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daser333 [38]
3 years ago
12

A 25,000 kg traveling east collides with a 2,000 kg truck standing still on the tracks. After the collision the train and truck

are moving together at 22.3 m/s. How fast was the train going to begin with? Assume there is no friction. My frame of reference sets east a positive​
Physics
1 answer:
Elis [28]3 years ago
3 0

Answer:

24.084 m/s

Explanation:

From the law of conservation of linear momentum

Total momentum before collision equals to the total momentum after collision

Since momentum=mv where m is mass and v is velocity

M_{truck}V_{truck}=V_{common}*(M_{truck} +M_{standing}) where M_{truck} is the mass of the truck, V_{truck} is velocity of the truck, V_{common} is the common velocity of moving and standing truck after collision and M_{standing} is the mass of the standing truck

Making V_{truck} the subject we obtain

V_{truck}=\frac { V_{common}*(M_{truck} +M_{standing})}{M_{truck}}

Substituting M_{truck} as 25000 Kg, V_{common} as 22.3 m/s, M_{standing} as 2000 Kg we obtain

V_{truck}=\frac { 22.3 m/s *(25000 Kg +2000 Kg)}{25000}= 24.084 m/s

Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive

The truck was moving at 24.084 m/s

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Alexeev081 [22]

The equation expressing the statement "y varies directly as x"  is y=kx.

Explanation

The statement that y varies directly as x is analogous to saying that  the ratio of y to x is constant. In other words, when x increases, y likewise increases and that when x decreases, y decreases proportionally.

Mathematically, we express the relationship that the ration of y is to x is constant is expressed as; \frac{y}{x} =k where k is the constant of proportionality.

We can then solve the relationship for y to determine the correct form of the relationship as shown below,

\frac{y}{x} =k\\\rightarrow y=kx


6 0
3 years ago
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An insulated rigid tank contains 3 kg of H2O in the form of a saturated mixture of liquid and vapor at a pressure of 150 kPa and
Andre45 [30]

Answer:

change in entropy is 1.44 kJ/ K

Explanation:

from steam tables

At 150 kPa

specific volume

Vf = 0.001053 m^3/kg

vg = 1.1594 m^3/kg

specific entropy values are

Sf = 1.4337 kJ/kg K

Sfg = 5.789 kJ/kg

initial specific volume is calculated as

v_1 = vf + x(vg - vf)

      = 0.001053 + 0.25(1.1594 - 0.001053)

v_1 = 0.20964  m^3/kg

s_1 = Sf + x(Sfg)

     = 1.4337 + 0.25 \times 5.7894 = 2.88 kJ/kg K

FROM STEAM Table

at 200 kPa

specific volume

Vf = 0.001061 m^3/kg

vg = 0.88578 m^3/kg

specific entropy values are

Sf = 1.5302 kJ/kg K

Sfg = 5.5698 kJ/kg

constant volume  sov_1 -  v_2  = 0.29064 m^3/kg

v_2 = v_1 = vf + x(vg - vf)

       =0.29064 = x_2(0.88578 - 0.001061)

x_2 = 0.327

s_2 = 1.5302 + 0.32 \times 5.5968 = 3.36035 kJ/kg K

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8 0
3 years ago
4.0 kg objects mive a distance of 7.8 m under the action of a cinstant gorce of 5.6 N. how much work is dine on object?
alexdok [17]

Answer:

43.68 J

Explanation:

Distance moved= 7.8 m

Force = 5.6 N

Work Done= Distance moved * Force

                   = 7.8 *5.6

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6 0
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How much energy ( in joules ) is released when 0.06 kilograms of mercury is condensed to a liquid at the same temperature ?
aleksley [76]

Answer:

B. 17,705.1 J

Explanation:

The hear released when the mercury condenses into a liquid is given by:

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\lambda_v is the latent heat of vaporization

For mercury, the latent heat of vaporization is \lambda_v = 296 kJ/kg, so the heat released during the process is:

Q=(0.06 kg)(296 kJ/kg)=17.76 kJ = 17,760 J

So, the closest option is

B. 17,705.1 J

5 0
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storchak [24]
$34.75 per month

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8 0
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