Answer:
I think it's 250
Explanation:
If the car is traveling 50 km/hr that means every hour, the car drives 50 km. So if you want to know how far it will go in 5 hours you do 50x5.
Answer:
Δ h = 52.78 m
Explanation:
given,
Atmospheric pressure at the top of building = 97.6 kPa
Atmospheric pressure at the bottom of building = 98.2 kPa
Density of air = 1.16 kg/m³
acceleration due to gravity, g = 9.8 m/s²
height of the building = ?
We know,
Δ P = ρ g Δ h
(98.2-97.6) x 10³ = 1.16 x 9.8 x Δ h
11.368 Δ h = 600
Δ h = 52.78 m
Hence, the height of the building is equal to 52.78 m.
Answer:
a) t1 = v0/a0
b) t2 = v0/a0
c) v0^2/a0
Explanation:
A)
How much time does it take for the car to come to a full stop? Express your answer in terms of v0 and a0
Vf = 0
Vf = v0 - a0*t
0 = v0 - a0*t
a0*t = v0
t1 = v0/a0
B)
How much time does it take for the car to accelerate from the full stop to its original cruising speed? Express your answer in terms of v0 and a0.
at this point
U = 0
v0 = u + a0*t
v0 = 0 + a0*t
v0 = a0*t
t2 = v0/a0
C)
The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v0 again? Express the separation distance in terms of v0 and a0 . Your answer should be positive.
t1 = t2 = t
Distance covered by the train = v0 (2t) = 2v0t
and we know t = v0/a0
so distanced covered = 2v0 (v0/a0) = (2v0^2)/a0
now distance covered by car before coming to full stop
Vf2 = v0^2- 2a0s1
2a0s1 = v0^2
s1 = v0^2 / 2a0
After the full stop;
V0^2 = 2a0s2
s2 = v0^2/2a0
Snet = 2v0^2 /2a0 = v0^2/a0
Now the separation between train and car
= (2v0^2)/a0 - v0^2/a0
= v0^2/a0
Answer:
Explanation:
Since 
, we calculate the resistance rate by deriving this formula with respect to time:
Deriving what is left (remember that 
):
So we have:
Which for our values is (the rate of <em>I(t)</em> is decreasing so we put a negative sign):
