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2 years ago

9

The electric motor of a model train accelerates the train from rest to 0.540 m/s in 27.0 ms. The total mass of the train is 610

g. Find the average power delivered to the train during its acceleration.

2 answers:

slega [8]2 years ago

6 0

Answer:

3.29 Watts

Explanation:

Given that

Initial velocity, u = 0 m/s

Final velocity, v = 0.54 m/s

Change in time, Δt = 27 ms

Mass of the train, m = 610 g = 0.61 kg

Power of any object is given as

P = Work done / time

P = W / t

Work done in this question, W is

W = 1/2mv²

W = 1/2 * 0.61 * 0.54²

W = 305 * 0.2916

W = 0.0889 kgm²/s²

Plugging this value into the equation we'd stated earlier, we have

P = W / t

P = 0.0889 / 27*10^-3

P = 3.293 Watts

Therefore, the average power delivered to the train during its acceleration is 3.29 Watts

Gnesinka [82]2 years ago

5 0

Answer:

The value is $P =3.294 \ W$

Explanation:

From the question we are that

The velocity $v = 0.540 \ m/s$

The time taken is $t = 27.0 ms = 27.0 *10^{-3} \ s$

The total mass of the train is $m = 610 \ g = 0.610 \ kg$

Generally the average power delivered is mathematically represented as

$P =\frac{KE }{t}$

$P =\frac{ \frac{1}{2} * m * v^2 }{t}$

=> $P =\frac{ \frac{1}{2} * 0.610 * 0.540 ^2 }{ 27.0 *10^{-3}}$

=> $P =3.294 \ W$

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so in our problem we have
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Answer:

a) C.M $=(\bar x, \bar y)=(0.767,0.7)m$

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Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

$\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}$

$\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}$

Where M represent the sum of all the masses on the system.

And the center of mass C.M $=(\bar x, \bar y)$

Part a

$m_1= 3 kg, m_2=5kg,m_3=7kg$ represent the masses.

$(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5)$ represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m$

C.M $=(\bar x, \bar y)=(0.767,0.7)m$

Part b

For this case we have an additional mass $m_4=6kg$ and we know that the resulting new center of mass it at the origin C.M $=(\bar x, \bar y)=(0,0)m$ and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m$

If we solve for a we got:

$(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0$

$a=-\frac{(5kg*2.3m)}{6kg}=-1.917m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m$

$(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0$

And solving for b we got:

$b=-\frac{(7kg*1.5m)}{6kg}=-1.75m$

So the coordinates for this new particle are:

$(x_4,y_4)=(-1.917,-1.75)m$

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borishaifa [10]

Answer:

7.5s

Explanation:

Given parameters:

Velocity = 30m/s

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Unknown:

Time it takes for the car to come to complete rest = ?

Solution:

To solve this problem, we use the kinematics expression below:

v = u + at

Since this is a deceleration

v = u - at

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u is the initial velocity

a is the acceleration

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v - u = -at

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~~

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Answer:

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