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Blababa [14]
3 years ago
10

A 2,537-kg truck moving at 14 m/s strikes a car waiting at a traffic light, hooking bumpers. The two continue to move together a

t 8 m/s. What was the mass (in kg) of the struck car
Physics
1 answer:
slavikrds [6]3 years ago
5 0

Answer:

1902.75 kg

Explanation:

From Law of conservation of momentum,

m₁u₁ + m₂u₂ = V (m₁ + m₂).................... Equation 1

make m₂ the subject of the equation,

m₂ = (m₁V - m₁u₁)/(u₂-V)..................... Equation 2

Where m₁ = mass of the truck, m₂ = mass of the car, u₁ initial velocity of the truck, u₂ = initial velocity of the car V = common velocity

Given: m₁ = 2537 kg, u₁ = 14, V= 8 m/s, u₂ = 0 m/s ( as the car was at rest waiting at a traffic light)

Substituting into equation 2.

m₂ =[2537(8) - 2537(14)]/(0-8)

m₂ = (20296-35518)/-8

m₂ = -15222/-8

m₂ = 1902.75 kg.

Thus the mass of the car = 1902.75 kg

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3 years ago
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Using newtons second law of motion, how fast for 100 KG object accelerates 350 N of force is applied to
fenix001 [56]

Answer:

3.5m/s^2

Explanation:

From Newton's second Law of Motion

F = ma

Where F is the applied force, m is the mass of the object and a is the acceleration.

F = 350 N

Mass = 100kg

350N = 100×a

a = 350/100

a = 3.5m/s^2

The acceleration of the object will be 3.5m/s^2

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Compressing the air by squeezing the bottle was accompanied by a(n) ________ in the temperature of air inside the bottle
Georgia [21]
Increase in temprature
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3 years ago
The force exerted by the wind on the sails of a sailboat is Fsail = 330 N north. The water exerts a force of Fkeel = 210 N east.
Elena L [17]

Answer:

The magnitude of the acceleration is a_r = 1.50 \ m/s^2

The direction is  \theta =  32.5 6^o north of  east

Explanation:

From the question we are told that

   The force exerted by the wind is  F_{sail} =  (330 ) \ N \ north

   The force exerted by water is  F_{keel} =  (210  ) \ N \ east

      The mass of the boat(+ crew) is  m_b  =  260  \ kg

Now Force is mathematically represented as

      F =  ma

Now the acceleration towards the north is mathematically represented as

      a_n  =  \frac{F_{sail}}{m_b}

substituting values

       a_n  =  \frac{330 }{260}

      a_n  =  1.269 \ m/s^2

Now the acceleration towards the east is mathematically represented as

       a_e = \frac{F_{keel}}{m_b }

substituting values

      a_e = \frac{210}{260}

      a_e =0.808 \ m/s^2

The resultant acceleration is  

      a_r =  \sqrt{a_e^2 + a_n^2}

substituting values

     a_r =  \sqrt{(0.808)^2 + (1.269)^2}

      a_r = 1.50 \ m/s^2

The direction with reference from the north is evaluated as

Apply SOHCAHTOA

        tan \theta =  \frac{a_e}{a_n}

       \theta = tan ^{-1} [\frac{a_e}{a_n } ]

substituting values

     \theta = tan ^{-1} [\frac{0.808}{1.269 } ]

    \theta = tan ^{-1} [0.636 ]

   \theta =  32.5 6^o

     

   

       

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1 year ago
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