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vitfil [10]
3 years ago
6

Object A has a mass m and a speed v, object B has a mass m/2 and a speed 4v, and object C has a mass 3m and a speed v/3. Rank th

e objects according to the magnitude of their momentum.
Physics
1 answer:
borishaifa [10]3 years ago
5 0
<h2>The rank is : P_B>P_A=P_C .</h2>

We need to rank the objects according to the magnitude of their momentum.

We know momentum ,P = m\times v{ here m is mass and v is velocity }

Momentum of object A , P_A = m v.

Momentum of object B , P_B= \dfrac{m}{2}\times4v=2\ mv.

Momentum of object C  , P_C= 3m\times\dfrac{v}{3}=mv.

Now, we can rank them in order of  their magnitude of momentum .

P_B>P_A=P_C.

Hence, this is the required solution.

Learn More :

Momentum

brainly.com/question/7957458

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Answer:

a)   x₁ = 290.50 feet ,  x₂ = 169.74 feet , b)  v_max= 41 mph

Explanation:

For this exercise we will work in two parts, the first with Newton's second law to find the acceleration of vehicles

X Axis          fr = m a

Y Axis          N-W = 0

                    N = W = mg

The force of friction has the expression

                  fr = μ N

We replace

                 μ mg = ma

                 a = μ g

                 g = 32 feet / s²

Let's calculate the acceleration for each coefficient and friction

μ              a (feet / s2)

0.599       19.168

0.536       17,152

0.480       15.360

0.350        11.200

These are the acceleration values, for the maximum distance we use the minimum acceleration (a₁ = 11,200 feet / s²) and for the minimum braking distance we use the maximum acceleration (x₂ = 19,168 feet / s²)

                 v² = v₀² - 2 a x

When the speed stops it is zero

                 x₁ = v₀² / 2 a₁

                         

Let's reduce speed

            v₀ = 55mph (5280 foot / 1 mile) (1h / 3600s) = 80,667 feet / s²

Let's calculate the maximum braking distance

            x₁ = 80.667² / (2 11.2)

            x₁ = 290.50 feet

The minimum braking distance

            x₂ = 80.667² / (2 19.168)

            x₂ = 169.74 feet

b) maximum speed to stop at distance x = 155 feet

            0 = v₀² - 2 a x

            v₀ = √2 a x

We calculate the speed for the two accelerations

             v₀₁ = √ (2 11.2 155)

             v₀₁ = 58.92 feet / s

       

             v₀₂ = √ (2 19.168 155)

             v₀₂ = 77.08 feet / s

To stop at the distance limit in the worst case the maximum speed must be 58.92 feet / s = 40.85 mph = 41 mph

5 0
3 years ago
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