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2 years ago

As shown in the figure below, Justin walks from the house to his truck on a windy day. He walks 20 m toward

Physics

1 answer:

juin [17]2 years ago

8 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

The velocity is $v =0.333 \ m/s$ in positive x -direction

The speed is $s = 0.733 \ m/s$

Explanation:

From the question we are told that

The distance from the house to truck is D = 20 m

The distance traveled back to retrieve wind-blown hat is d = 15

The distance from the wind-blown hat position too the truck is k = 20 m

The total time taken is t = 75 s

Generally when calculating the displacement the Justin's backward movement to collect his wind - blown hat is taken as negative

Generally Justin's displacement is mathematically represented as

$L = 20 - 15 + 20$

=> $L = 25 \ m$

Generally the average velocity is mathematically represented as

$v = \frac{L}{t}$

=> $v = \frac{25}{75}$

=> $v =0.333 \ m/s$

Generally the distance covered by Justin is mathematically represented as

$R = D+ d + k$

=> $R = 20 + 15 +20$

=> $R = 55 \ m$

Generally Justin's average speed over a 75 s period is mathematically represented as

$s = \frac{R}{ t}$

=> $s = \frac{55}{ 75}$

=> $s = 0.733 \ m/s$

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Volume H2 at STP = 0.005 mol H2 ( 22.4 L / 1 mol ) = 0.102 L of H2

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2 years ago

In very cold weather, a significant mechanism for heat loss by the human body is energy expended in warming the air taken into t

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Answer:

A) $Q_a=74256\ J$

B) $Q=93562560\ J$

Explanation:

Given:

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temperature of lungs, $T_l=37+273=310\ K$

specific Heat exchanged from the lungs , $c_l=0.47\ J.kg^{-1}.K^{-1}$

specific heat of air, $c_a=1020\ J.kg^{-1}.K^{-1}$

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breath rate, $b=21\ breath.min^{-1}$

Now,

amount of heat needed to warm the air of lungs to the body temperature:

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$Q_a=74256\ J$

Amount of heat lost per hour:

No. of breaths per hour:

$B=b.60$

$B=21\times 60$

$B=1260$

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$Q=Q_a.B$

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2 years ago

The human eye can respond to as little as 10^−18 J of light energy. For a wavelength at the peak of visual sensitivity, 550 nm,

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2 years ago

Read 2 more answers

The Bernoulli equation is valid for steady, inviscid, incompressible flows with a constant acceleration of gravity. Consider flo

irina1246 [14]

Answer:

$p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

Explanation:

first write the newtons second law:

F $_{s}$ =δma $_{s}$

Applying bernoulli,s equation as follows:

∑ $δp+\frac{1}{2} ρδV^{2} +δγz=0\\$

Where, $δp$ is the pressure change across the streamline and $V$ is the fluid particle velocity

substitute $ρg$ for {tex]γ[/tex] and $g_{0}-cz$ for $g$

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integrating the above equation using limits 1 and 2.

$\int\limits^2_1 \, dp +\int\limits^2_1 {(\frac{1}{2}ρV^{2} )} \, +ρ \int\limits^2_1 {(g_{0}-cz )} \,dz=0\\p_{1}^{2}+\frac{1}{2}ρ(V^{2})_{1}^{2}+ρg_{0}z_{1}^{2}-ρc(\frac{z^{2}}{2})_{1}^{2}=0\\p_{2}-p_{1}+\frac{1}{2}ρ(V^{2}_{2}-V^{2}_{1})+ρg_{0}(z_{2}-z_{1})-\frac{1}{2}ρc(z^{2}_{2}-z^{2}_{1})=0\\p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

there the bernoulli equation for this flow is $p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

note: $ρ$ =density(ρ) in some parts and change(δ) in other parts of this equation. it just doesn't show up as that in formular

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2 years ago

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Strike441 [17]

Answer:

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2 years ago