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Alex17521 [72]
3 years ago
7

A 300-kg truck moving rightward with a speed of 5 km/hr collides head on with a 1000kg car moving leftward with a speed of 100 k

m/hr. The two vehicles stick together and move with the same velocity after the collision determine the post - collision speed of the car and truck
Physics
1 answer:
igor_vitrenko [27]3 years ago
7 0

Answer:

The final velocity of the vehicles will be -75km per hour (leftwards).

Explanation:

Let us call m_1, v_1 mass and velocity of the truck , m_2, v_2 mass and velocity of the car, and v_f the final velocity of the vehicles.

The law of conservation of momentum states that

m_1v_1-m_2v_2= v_f(m_1+m_2)

solving for v_f we get:

v_f = \dfrac{m_1v_1-m_2v_2}{m_1+m_2}

putting in numerical values

m_1 =300 kg \\v_1 =5km/hr = 1.39m/s\\m_2 = 1000kg\\v_2 = 100km/hr = 27.78m/s

v_f = \dfrac{300*(1.39)-1000*(27.78)}{300+1000}

\boxed{v_f = -21.0m/s } (<em>the negative sign indicates that the vehicles are moving in the leftward direction)</em>

or -75.8 km per hour.<em> </em>

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The current through a 10 ohm resistor connected to a 120 volt power supply is
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Explanation:

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When does a rubber band, which has been shot at a wall, have the most potential energy?
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A mass weighing pounds is attached to a spring whose constant is lb/ft. The medium offers a damping force that is numerically eq
andreyandreev [35.5K]

Answer:

hello your question has some missing values attached below is the complete question with the missing values

answer :

a) 0.083 secs

b) 0.33 secs

c)  3e^-4/3

Explanation:

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g = 32 ft/s^2 ,  spring constant ( k ) = 2 Ib/ft

initial displacement = 1 ft above equilibrium

mass = weight / g = 4/32 = 1/8

damping force = instanteous velocity  hence  β = 1

a<u>)Calculate the time at which the mass passes through the equilibrium position.</u>

time mass passes through equilibrium = 1/12 seconds = 0.083

<u>b) Calculate the time at which the mass attains its extreme displacement </u>

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<u>c) What is the position of the mass at this instant</u>

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attached below is the detailed solution to the given problem

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2 years ago
If the humidity in a room of volume 410 m3 at 25 ∘C is 70%, what mass of water can still evaporate from an open pan? Express you
Keith_Richards [23]

Answer:

m=1864.68\ g

Explanation:

Given:

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putting T=298 K we have

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<u>The no. of moles of water molecules that this volume of air can hold is:</u>

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n=\frac{P_{sw}.V}{R.T}

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Currently we have 80% of n, so the mass of 20% of n:

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where;

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m=1864.68\ g is the mass of water that can vaporize further.

3 0
2 years ago
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