Motorcycle helmets are padded to give the rider the protection they require in case of accidents. Any hit to the head is absorbed by the padding and the rider is saved from any kind of fatal injury. Whenever a motorcycle rider falls from his motorcycle, there is every possibility that the head will hit the ground first. If the padding was not there, then the rider would get the direct impact of hitting. As most of the pressure of the hit gets absorbed by the inner padding of the helmet, so the rider is saved from the fatal accident. It does not mean that the driver will not have minor injuries but by all chance his life would be saved.
In order to be considered a vector, a quantity must include Magnitude (A) and Direction (D).
Answer:
Jet stream would be displaced southwards causing heavy rain and flooding.
Explanation:
The other options of the question were A) Jet stream would be displaced northwards causing drought. B) Jet stream would be displaced southwards causing drought. D) Jet stream would be displaced northwards causing heavy rain and flooding,
The statement that is a likely impact of stronger than normal trade winds in the Pacific Northwest to the United States is "Jet stream would be displaced southwards causing heavy rain and flooding."
We are talking about climate or weather terminology. In this case, we are referring to the "El Niño" (the Children) effect. Its presence affects the weather in North America. This phenomenon combines with the "La Niña) effect and it presents itself every two to seven years, ad they last from 8 to 12 months, affecting the weather conditions of the region.
1.Life science involves fields of discipline catering to living organisms such as we humans while physical science caters to non-living organisms.
2.Life science has more fields of discipline than physical science.
3.Physical science relies on laws and theories to explain concepts while life science relies on biological explanations and can also rely on theories.
Answer:
18.1 × 10⁻⁶ A = 18.1 μA
Explanation:
The current I in the wire is I = ∫∫J(r)rdrdθ
Since J(r) = Br, in the cylindrical wire. With width of 10.0 μm, dr = 10.0 μm. r = 1.20 mm. We have a differential current dI. We integrate first by integrating dθ from θ = 0 to θ = 2π.
So, dI = J(r)rdrdθ
dI/dr = ∫J(r)rdθ = ∫Br²dθ = Br²∫dθ = 2πBr²
Now I = (dI/dr)dr at r = 1.20 mm = 1.20 × 10⁻³ m and dr = 10.0 μm = 0.010 mm = 0.010 × 10⁻³ m
I = (2πBr²)dr = 2π × 2.00 × 10⁵ A/m³ × (1.20 × 10⁻³ m)² × 0.010 × 10⁻³ m = 0.181 × 10⁻⁴ A = 18.1 × 10⁻⁶ A = 18.1 μA
