Answer:four times
Explanation:
Given
mass of both cars A and B are same suppose m
but velocity of car B is same as of car A
Suppose velocity of car A is u
Velocity of car B is 2 u
A constant force is applied on both the cars such that they come to rest by travelling certain distance
using to find the distance traveled
where, v=final velocity
u=initial velocity
a=acceleration(offered by force)
s=displacement
final velocity is zero
For car A


For car B


divide 1 and 2 we get

thus 
distance traveled by car B is four time of car A
Answer:
a=2.378 m/s^2
Explanation:
a=Δv/Δt------eq(1)
Δv=Vf-Vi=120 km/h-0 km/h=120 km/h
or Δv=33.3 m/sec
or time=t=14s
putting values in eq(1)
a=33.3/14
a=2.378 m/s^2
52800000000000000000000000000000000000000000
Answer:
1069.38 gallons
Explanation:
Let V₀ = 1.07 × 10³ be the initial volume of the gasoline at temperature θ₁ = 52 °F. Let V₁ be the volume at θ₂ = 97 °F.
V₁ = V₀(1 + βΔθ) β = coefficient of volume expansion for gasoline = 9.6 × 10⁻⁴ °C⁻¹
Δθ = (5/9)(97°F -52°F) °C = 25 °C.
Let V₂ be its final volume when it cools to 52°F in the tank is
V₂ = V₁(1 - βΔθ) = V₀(1 + βΔθ)(1 - βΔθ) = V₀(1 - [βΔθ]²)
= 1.07 × 10³(1 - [9.6 × 10⁻⁴ °C⁻¹ × 25 °C]²)
= 1.07 × 10³(1 - [0.024]²)
= 1.07 × 10³(1 - 0.000576)
= 1.07 × 10³(0.999424)
= 1069.38 gallons