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Svetllana [295]
3 years ago
8

A particle of mass m=375 g is launched with velocity of A =4 moves with a length AB=2.5m under the action of tractive force F=3.

46 N making an angle 30. during its motion between A and B which is subjected to a frictional force f=1.5 N
calculate V of B by applying the kinetic energy theorem​
Physics
1 answer:
DiKsa [7]3 years ago
3 0

Answer:

The final speed is 5.78 m/s.

Explanation:

mass, m = 375 g = 0.375 kg

initial velocity, u = 4 m/s

Distance, s = 2.5 m

Angle, A = 30 degree

Force, F = 1.5 N

let the final velocity is v.

Use the work energy theorem

Work done = change in kinetic energy

W= 0.5 m(v^2 - u^2)\\\\F s cos A= 0.5 m (v^2 - u^2)\\\\1.5\times 2.5\times cos30= 0.5\times 0.375\times (v^2 - 16)\\\\v = 5.78 m/s

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It is correct to say that impulse is equal toA) momentum.B) the change in momentum.C) the force multiplied by the distance the f
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An electron enters a region between two large parallel plates made of aluminum separated by a distance of 2.0 cm and kept at a p
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Answer:

a)v=1.77\times 10^6\ m/s

b)v=3.872\times 10^6\ m/s

c)v=5.5\times 10^6\ m/s

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Given that

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F = E q

F = m a

E = V/d

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m a =  q .V/d b            

a=\dfrac{q.V}{m.d}                 ---------1

The mass of electron

m=9.1\times 10^{-31}\ kg

The charge on electron

q=1.6\times 10^{-19}\ C

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a=\dfrac{1.6\times 10^{-19}\times 200}{9.1\times 10^{-31}\times 0.02}\ m/s^2

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v^2=u^2+2as

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s = 0.1 cm

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v=\sqrt{3.16\times 10^{12}}\ m/s

v=1.77\times 10^6\ m/s

b)

s = 0.5 cm

v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 0.5\times 10^{-2}

v=\sqrt{1.5\times 10^{13}}\ m/s

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s = 1 cm

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s = 1.5 cm

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e)

s = 2 cm

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v=\sqrt{6.06\times 10^{13}}\ m/s

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Answer:

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