Answer:
v = 14 m/s
= 31.3 mph
The answer would be the same if the mass of the car were 2000 kg
Explanation:
Let V be the final velocity of the car after skidding, and v be the initial velocity of the car. Let a be the acceleration of the car and Δx be the distance the car travels after applying brakes (length of the skid marks). Let Fk be the force of friction between the tyres and the road. Let N be the normal force exerted on the car and μ be the co efficient of kinetic friction.
V^2 = v^2 + 2×a×Δx
Now V, the final velocity is zero as the car stops
0 = v^2 + 2×a×Δx
v^2 = -2×a×Δx
v =√-2×a×Δx .....*
Now applying Newton's Second Law
Fnet = m×a
-Fk = m×a
-μ×N = m×a
-μ×m×g = m×a (The mass cancels out)
a = -μ×g
Substituting the value of a back to equation *
v = √-2×(-μ×g)×Δx
v = √-2×(-0.5×9.8)×20
v = 14 m/s
Therefore the speed the car was travelling with v = 14 m/s
which is equal to 31.3 mph
Now if you were to change the mass of the car to 2000 kg the value for v would still be the same. As it is seen above mass cancels out so it does not influence or affect the value of the velocity obtained.
Answer:
Consider the followig calculation
Explanation:
a) use deal equation:
PV = nRT
ρ = m/V,= ==> V = m/ρ
therefore,
ρ = Pm/RT
convert 95 oF in degree
95 oF = 308.15 K
1 atm = 1.013 * 105 pascal
ρ = 1.013*105 * 29 * /8.314 * 308.15
= 1.146 kg/m3
b) again use ideal gas equation:
ρ = Pm/RT
T = 50 oF = 283.15 K
1 atm = 1.013 * 105 pascal
molar mass will be same
ρ = 1.013 * 105 * 29 / 8.314 * 283.15
ρ = 1.248 kg / m3
So,
c) . more than density of the hot, dry air computed in part (a)
Answer:
There are three ways an object can accelerate: a change in velocity, a change in direction, or a change in both velocity and direction.
Explanation:
15,000,000/300,000=50
It will take 50 seconds.
