Answer:
d. 100.0 J
Explanation:
To solve this problem we must use the theorem of work and energy conservation. This tells us that the mechanical energy in the final state is equal to the mechanical energy in the initial state plus the work done on a body. In this way we come to the following equation:
E₁ + W₁₋₂ = E₂
where:
E₁ = mechanical energy at state 1. [J] (units of Joules)
E₂ = mechanical energy at state 2. [J]
W₁₋₂ = work done from 1 to 2 [J]
We have to remember that mechanical energy is defined as the sum of potential energy plus kinetic energy.
The energy in the initial state is zero, since there is no movement of the hockey puck before imparting force. E₁ = 0.
The Work on the hockey puck is equal to:
W₁₋₂ = 100 [J]
100 = E₂
Since the ice rink is horizontal there is no potential energy, there is only kinetic energy
Ek = 100 [J]
It can be said that the work applied on the hockey puck turns into kinetic energy
Answer:
Final velocity of both goalie & puck = 0.018116 m/s
Explanation:
M1U1 + M2U2 = (M1+M2)V
70 x 0 + 0.17 x 33.5 = (70+0.17)V
V = 0.08116m/s
To answer this question, it helps enormously if you know
the formula for momentum:
Momentum = (mass) x (speed) .
Looking at the formula, you can see that momentum is directly
proportional to speed. So if speed doubles, so does momentum.
If the car's momentum is 20,000 kg-m/s now, then after its speed
doubles, its momentum has also doubled, to 40,000 kg-m/s.
Answer:
The current is 
Explanation:
From the question we are told that
The radius is 
The current density is 
The distance we are considering is 
Generally current density is mathematically represented as
Where A is the cross-sectional area represented as
=> 
=> 
Now the change in current per unit length is mathematically evaluated as
Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows
![I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.](https://tex.z-dn.net/?f=I%20%20%3D%202%5Cpi%20%2A%289.0%2A10%5E%7B6%7D%29%20%5B%5Cfrac%7Br%5E4%7D%7B4%7D%20%5D%20%20%7C%20%5Cleft%20%20%20%200.001585%7D%20%5Catop%200%7D%7D%20%5Cright.)
![I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ]](https://tex.z-dn.net/?f=I%20%20%3D%202%5Cpi%20%2A%289.0%2A10%5E%7B6%7D%29%20%5B%20%5Cfrac%7B0.001585%5E4%7D%7B4%7D%20%5D)
substituting values
![I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ]](https://tex.z-dn.net/?f=I%20%20%3D%202%20%2A%20%203.142%20%20%2A%20%209.00%20%2A10%5E6%20%2A%20%20%20%5B%20%5Cfrac%7B0.001585%5E4%7D%7B4%7D%20%5D)
2: It's not just the capillary action, but the pull from transpiration (the evaporation of water from the tree) that is used to pull water up from the roots.
<span>The second question needs context. Strong bonds alone won't cause tension. I don't see how adhesion is different. High vapour pressure could do it, but it's the difference in pressures that'd cause tension (and the resistance of that pressure by the surface). So, a low and high pressure would be needed. Poorly worded question :( </span>
<span>1: "Adhesion is the tendency of certain dissimilar molecules to cling together due to attractive forces." [1] </span>
<span>3: The other three answere would not work. Think of a boat. </span>
<span>3: If you push gas, it will be compressed(get smaller). If you push liquid it will push something else. Thus, liquids are good for transferring force. This is a hydraulic system.</span>
