Answer:
an air mass is a volume of air defined by its temperature and water vapor content. Air masses cover many hundreds or thousands of miles, and adapt to the characteristics of the surface below them. They are classified according to latitude and their continental or maritime source regions. Colder air masses are termed polar or arctic, while warmer air masses are deemed tropical. Continental and superior air masses are dry while maritime and monsoon air masses are moist. Weather fronts separate air masses with different density (temperature and/or moisture) characteristics. Once an air mass moves away from its source region, underlying vegetation and water bodies can quickly modify its character.When winds move air masses, they carry their weather conditions (heat or cold, dry or moist) from the source region to a new region. When the air mass reaches a new region, it might clash with another air mass that has a different temperature and humidity. This can create a severe storm.
Air masses can affect the weather because of different air masses that are different in temperature, density, and moisture. When two different air masses meet a front forms. This is one way air masses effect our weather.
Answer:
E.two angles are vertical angles if, and only if they are not adjacent angles
When going round a corner your direction changes which means your velocity changes which means there is an acceleration.
Answer:
Explanation:
Since the roundabout is rotating with uniform velocity ,
input power = frictional power
frictional power = 2.5 kW
frictional torque x angular velocity = 2.5 kW
frictional torque x .47 = 2.5 kW
frictional torque = 2.5 / .47 kN .m
= 5.32 kN . m
= 5 kN.m
b )
When power is switched off , it will decelerate because of frictional torque .
Answer:
0.011 m.
Explanation:
Energy stored in the spring = Energy of the projectile.
1/2ke² = mgh ................ Equation 1
Where k = spring constant, e = extension or compression, m = mass of the projectile, g = acceleration due to gravity, h = height.
make e the subject of the equation
e = √(2mgh/k)............................. Equation 2
Given: k = 12 N/cm = 1200 N/m, m = 15 g = 0.015 kg, h = 5.0 m
Constant: g = 9.8 m/s²
Substitute into equation 2
e = √(2×0.015×5/1200)
e = √(0.15/1200)
e = √(0.000125)
e = 0.011 m.
