Answer:
Explanation:
It is frictional force of the ground that helps in bringing the vehicle to stop . In the process of stopping , negative work is done on the car by friction force to overcome its kinetic energy .
At levelled road , for stoppage
Kinetic energy of vehicle = Work done by frictional force . = friction force x displacement .
At upward slopping road , gravitational force acting downward also helps the vehicle to stop do friction has to do less work .
At upward inclined road , for stoppage
Kinetic energy of vehicle = Work done by frictional force + work done by gravitational force = (friction force + gravitational force ) x displacement .
Hence displacement is less .
At downward slopping road , friction has to do more work because friction has to do work against gravitational force acting downwards wards and kinetic energy of the vehicle also .
At downward inclined road , for stoppage
Kinetic energy of vehicle + work done by gravitational force = Work done by frictional force = friction force x displacement .
Hence displacement is more .
Flourine:
There are 9 protons, 10 neutrons, and 9 electrons.
Bromine: Ther are 35 protons, 45 neutrons, and 35 electrons.
Answer:
v₃ = 22.67 [m/s]
Explanation:
In order to solve this problem, we must use the principle of conservation of the quantity of linear momentum. Where momentum is conserved before and after the collision, i.e. remains the same.
The terms on the left of the equation represent the amount of linear momentum before the collision and the members on the right represent the momentum after the collision.
where:
m₁ = mass of the truck = 3000 [kg]
m₂ = mass of the car = 1000 [kg]
v₁ = velocity of the truck before the coliision = 25 [m/s]
v₂ = velocity of the car parked = 0 (without movement)
v₃ = velocity of the truck after the collision [m/s]
v₄ = velocity of the car after the collision = 7 [m/s]
Now replacing:
![(3000*25)+(1000*0)=(3000*v_{3})+(1000*7)\\75000-7000 = 3000*v_{3}\\v_{3}=22.67 [m/s]](https://tex.z-dn.net/?f=%283000%2A25%29%2B%281000%2A0%29%3D%283000%2Av_%7B3%7D%29%2B%281000%2A7%29%5C%5C75000-7000%20%3D%203000%2Av_%7B3%7D%5C%5Cv_%7B3%7D%3D22.67%20%5Bm%2Fs%5D)
Answer:
24 N
Explanation:
Given,
Unstretched length or original length , x1 = 12cm
Stiffness constant, k = 8 N/cm
Load required to take spring to a length , x2 of 15cm
Recall the relation :
F = Ke
Where, e = extension
e = x2 - x1
e = (15 - 12) = 3
F = ke
F = 8 N/cm * 3cm
F = 24 N/cm*cm
F = 24 N
Hence, required load = 24 N
