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MissTica
3 years ago
9

Which object has the greatest momentum...and why

Physics
2 answers:
damaskus [11]3 years ago
7 0
The formula for momentum is p=mv where p is the momentum (kgms-1), m is the mass (kg) and v is the velocity (ms-1). So, to work out the momentum, we just multiply these numbers together and work out which is the largest.

1) 12kgms-1
2) 10kgms-1
3) 27kgms-1
4) 16kgms-1

Therefore the object with the greatest momentum is 3 - a 9kg mass moving at 3m/s
kap26 [50]3 years ago
6 0

Answer:

Object 3 has the greatest momentum, because the mass is small enough compared with the velocity of the object.

Explanation:

The momentum is defined as:

p = mv  (1)

                             

Where m is the mass and v is the velocity.

Therefore, the value for the mass and velocity of each object can be replaced in equation 1:

                                 

<em>1) Case for m = 12  kg and v= 1 m/s</em>

p = (12kg)(1m/s)

p = 12kg.m/s

<em>2) Case for m = 5  kg and v= 2 m/s</em>

p = (5kg)(2m/s)

p = 10kg.m/s

<em>3) Case for m = 9  kg and v= 3 m/s</em>

       

p = (9kg)(3m/s)

     

p = 27kg.m/s

<em>4) Case for m = 4  kg and v= 4 m/s</em>

p = (4kg)(4m/s)

p = 16kg.m/s

                 

Hence, Object 3 has the greatest momentum (27kg.m/s), because the mass is small enough compared with the velocity of the object.  

<em>Summary:</em>

Remember that the momentum establishes the quantity of movement that any object can have as a consequence of its mass and its velocity.

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Explanation:

HOPE THAT THIS IS HELPFUL.

HAVE A GREAT DAY.

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It took a crew 9 h 36 min to row 8 km upstream and back again. If the rate of flow of the stream was 2 km/h, what was the rowing
babunello [35]

Answer:

3 km/h

Explanation:

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Rowing speed in upstream is: x - 2 km/h

Rowing speed in downstream is: x + 2 km/h

It took a crew 9 h 36 min ( = 9 3/5 = 48/5) to row 8 km upstream and back again. Therefore:

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Multiplying by x² - 2², which is equivalent to (x-2)*(x+2)

8*(x+2) + 8*(x-2) =  (48/5)*(x² - 4)

Dividing  by 8

(x+2) + (x-2) = (6/5)*(x² - 4)

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0 =  (6/5)*x² - 2*x - 24/5

Using quadratic formula

x = \frac{2 \pm \sqrt{(-2)^2 - 4(6/5)(-24/5)}}{2(6/5)}

x = \frac{2 \pm 5.2}{2.4}

x_1 = \frac{2 + 5.2}{2.4}

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7 0
3 years ago
classic physics problem states that if a projectile is shot vertically up into the air with an initial velocity of 128 feet per
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Answer:

t_1 = 0.28 s

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Explanation:

Given that height of the projectile as a function of time is

h = -16 t^2 + 128 t + 112

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h = 147 ft

so from above equation

147 = -16 t^2 + 128 t + 112

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now by solving above quadratic equation we know that

t_1 = 0.28 s

t_2 = 7.72 s

6 0
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