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4 years ago

12

At the starting gun a runner accelerates from rest at 1.7 m/s^2 for 2.2 What is the runner's speed 1.3 s after she starts runnin

g?

1 answer:

kodGreya [7K]4 years ago

4 0

Answer:

The runner´s speed is 2.2 m/s after 1.3 s.

Explanation:

The acceleration is the variation of velocity over time. If the acceleration is constant, then:

dv/dt = a

Separating variables:

dv = a · dt

integrating from initial velocity (v0) to v and from t = 0 to t:

v - v0 = a · t

v = v0 + a · t

Then usign this equation, we can calcualte the speed at t = 1.3 s:

v = 0 m/s + 1.7 m/s² · 1.3 s

v = 2.2 m/s

The runner´s speed is 2.2 m/s after 1.3 s.

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Answer:

The distance travel before stopping is 1.84 m

Explanation:

Given :

coefficient of kinetic friction $\mu_{k} = 0.250$

Zak's speed $v = 3 \frac{m}{s}$

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Work done by frictional force is given by,

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$\mu _{k} mg d = \frac{1}{2} m v^{2}$

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natali 33 [55]

Complete Question

The complete question is shown on the uploaded image

Answer:

The tension on the shank is $T =8391.6 N$

Explanation:

From the question we are told that

The strain on the strain on the head is $\Delta l = 0.1 mm/mm = \frac{0.1}{1000} = 0.1 *10^{-3} m/m$

The contact area is $A = 2.8 mm^2 = 2.8* (\frac{1}{1000} )^2 = 2.8*10^{-6} m^2$

Looking at the first diagram

At 600 MPa of stress

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At 450 MPa of stress

The strain is $0.0015 mm/mm$

To find the stress at $\Delta l$ we use the interpolation method

$\frac{\sigma_{\Delta l} - \sigma_{0.0015} }{ \sigma _ {0.3} - \sigma_{0.0015} } = \frac{e_{\Delta l } - e_{0.0015}}{e_{0.3} - e_{ 0.0015}}$

Substituting values

$\frac{\sigma _{\Delta l} - 450}{600 - 450} = \frac{0.1 -0.0015}{0.3 - 0.0015}$

$\sigma _{\Delta l} -450 = 49.50$

$\sigma _{\Delta l} = 499.50 MPa$

Generally the force on each head is mathematically represented as

$F = \sigma_{\Delta l} * A$

Substituting values

$F = 499.50*10^{6} * 2.8*10^{-6}$

$=1398.6N$

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$T = 6 * F$

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4 years ago