Two parallel plates are charged

What is the difference between occlumency and legilimency?

Anastasy [175]

One is legitimate and the other is not.

3 0

3 years ago

Why material selection is important to design and manufacturing?

Darina [25.2K]

Answer:

You want your product to be as strong and as long lasting as possible. There are also the safety implications to consider. You see, dangerous failures arising from poor material selection are still an all too common occurrence in many industries.

Explanation:

5 0

3 years ago

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A trolley of mass 4 kg moves with a velocity of 0.5 meter per second It colides with a stationary trolley of mass 3 kg. If the t

Luda [366]

Answer:

Approximately $0.29\; {\rm m \cdot s^{-1}}$ .

Explanation:

Make use of the fact that total momentum is conserved in collisions.

The momentum of an object of mass $m$ and velocity $v$ is $p = m\, v$ .

The momentum of the two trolleys before the collision would be:

$4\; {\rm kg} \times 0.5\; {\rm m \cdot s^{-1}} = 2\; {\rm kg \cdot m \cdot s^{-1}}$ .
$3\; {\rm kg} \times 0\; {\rm m\cdot s^{-1}} = 0\; {\rm kg \cdot m \cdot s^{-1}}$ .

Thus, the total momentum of the two trolleys right before the collision would be $2\; {\rm kg \cdot m \cdot s^{-1}}$ .

Since the two trolleys are stuck to one another after the collision, they could modelled as one big trolley of mass $m = 3\; {\rm kg} + 4\; {\rm kg} = 7\; {\rm kg}$ .

The momentum of the two trolleys, combined, is conserved during the collision. Thus, the total momentum of the new trolley of mass $m = 7\; {\rm kg}$ would continue to be $v = 2\; {\rm kg \cdot m \cdot s^{-1}}$ shortly after the collision.

Rearrange the equation $p = m\, v$ to find the velocity of the two trolleys combined:

$\begin{aligned}v &= \frac{p}{m} \\ &= \frac{2\; {\rm kg \cdot m \cdot s^{-1}}}{7\; {\rm kg}} \\ &\approx 0.29\; {\rm m \cdot s^{-1}}\end{aligned}$ .

6 0

2 years ago

A car moving at 95 km/h passes a 1.00-km-long train traveling in the same direction on a track that isparallel to the road. If t

victus00 [196]

Answer:

Time to pass the train=0.05 h

How far the car traveled in this time=4.75 Km

Explanation:

We have that the train and the car are moving in the same direction, the difference between the speed of the vehicles is:

$\Delta V=V_{car}-V_{train}=95km/h-75km/h=20km/h$

We will use this difference in the speed of the car an train to calculate how much time take the car to pass the train. For this we have that the train is 1km long and the car is moving with a speed of 20km/h (we use this value because is the speed that the car have in advantage of the train) then for a movement with a constant speed we have:

$V=\dfrac{x}{t}$

Where x is the distance, t is the time and v is the speed. using the data that we have:

$V=\dfrac{x}{t}=\dfrac{1km}{20km/h}=0.05h$

This is the time that the car take to pass the train. Now to calculate how far the car have traveled in this time we have to considered the speed of 95Km/h of the car, then:

$V=\dfrac{x}{t}\\x=v\cdot t\\x=95km/h\cdot 0.05h\\x=4.75km$