Wind speed and air temperature are used to calculate a windchill factor.
<u>Explanation:</u>
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Wind-chill factor is the reduction of body temperature due to the passing flow of lower-temperature air.
The air temperature value is always higher than the wind chill numbers. the heat index will be used if the apparent temperature is higher than the air temperature.So, Wind speed and air temperature are mainly used to calculate a windchill factor.
There are many ways, the surface loses its heat through conduction, evaporation,radiation, and convection.The rate of convection depends on the difference in temperature between the surface and the fluid surrounding the surface and the velocity of that fluid with respect to the surface. The air around the warm surface will be heated, an insulating layer of warm air forms against the surface.The layer becomes a boundary between two. As the wind speed is high the surface cools down rapidly.
Answer:
a = - 9.8 j ^ m/s²
Explanation:
This is a projectile launch problem, they give us the initial velocity in the two components
v₀ₓ = 17.1 m / s
= 14.7 m / s
They indicate that the only acceleration that exists is the acceleration of gravity, which acts in the direction towards the center of the Earth, in general in a coordinate system it coincides with the direction of the y axis.
a = - g j ^
a = - 9.8 j ^ m /s²
Answer:
the number density of the protons in the beam is 3.2 × 10¹³ m⁻³
Explanation:
Given that;
diameter D = 2.0 mm
current I = 1.0 mA
K.E of each proton is 20 MeV
the number density of the protons in the beam = ?
Now, we make use of the relation between current and drift velocity
I = MeAv ⇒ 1 / eAv
The kinetic energy of protons is given by;
K = 
v²
v = √( 2K / )
lets relate the cross-sectional area A of the beam to its diameter D;
A = 
πD²
now, we substitute for v and A
n = I / πeD² ×√( 2K / )
n = 4I/π eD² × √( / 2K )
so we plug in our values;
n = ((4×1.0 mA)/(π(1.602×10⁻¹⁹C)(2mm)²) × √(1.673×10⁻²⁷kg / 2×( 20 MeV)(1.602×10⁻¹⁹ J/ev )
n = 1.98695 × 10¹⁸ × 1.6157967 × 10⁻⁵
n = 3.2 × 10¹³ m⁻³
Therefore, the number density of the protons in the beam is 3.2 × 10¹³ m⁻³
The most basic and perhaps powerful example is the fin-tailed fish, whose powerful tail is crucial for mobility, agility, and speed underwater. These adaptations have been underway for millions of years.
Answer:
The correct option is;
(b) The end A of the solenoid behaves like a north pole
Explanation:
According to Lenz's law we have that the induced emf direction in the solenoid due to the rapid introduction of the bar magnet will be such that the electric current induced will have a resultant magnet field that will oppose to the movement of the north pole of the bar magnet that resulted in the magnetic field
Therefore, the opposing magnetic pole to the north pole of a magnet is a north pole and the solenoid end A will act like the north pole.
