The air mass that originates over land in Canada is most likely to be cold and dry
<u>Explanation:</u>
- The air masses overland from a continent is usually Dry. The air masses are formed by the uneven warming and cooling of the surface of the earth by the sun gives rise to air masses.
- Canada is affected by the five air masses they are Continental Arctic, Maritime Arctic, Maritime polar, Maritime tropical, and Continental Tropic.
- As Canada is most likely affected by the Continental Arctic the air masses are cold and dry.
3 seconds (330 m) / second =
990 meter
Answer: It has a range of visible and invisible forms of radiation.
Explanation:
Electromagnetic wave is defined as the wave which is associated with both electrical and magnetic component associated with them. They can travel in vacuum as well and travel with the speed of light i.e 
The electromagnetic radiations consist of radio waves, microwaves, infrared ,Visible , ultraviolet, X rays and gamma rays arranged in order of increasing frequency and decreasing wavelengths.
Only visible light is visible to naked eyes.
The relationship between wavelength and frequency of the wave follows the equation:

where,
= frequency of the wave
c = speed of light
= wavelength of the wave
From the above relation, it is visible that wavelength and frequency follow inverse relation. For increase in wavelength, the value of frequency decreases and vice-versa.
Answer:
The magnetic force is 
Explanation:
From the question we are told that
The charge is 
The velocity is 
The magnetic field is 
Generally the magnetic force is mathematically represented as

=> 
=> 
=>
N/B - Applied cross - product of unit vector
Answer:
Tension= 21,900N
Components of Normal force
Fnx= 17900N
Fny= 22700N
FN= 28900N
Explanation:
Tension in the cable is calculated by:
Etorque= -FBcostheta(1/2L)+FT(3/4L)-FWcostheta(L)= I&=0 static equilibrium
FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)
Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)
Ftorque= 2/3FBcostheta+ 4/3FWcostheta
Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°
Ftorque= 21900N
b) components of Normal force
Efx=FNx-FTcos(90-theta)=0 static equilibrium
Fnx=21900cos(90-55)=17900N
Fy=FNy+ FTsin(90-theta)-FB-FW=0
FNy= -FTsin(90-55)+FB+FW
FNy= -21900sin(35)+(1350+2250)×9.81=22700N
The Normal force
FN=sqrt(17900^2+22700^2)
FN= 28.900N