Temperature is how hot or cold something is,barometric is air pressure(when the pressures high,the weather is dry)humidity is how moist the air is and how many water particles are there, wind speed and direction is how the hot and cold air is moving and how fast,which makes wind (high to low, hot to cold) precipitation is how much it is raining at that point.
Answer:
0.5A
Explanation:
Using 
,
R is the resistance (in Ohms)
V is the voltage (in V)
I is the current (in A)
I = 0.5A
Answer:
The average velocity has magnitude = 10 km/h , direction: east
Explanation:
In order to find the average velocity of the car we need to know the final and initial positions, and the time that took to get from one to the other.
Notice that since its movement was 60 km straight east and then from there 40 km straight west, the car is positioned at 20 km to the east of its initial departure point. therefore the vector change in position is a vector 20 km in magnitude, and direction towards the east.
Since it took the car a total of 1.33 hours plus 0.67 hours to reach its final position, the total time elapsed is: 1.33 + 0.67 hours = 2 hours.
Then,the velocity vector has magnitude; 20 km / 2 hours = 10 km/hour
As we mentioned above. the direction of the velocity vector is east.
Answer:
54%
Explanation:
So, we have that the "magnitude of its displacement from equilibrium is greater than (0.66)A—''. Thus, the first step to take in answering this question is to write out the equation showing the displacement in simple harmonic motion which is = A cos w×t.
Therefore, we will have two instances t the displacement that is to say at a point 2π/w - a2 and the second point at a = a2.
Let us say that 2π/w = A, then, we have that a = A cos ^-1 (0.66)/2π. Also, we have that a2 = A/2 - A cos^- (0.66) / 2π.
The next thing to do is to calculate or determine the total length of of the required time. Thus, the total length is given as:
2a1 + ( A - 2a2) = 2A{ cos^-1 (0.66)}/ π.
Therefore, the total percentage of the period does the mass lie in these regions = 100 × {2a1 + ( A - 2a2) }/A = 2 { cos^-1 (0.66)}/ π × 100 = 54%.
Thus, the total percentage of the period does the mass lie in these regions = 54%.
