Answer:
carbon, hydrogen, nitrogen and oxygen
Answer:
30.96 m
Explanation:
If the particle has a lifetime of 129 ns as measured by observer A, and has a speed of 0.8c as measured by observer A, the distance between the markers will be:
d = v * Δt
v = 0.8*c = 0.8 * 3e8 = 2.4e8
Δt = ζ = 129 ns = 1.29e-7 s
d = 2.4e8 * 1.29e-7 = 30.96 m
This is the distance as measured by observer A.
Answer:
The distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.
Explanation:
Given;
speed of the faster car, v₁ = 60 mi/h
speed of the slower car, v₂ = 55 mi/h
Let the distance traveled by the faster car when it is 15 mins ahead of the slower car = x miles

Note: divide 15 mins by 60 to convert to hours for consistency in the units.

Therefore, the distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.
You just need to replace x with 5 in each function
.5^5 - 11
-5-3
.5 ^-6
-8
64 - 8 = 56 A Celcius
Hope this helps
Your answer is C)
a)t=2.78 sec
b)R=835.03 m
c)
Explanation:
Given that
h= 38 m
u=300 m/s
here given that
The finally y=0
So
t=2.78 sec
The horizontal distance,R
R= u x t
R=300 x 2.78
R=835.03 m
The vertical component of velocity before the strike