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3 years ago

A train moves with a constant velocity for 15s for 155m. How fast is the train moving?

Physics

1 answer:

bonufazy [111]3 years ago

5 0

Answer: 10.3m/s

Explanation:

In theory and for a constant velocity the physics expression states that:

Eq(1): distance = velocity times time <=> d = v*t for v=constant.

If we solve Eq (1) for the velocity (v) we obtain:

Eq(2): velocity = distance divided by time <=> v = d/t

Substituting the known values for t=15s and d=155m we get:

v = 155 / 15 <=> v = 10.3

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A constant force = 2.00 n + 3.00 n acts on a 5.00 kg object as it moves in a straight line from the position1 = 1.00 m + 1.00 m

nata0808 [166]

force acting on the object is given as

$F = 2.00\hat i + 3.00 \hat j$

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$r_i = 1.00 \hat i + 1.00\hat j$

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now the displacement is given as

$d = r_f - r_i$

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$d = 3 \hat i - 2\hat j$

now the work done is given as

$W = F.d$

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4 years ago

Read 2 more answers

Water flows past a flat plate that is oriented parallel to the flow with an upstream velocity of 0.4 m/s. (a) Determine the appr

Trava [24]

Answer:

1.12 m

0.08291 m

Explanation:

u = Upstream velocity = 0.4 m/s

Re = Reynold's number = $5\times 10^5$ (turbulent)

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Here the flow is turbulent so we have the relation

$Re_{xcr}=\frac{ux_{cr}}{\nu}\\\Rightarrow x_{cr}=\frac{Re_{xcr}\nu}{u}\\\Rightarrow x_{cr}=\frac{5\times 10^5\times 1.12\times 10^{-6}}{0.4}\\\Rightarrow x_{cr}=1.4\ m$

The approximate location downstream from the leading edge where the boundary layer becomes turbulent is 1.4 m

Boundary layer thickness relation is given by

$\delta={\frac{\nu x}{u}}^{\frac{1}{5}}\\\Rightarrow \delta={\frac{1.12\times 10^{-6}\times 1.4}{0.4}}^{\frac{1}{5}}\\\Rightarrow \delta=0.08291\ m$

The boundary layer thickness is 0.08291 m

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