Answer:
The answer is 5.05 hours.
Explanation:
If the plane has an airspeed of 203 km/h which only applies for air and not the ground speed, we can subtract the speed of the wind since it is a headwind in the directly opposite direction.
So the speed of the plane becomes 203 - 53.5 = 149.5 km/h which will give us the true airspeed of the plane and the ground speed as well.
From here we can calculate the time it will take to reach the city as
756 km / 149.5 km/h = 5.05 hours.
I hope this answer helps.
Answer:
B. 0.98 m/s
Explanation:
This is because we use the simple formula of dividing the distance by the time. In which case would be 13.69m (distance) divided by 13.92s (time) and we will get 0.983477011 or 0.98m/s (your answer)
I hope this made sense and hoped it helped. Good luck with your test luv :)
Answer:
C. 157 bar/2270 psi
Explanation:
Calculation to determine what we should head back when either of our SPGs read
SPGs=200 bar -[200 bar-(50 bar + 20 bar)]÷1/3]
SPGs=200 bar-[(200 bar-70 bar)÷1/3]
SPGs=200 bar-(130 bar÷1/3)
SPGs=200 bar-43 bar
SPGs=157 bar/2270 psi
Therefore based on the above calculation we should head back when either of our SPGs read 157 bar/2270 psi
Answer:
approximately 30 degrees
Explanation:
If it takes the cannonball 2 seconds to reach the maximum height, we can use the analysis of the vertical component of the velocity and the fact that the acceleration of gravity is the one acting opposite to this initial vertical component 
of the velocity. We know as well that at the top of the trajectory, the vertical component of the velocity is zero, and then the movement starts going down in it trajectory. So, the final velocity for the first part of the ascending movement is zero, giving us the following equation for the velocity under an accelerated movement (with acceleration of gravity "g" acting):
By knowing the vertical component of the initial velocity (19.6 m/s), and the actual magnitude of the total initial velocity (40 m/s), we can calculate what angle was the initial velocity vector forming above the horizontal. We use for such the fact that the sine of the angle relates the opposite side of a right angle triangle with the hypotenuse, and solve for the angle using the arcsin function:
which tells us that the closer answer shown is 
Answer:
4.9 eV .
Explanation:
It is the case of discharge through mercury tube light . In it , mercury atoms are exited due to which electrons are sent to higher energy level . Here current drops to zero because electrons are excited to higher level . Energy are absorbed in quantised manner . Energy absorbed by electrons will be 4.9 eV . That means , difference in energy between two energy level is 4.9 eV .
