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Mila [183]
3 years ago
12

A bionic man running at 6.5 m/sec , east is acceleration at a uniform rate of 1.5 m/sec^2 east over a displacement of 100.0 m ea

st . what is the velocity he reaches after traveling this displacement
CAN YOU PLEASE SHOW WORK
Physics
1 answer:
tangare [24]3 years ago
5 0
Given:
u = 6.5 m/s, initial velocity 
a = 1.5 m/s², acceleration
s = 100.0 m, displacement

Let v =  the velocity attained after the 100 m displacement.
Use the formula
v² = u² + 2as
v² = (6.5 m/s)² + 2*(1.5 m/s²)*(100 m) = 342.25 (m/s)²
v = 18.5 m/s

Answer: 18.5 m/s
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A block, M1=10kg, slides down a smooth, curved incline of height 5m. It collides elastically with another block, M2=5kg, which i
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Answer:

2.86 m

Explanation:

Given:

M₁ = 10 kg

M₂ = 5 kg

\mu_k = 0.5

height, h = 5 m

distance traveled, s = 2 m

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now,

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or

u₁ = √(2 × 9.8 × 5)

or

u₁ = 9.89 m/s

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M₁ × u₁ = M₂ × v₂

on substituting the values, we get

10 × 9.89 = 5 × v₂

or

v₂ = 19.79 m/s

now,

let the velocity of mass 2 when it reaches the spring be v₃

from the work energy theorem,  we have

Work done by the friction force = change in kinetic energy of the mass 2

or

0.5\times5\times9.8\times2 = \frac{1}{2}\times5\times( v_3^2-19.79^2)

or

v₃ = 20.27 m/s

now, let the spring is compressed by the distance 'x'

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we have

Energy of the spring =  Kinetic energy of the mass 2

or

\frac{1}{2}kx^2=\frac{1}{2}mv_3^2

on substituting the values, we get

\frac{1}{2}\times250\times x^2=\frac{1}{2}\times5\times20.27^2

or

x = 2.86 m

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