Answer:
b) 2ft/s
Explanation:
A scalar has only magintude, not direction
6.2m, 3kg, and -100 o C are all scalars because they only have magnitude.
2ft/s is not a scalar because it has a direction.
The options are;
a. V2 equals 2V1.
b. V2 equals (V1)/2.
c. V2 equals V1.
d. V2 equals (V1)/4.
e. V2 equals 4V1.
Answer:
Option A: V2 equals 2V1
Explanation:
Since the flow is steady, then we can say;
mass flow rate at input = mass flow rate at output.
Formula for mass flow rate is;
m' = ρVA
Thus;
At input;
m'1 = ρ1•V1•A1
At output;
m'2 = ρ2•V2•A2
So, m'1 = m'2
Now, we are told that the density of the fluid decreases to half its initial value.
Thus; ρ2 = (ρ1)/2
Since m'1 = m'2, then;
ρ1•V1•A1 = (ρ1)/2•V2•A2
Now, the pipe is uniform and thus the cross section doesn't change. Thus;
A1 = A2
We now have;
ρ1•V1•A1 = (ρ1)/2•V2•A1
A1 and ρ1 will cancel out to give;
V1 = (V2)/2
Thus, V2 = 2V1
Answer:

Explanation:
To solve this exercise it is necessary to take into account the concepts related to gravitational potential energy, as well as the concept of perigee and apogee of a celestial body.
By conservation of energy we know that,

Where,

Replacing


Our values are given by,





Replacing at the equation,


Therefore the Energy necessary for Sputnik I as it moved from apogee to perigee was 
Answer:
reaching a speed of 30 meters per second in 6 seconds.
During the 6 seconds, the car has traveled
a distance of
1. 180 m
2. 60 m
3. 15 m
4. 30 m
5. 90 m
Explanation:
Answer: The given statement is false.
Explanation:
Lightening is defined as steady expansion of air within and surrounding path due to the sudden increase in pressure and temperature that leads to emission of lightning.
Basically, lightning is a state of plasma in which the molecules are very rapidly striking with each other as it also contains positively charged ions and negatively charged electrons.
Lightning always leads to the production of thunder.
Therefore, we can conclude that the statement not every lightning strike produces thunder, is false.