Answer:
3.9 m/s
Explanation:
We are given that
Mass of car,m=
Initial velocity,u=0
Distance,s=5.9 m

Average friction force,f=
We have to find the speed of the car at the bottom of the driveway.
Net force,
Where 
Acceleration,


v=3.9 m/s
Wow ! This is not simple. At first, it looks like there's not enough information, because we don't know the mass of the cars. But I"m pretty sure it turns out that we don't need to know it.
At the top of the first hill, the car's potential energy is
PE = (mass) x (gravity) x (height) .
At the bottom, the car's kinetic energy is
KE = (1/2) (mass) (speed²) .
You said that the car's speed is 70 m/s at the bottom of the hill,
and you also said that 10% of the energy will be lost on the way
down. So now, here comes the big jump. Put a comment under
my answer if you don't see where I got this equation:
KE = 0.9 PE
(1/2) (mass) (70 m/s)² = (0.9) (mass) (gravity) (height)
Divide each side by (mass):
(0.5) (4900 m²/s²) = (0.9) (9.8 m/s²) (height)
(There goes the mass. As long as the whole thing is 90% efficient,
the solution will be the same for any number of cars, loaded with
any number of passengers.)
Divide each side by (0.9):
(0.5/0.9) (4900 m²/s²) = (9.8 m/s²) (height)
Divide each side by (9.8 m/s²):
Height = (5/9)(4900 m²/s²) / (9.8 m/s²)
= (5 x 4900 m²/s²) / (9 x 9.8 m/s²)
= (24,500 / 88.2) (m²/s²) / (m/s²)
= 277-7/9 meters
(about 911 feet)
Because the temperature remains constant, we can apply Boyle's Law which states that
pV = constant
where
p = pressure
V = volume
Define the two states of the gas.
State 1
Pressure = p₁
Volume = 1000 ml
State 2
Pressure = p₂
Volume = 500 ml
Apply Boyle's law.
1000p₁ = 500p₂
2 = p₂/p₁
By halving the volume, the pressure doubles.
Answer:
The pressure increases by a factor of 2.
Answer:
Please find the answer in the explanation
Explanation:
When you calculate the SLOPE of a line segment, what does the SLOPE represent? (Choose all that apply) the Distance traveled the Displacement the Velocity the Acceleration None of the above
The slope of any time graph can not give you distance or displacement except for position - time graph.
When you plot either distance or displacement against time, that is, distance time graph or displacement time graph, you can get speed or velocity as the slope of the line segment.
You can only acceleration as a slope in a line of best fit if velocity is plotted against time. That is, in a velocity time graph.