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nata0808 [166]
3 years ago
8

a net force of 100 Newton's is applied to a 20-kilogram cart that is already moving at 3 meters per second. The final speed of t

he cart was 8 meters per second. For how long was the force applied?​
Mathematics
1 answer:
sergeinik [125]3 years ago
4 0

Answer:

The Force is applied for 1 sec

Step-by-step explanation:

Given as :

The Total force applied on cart = 100 Newton

The mass of cart                         = 20 kg

The initial velocity of  cart ( v1)   = 3 meter per second

The final velocity of cart (v2)      = 8 meter per second

∵ Force = mass × acceleration

Or, 100 N = 20 kg × acceleration

Or,  Acceleration = \frac{100}{20}

Or,  Acceleration = 5 Newton per kg

Again ∵   Acceleration = Rate of change of velocity

I.e           Acceleration = \frac{v2 - v1}{t}

Or,           5 Newton per kg =   \frac{8 -3 }{t}

Or,           Time     =   \frac{5}{5} = 1 sec

Hence The Force is applied for 1 sec   Answer

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Find the distance between the pair of points. Round your answer to the nearest hundredth. (-8,19) and (3,5)
Amiraneli [1.4K]

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<h3>The answer is 17.80 units</h3>

Step-by-step explanation:

The distance between two points can be found by using the formula

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From the question the points are

(-8,19) and (3,5)

The distance between them is

d =  \sqrt{ ({ - 8 - 3})^{2}  +  ({1 9 - 5})^{2} }  \\  =  \sqrt{( { - 11})^{2}  +  {14}^{2} }  \\  =  \sqrt{121 + 196}  \\  =  \sqrt{317}  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\  = 17.804493...

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