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.
Please show picture of diagrams
Answer:
force (tension) of 29.4 N (upward) in 100 cm
force (tension) of 58.4 N (upward) in 200 cm
Explanation:
Given:
Length of tube = 5 m (500 cm)
Mass of tube = 9
Suspended vertically from 150 cm and 50 cm.
Computation:
Force = Mass × gravity acceleration.
Force = 9.8 x 9
Force = 88.2 N
So,
Upward forces = Downward forces
D1 = 150 - 50 = 100 cm
D2 = 150 + 50 = 200 cm
And F1 = F2
F1 x D1 = F2 x D2
F1 x 100 = F2 x 200
F = 2F
Total force = Upward forces + Downward forces
3F = 88.2
F = 29.4 and 2F = 58.8 N
force (tension) of 29.4 N (upward) in 100 cm
force (tension) of 58.4 N (upward) in 200 cm
The ratio of concentration of ionized acid to the initial concentration of acid multiplied by 100 will give the percent ionization of a weak acid in water increases as the concentration of acid decreases.
Explanation:
Percent ionization is used for quantifying the number of ions present in the weak acid when dissolved in a solution. So it is similar to the pKa value. The percent ionization value can be determined as negative log of dissociation constant. Also the as the number of ions increases in weak acid, the concentration of acid will be decreasing . It can be calculated using the formula for percent ionization as follows:
As the water volume or concentration increases, the acid will get diluted much more thus leading to decrease in the concentration of acid.
So the ratio of concentration of ionized acid to the initial concentration of acid multiplied by 100 will give the percent ionization of a weak acid in water increases as the concentration of acid decreases.
The momentum p of a moving particle is the product between its mass, m, and tis velocity, v:
In our problem, we know
and
, and using the relationship mentioned above, we can find the mass m of the particle:
