0.8g/m³ in a reservoir of volume 2798740.647m³, F- of mass = 0.8g/m³ × 2798740.647m³ = 2.23 × 10^6 g must be added to the tank.
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Answer:
D. Newton's first law
Explanation:
Newton's first law of inertia says that an object will remain how it is, unless affected by an outside force. In this case, the plates want to remain stationary(not moving). Therefore, if you pull the table cloth fast enough, the force of friction produced will be small enough so that the Inertia of the plates will overcome the force of friction.
V = f x lamda
Wave speed = frequency x wavelength
2x1 = 2 so wave speed = 2m/s
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Answer:
3. 0.5 sec.
Explanation:
A bullet fired horizontally follows a projectile motion, which consists of two independent motions:
- A horizontal motion with constant speed
- A vertical motion with constant acceleration, g = 9.8 m/s^2, towards the ground
The time taken for the bullet to reach the ground can be calculated just by considering the vertical motion:
where y is the vertical position at time t, h is the initial height, and 
is the initial vertical velocity of the bullet.
Since the bullet is fired horizontally, 
. So the equation becomes
And the time that the bullet takes to reach the ground can be found by requiring y=0 and solving for t:
As we can see, in this equation there is no dependance on the initial speed of the bullet: therefore, if the bullet is fired still horizontally but with a different speed, it will still take the same time (0.5 s) to reach the ground.
