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Anton [14]
3 years ago
5

A train locomotive is pulling two cars of the same mass behind it. Determine the ratio of the tension in the coupling (think of

it as a cord) between the locomotive and the first car (FT1) to that between the first car and the second car (FT2), for any nonzero acceleration of the train

Physics
1 answer:
Anna007 [38]3 years ago
7 0

Answer:

The ratio is  \frac{F_{T1}}{F_{T2}}  =  2

Explanation:

The diagram for this question is shown on the first uploaded image

Here we are assume the acceleration of the train is a

which makes the acceleration of each car a

From the question we are told that

      Considering the second car

 The force causing it s movement  is mathematically represented as

       F_{T2} =  ma

 Considering the first car

 The force causing it s movement  is mathematically represented as

      F  = F_{T1} -F_{T2} = ma

=>   F_{T1} -ma  = ma

=>   F_{T1} =  2 ma

=> \frac{F_{T1}}{ma}  =  2

=> \frac{F_{T1}}{F_{T2}}  =  2

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Your question is not necessarily a question. It's also a statement, You could write

"<em>Watt's the basic unit of the measurement of power</em>".

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Problem 3.
Korolek [52]

Answer:

143.352 watt.

Explanation:

So, in the question above we are given the following parameters or data or information that is going to assist us in answering the question above efficiently. The parameters are:

"A 1.8 m wide by 1.0 m tall by 0.65m deep home freezer is insulated with 5.0cm thick Styrofoam insulation"

The inside temperature of the freezer = -20°C.

Thickness = 5.0cm = 5.0 × 10^-2 m.

Step one: Calculate the surface area of the freezer. That can be done by using the formula below:

Area = 2[ ( Length × breadth) + (breadth × height) + (length × height) ].

Area = 2[ (1.8 × 0.65) + (0.65 × 1.0) + (1.8 × 1.0)].

Area = 7.24 m^2.

Step two: Calculate the rate of heat transfer by using the formula below;

Rate of heat transfer =[ thermal conductivity × Area (T1 - T2) ]/ thickness.

Rate of heat transfer = 0.022 × 7.24(25+20)/5.0 × 10^-2 = 143.352 watt.

8 0
3 years ago
Four point masses of 3.0 kg each are arranged in a square on masslessrods. The length of a side of the square is 0.50m. What is
Zigmanuir [339]

Answer:

Part a)

I = 1.5 kg m^2

Part b)

I = 0.75 kg m^2

Part c)

I = 1.5 kg m^2

Explanation:

Part a)

Moment of inertia of the system about an axis passing through B and C is given as

I = mL^2 + mL^2 + m(0) + m(0)

I = 2mL^2

I = 2(3 kg)(0.50^2)

I = 1.5 kg m^2

Part b)

Moment of inertia of the system about an axis passing through A and C is given as

I = m(0^2) + m(\frac{L}{\sqrt2})^2 + m(0) + m(\frac{L}{\sqrt2})^2

I = 2m\frac{L^2}{2}

I = (3 kg)(0.50^2)

I = 0.75 kg m^2

Part c)

Moment of inertia of the system about an axis passing through the center of the square and perpendicular to the plane of the square

I = m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2

I = 4m\frac{L^2}{2}

I = 2(3 kg)(0.50^2)

I = 1.5 kg m^2

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gregori [183]
The answer is is b hope this helped
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3 years ago
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