Answer:
The ball has an initial linear kinetic energy and initial rotational kinetic energy which can both be converted into gravitational potential energy. Therefore the hill with friction will let the ball reach higher.
Explanation:
The ball has an initial linear kinetic energy and initial rotational kinetic energy which can both be converted into gravitational potential energy. Therefore the hill with friction will let the ball reach higher.
This is because:
If we consider the ball initially at rest on a frictionless surface and a force is exerted through the centre of mass of the ball, it will slide across the surface with no rotation, and thus, there will only be translational motion.
Now, if there is friction and force is again applied to the stationary ball, the frictional force will act in the opposite direction to the force but at the edge of the ball that rests on the ground. This friction generates a torque on the ball which starts the rotation.
Therefore, static friction is infact necessary for a ball to begin rolling.
Now, from the top of the ball, it will move at a speed 2v, while the centre of mass of the ball will move at a speed v and lastly, the bottom edge of the ball will instantaneously be at rest. So as the edge touching the ground is stationary, it experiences no friction.
So friction is necessary for a ball to start rolling but once the rolling condition has been met the ball experiences no friction.
145 Grams!
It asks for the “Total Mass” basically asking to add, If you add 20 to 125, you get 145! Correct me if im wrong
Answer:
There's a decrease in width of 2.18 × 10^(-6) m
Explanation:
We are given;
Shear Modulus;E = 207 GPa = 207 × 10^(9) N/m²
Force;F = 60000 N.
Poisson’s ratio; υ =0.30
We are told width is 20 mm and thickness 40 mm.
Thus;
Area = 20 × 10^(-3) × 40 × 10^(-3)
Area = 8 × 10^(-4) m²
Now formula for shear modulus is;
E = σ/ε_z
Where σ is stress given by the formula Force(F)/Area(A)
While ε_z is longitudinal strain.
Thus;
E = (F/A)/ε_z
ε_z = (F/A)/E
ε_z = (60,000/(8 × 10^(-4)))/(207 × 10^(9))
ε_z = 3.62 × 10^(-4)
Now, formula for lateral strain is;
ε_x = - υ × ε_z
ε_x = -0.3 × 3.62 × 10^(-4)
ε_x = -1.09 × 10^(-4)
Now, change in width is given by;
Δw = w_o × ε_x
Where w_o is initial width = 20 × 10^(-3) m
So; Δw = 20 × 10^(-3) × -1.09 × 10^(-4)
Δw = -2.18 × 10^(-6) m
Negative means the width decreased.
So there's a decrease in width of 2.18 × 10^(-6) m
Answer:
0.0000076 grams
Explanation:
We're given the half life of Tritium to be 12.3 years. In order to find out the amount of substabce remaining:
Let's first find how many 'half lives' are in 250 years.
Now what is half life? It means the time taken for a given quantity of an element to lose half it's mass.
So in 12.3 years we can find that The amount of 250 g of Tritium will be 250/2 = 125 g. In 24.6 years we'll have 125/2 = 62.5 g
So now we can devise a formula:
Where m is the remaining amount and n is th number of half lives in the time given.
Using this formula we can calculate:
Doing this calculation we get:
As we can see a very small value remains.
