B. The bus is going forward and suddenly stopped.
Answer:
The rocket's speed relative to Earth is 
.
Explanation:
Given that,
Speed of spaceship = 0.892c
Speed of rocket = 0.426c
We need to calculate the rocket's speed relative to Earth
Using formula of relative speed
Put the value into the formula
Hence, The rocket's speed relative to Earth is .
The solution for this problem:
PV = nRT
But we are looking for n so, alter the original formula:
n = PV / RT
= (1.00 atm) x (2.09 x 10^8 L) / ((0.08205746 L atm/K mol) x (25.1 + 273.15 K))
= 8539801.26 mol H2
total volume = (8539801.26 mol H2) x (-286 kJ/mol H2) = -2.44 x 10^9 k J
Answer:
Answer:
safe speed for the larger radius track u= √2 v
Explanation:
The sum of the forces on either side is the same, the only difference is the radius of curvature and speed.
Also given that r_1= smaller radius
r_2= larger radius curve
r_2= 2r_1..............i
let u be the speed of larger radius curve
now, \sum F = \frac{mv^2}{r_1} =\frac{mu^2}{r_2}∑F=
r
1
mv
2
=
r
2
mu
2
................ii
form i and ii we can write
v^2= \frac{1}{2} u^2v
2
=
2
1
u
2
⇒u= √2 v
therefore, safe speed for the larger radius track u= √2 v
Answer:
rise the air temperature is 0.179241 K
Explanation:
Given data
mass = 20000 kg
velocity = 18.5 m/s
long = 65 m
wide = 20 m
height = 12 m
density of the air = 1.20 kg/ m³
specific heat = 1020 J/(kg*K)
to find out
how much does the air temperature in the station rise
solution
we know here Energy lost by the train that is calculated by
loss in the kinetic energy that is = 1/2 m v²
loss in the kinetic energy = 0.5 × 20000 ×18.5²
loss in the kinetic energy is 3422500 J
and
this energy is used here to rise the air temperature that is KE / ( specific hat × mass )
so here
air volume = 65 ×20×12
air volume = 15600 m³
air mass = ρ × V = 1.2 × 15600
air mass = 18720 kg
so
rise the air temperature = 3422500 / ( 1020 × 18720)
rise the air temperature is 0.179241 K
