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.
Answer:
141.14098 secs
Explanation:
Time taken to see the lightning flash can be gotten from:
Velocity = distance/time
Time = distance/velocity
Time = (47 * 1000)/(3 * 10^8)
Time = 0.0001567 secs
Time taken to hear the thunder can be gotten from:
Velocity = distance/time
Time = distance/velocity
Time = (47 * 1000)/(333)
Time = 141.14114 secs
The time lapse between the lightning flash and the thunder will be:
141.14114 - 0. 0001567
= 141.14098 secs
Answer:
Vb = k Q / r r <R
Vb = k q / R³ (R² - r²) r >R
Explanation:
The electic potential is defined by
ΔV = - ∫ E .ds
We calculate the potential in the line of the electric pipe, therefore the scalar product reduces the algebraic product
VB - VA = - ∫ E dr
Let's substitute every equation they give us and we find out
r> R
Va = - ∫ (k Q / r²) dr
-Va = - k Q (- 1 / r)
We evaluate with it Va = 0 for r = infinity
Vb = k Q / r r <R
We perform the calculation of the power with the expression of the electric field that they give us
Vb = - int (kQ / R3 r) dr
We integrate and evaluate from the starting point r = R to the final point r <R
Vb = ∫kq / R³ r dr
Vb = k q / R³ (R² - r²)
This is the electric field in the whole space, the places of interest are r = 0, r = R and r = infinity
To solve this, we use the Wien's Displacement Law as shown in the attached picture. First, convert the temperature to Kelvin.
C to F:
C = (F - 32)*5/9
C = (325 - 32)*5/9 = 162.78 °C
C to K:
K = C + 273
K = 162.78 + 273 = 435.78 K
λmax = 2898/435.78 =
<em>6</em><em>.65 μm</em>
Is there a question? Because All your doing t explaining a british philosopher to us..
