Momentum = mass x velocity
12 = 4 x v | ÷ both sides by 4
12 ÷ 4 =v
v= 3 m/s
Answer:
At time 10.28 s after A is fired bullet B passes A.
Passing of B occurs at 4108.31 height.
Explanation:
Let h be the height at which this occurs and t be the time after second bullet fires.
Distance traveled by first bullet can be calculated using equation of motion
Here s = h,u = 450m/s a = -g and t = t+3
Substituting
Distance traveled by second bullet
Here s = h,u = 600m/s a = -g and t = t
Substituting
Solving both equations
So at time 10.28 s after A is fired bullet B passes A.
Height at t = 7.28 s
Passing of B occurs at 4108.31 height.
Answer: 0.86 × 10^14
Explanation:
Given the following :
Radius of proton = 1.2 × 10-15 m
Radius of hydrogen atom = 5.3 × 10-11 m
Density of proton could be calculated thus:
Mass of proton = 1.67 × 10^-27 kg
Using the formula :
(4/3) × pi × r^3
(4/3) × 3.142 × (1.2 × 10^-15)^3 = 7.24 × 10^-45
Density = mass / volume
Density = (1.67 × 10^-27) / ( 7.24 × 10^-45)
= 0.2306 × 10^18
Density of hydrogen atom:
Mass of hydrogen atom= 1.67 × 10^-27 kg
Using the formula :
(4/3) × pi × r^3
(4/3) × 3.142 × (5.3 × 10^-11)^3 = 6.24 × 10^-31
Density = mass / volume
Density = (1.67 × 10^-27) / ( 6.24 × 10^-31)
= 0.2676 × 10^4
Ratio is thus:
Density of proton / density of hydrogen atom
0.2306 × 10^18 / 0.2676 × 10^4 = 0.8617 × 10^14
Answer:
Honestly I cant understand that so please explain it someone
Explanation:
Radar waves are the waves with the lowest energy.
