The maximum force that the tires can exert on the road before slipping is 16200 N.
From the information in the question;
The coefficient of static friction = 0.9
The mass of the car = 1800 kg
Using the formula;
μ = F/R
μ = coefficient of static friction
F = force on the tires
R = the reaction force
But recall that the reaction is equal in magnitude to the weight of the car.
W=R
Hence; R = 1800 kg × 10 ms-2 = 18000 N
Making F the subject of the formula;
F = μR
Substituting values;
F = 18000 N × 0.9
F = 16200 N
Hence, the maximum force that the tires can exert on the road before slipping is 16200 N.
Learn more: brainly.com/question/18754989
Answer:
Explanation: BY ATOMIC NUMBER
Answer:
v = 719.2 m / s and a = 83.33 m / s²
Explanation:
This is a rocket propulsion system where the system is made up of the rocket plus the ejected mass, where the final velocity is
v - v₀ = ln (M₀ / M)
where v₀ is the initial velocity, v_{e} the velocity of the gases with respect to the rocket and M₀ and M the initial and final masses of the rocket
In this case, if fuel burns at 75 kg / s, we can calculate the fuel burned for the 10 s
m_fuel = 75 10
m_fuel = 750 kg
As the rocket initially had a mass of 3000 kg including 1000 kg of fuel, there are still 250 kg, so the mass of the rocket minus the fuel burned is
M = 3000 -750 = 2250 kg
let's calculate
v - 0 = 2500 ln (3000/2250)
v = 719.2 m / s
To calculate the acceleration, let's use the concept of the rocket thrust, which is the force of the gases on it. In the case of the rocket, it is
Push = v_{e} dM / dt
let's calculate
Push = 2500 75
Push = 187500 N
If we use Newton's second law
F = m a
a = F / m
let's calculate
a = 187500/2250
a = 83.33 m / s²
Answer:
Multiply the mass times the speed for both, then compare.I would expect the mass of the oil ship to be more than 10 times the mass of the car, in which case its momentum would be greater.
PLZ MARK BRAINLIEST
<span>Its the phosphorus cycle</span>