Warmer air has molecules more spread out, so I'd say a decrease in air pressure.
Answer:
The jp2003parker guy is extremely wrong
So he says that the size wont matter and a physical change should occur, but how would the size change without having a physical change occur.
Explanation:
1. it is difficult to search for it . Because infrared rays will never penetrate through earth atmosphere.
2. we are unaware of how it looks like and we only know it is red and will glow . A damaged star also looks like this.
3. Dust also makes is hard to detect Dyson spheres . So we will get confused between Dyson sphere and a star surrounded by dust.
Answer:
Explanation:
Given that,
Mass m = 6.64×10^-27kg
Charge q = 3.2×10^-19C
Potential difference V =2.45×10^6V
Magnetic field B =1.6T
The force in a magnetic field is given as Force = q•(V×B)
Since V and B are perpendicular i.e 90°
Force =q•V•BSin90
F=q•V•B
So we need to find the velocity
Then, K•E is equal to work done by charge I.e K•E=U
K•E =½mV²
K•E =½ ×6.64×10^-27 V²
K•E = 3.32×10^-27 V²
U = q•V
U = 3.2×10^-19 × 2.45×10^6
U =7.84×10^-13
Then, K•E = U
3.32×10^-27V² = 7.84×10^-13
V² = 7.84×10^-13 / 3.32×10^-27
V² = 2.36×10^14
V=√2.36×10^14
V = 1.537×10^7 m/s
So, applying this to force in magnetic field
F=q•V•B
F= 3.2×10^-19 × 1.537×10^7 ×1.6
F = 7.87×10^-12 N
