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Travka [436]
3 years ago
5

At time t = 1, a particle is located at position (x, y) = (5, 2). If it moves in the velocity field F(x, y) = xy − 1, y2 − 11 fi

nd its approximate location at time t = 1.02.
Physics
1 answer:
Amanda [17]3 years ago
3 0

Answer:

Its approx location is (5.18,1.9)

Explanation:

Using F( 5,2) = ( xy-1, y²-11)

= ( 5*2-¹, 2²-11)

= (9,-5)

= so at point t=1.02

(5,2)+(1.02-1)*(9,-5)

(5,2)+( 0.02)*(9,-5)

(5+0.18, 2-0.1)

= ( 5.18, 1.9)

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Answer:

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4 years ago
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tino4ka555 [31]

Answer:

Explanation:

This problem bothers on the energy stored in a spring in relation to conservation of energy

Given data

Mass of block m =200g

To kg= 200/1000= 0.2kg

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=1400N/m

Compression x= 10cm

In meter x=10/100 = 0.1m

Using energy considerations or energy conservation principles

The potential energy stored in the spring equals the kinetic energy with which the block move away from the spring

Potential Energy stored in spring

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Kinetic energy of the block

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Where v = velocity of the block

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1/2kx^2=1/mv^2

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Solving for v we have

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v^2= 70

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a. Distance moved if the ramp exerts no force on the block

Is

S= v^2/2gsinθ

Assuming g= 9. 81m/s^2

S= (8.36)^2/2*9.81*sin60

S= 69.88/19.62*0.866

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S= 4.11m

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