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2 years ago

What can happen to you if you are in a crash and not wearing a seat belt? Explain.

Engineering

2 answers:

Marizza181 [45]2 years ago

7 0

Explanation:

If you don't wear your seat belt, you could be thrown into a rapidly opening frontal air bag. Such force could injure or even kill you.

poizon [28]2 years ago

3 0

What can happen is you could get hurt really bad ! seats beats could literally save your life !

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I want to solve the question

DedPeter [7]

Answer:

yes.

Explanation:

5 0

3 years ago

The U.S. Coast Guard requires certain equipment be carried onboard a vessel based on the vessel's length. How must an operator m

nevsk [136]

Answer:

The answer is option B(:

Explanation:

3 0

2 years ago

Use the graph to determine which statement is true about the end behavior of f(x).

Airida [17]

Answer:

As the x-values go to negative infinity, the function’s values go to positive infinity.

Explanation:

if the ans choices are:

As the x-values go to negative infinity, the function’s values go to negative infinity.

As the x-values go to negative infinity, the function’s values go to positive infinity.

As the x-values go to positive infinity, the function’s values go to negative infinity.

As the x-values go to positive infinity, the function’s values go to zero.

the ans is the 2nd choice

4 0

3 years ago

Read 2 more answers

A non-inductive load takes a current of 15A at 125V. An inductor is then connected in series in order that the same current shal

Norma-Jean [14]

Answer:

The inductance of the inductor is 0.051H

Explanation:

From Ohm's law;

V = IR .................. 1

The inductor has its internal resistance referred to as the inductive reactance, X $_{L}$ , which is the resistance to the flow of current through the inductor.

From equation 1;

V = IX $_{L}$

X $_{L}$ = $\frac{V}{I}$ ................ 2

Given that; V = 240V, f = 50Hz, $\pi$ = $\frac{22}{7}$ , I = 15A, so that;

From equation 2,

X $_{L}$ = $\frac{240}{15}$

= 16Ω

To determine the inductance of the inductor,

X $_{L}$ = 2 $\pi$ fL

L = $\frac{X_{L} }{2 \pi f}$

= $\frac{16}{2*\frac{22}{7}*50 }$

= 0.05091

The inductance of the inductor is 0.051H.

4 0

3 years ago

Derive an expression for the specific heat difference of a substance whose equation of state is 1 2 ( ) RT a P b b T ν ν ν = − −

sergij07 [2.7K]

Answer:

Given data:

Equation of the state $p=\frac{RT}{v-b}-\frac{a}{v(v+b) T^{1/2} }$

Where p = pressure of fluid, pα

T = Temperature of fluid, k

V = Specific volume of fluid $m^{3} / k g$

R = gas constant , $j/k g k$

a, b = Constants

Solution:

Specific heat difference, $\begin{array}{c}c_{p}-c_{v}=-T\left(\frac{\partial v}{\partial T}\right)^{2} p \\\left(\frac{\partial P}{\partial v}\right)_{r}\end{array}$

According to cyclic reaction

$\left(\frac{\ dv}{\ dT}\right)_{p}=-\frac{\left(\frac{\ d P}{\ d T}\right)_{v}}{\left(\frac{\ d P}{\ d v}\right)_{v}}$

Hence specific heat difference is

$c_{p}-c_{v}=\frac{-T\left(\frac{\ d v}{\ d T}\right)_{p}^{2}}{\left(\frac{\ d p}{\ dv}\right)_{v}}$

Equation of state, $p=\frac{R T}{v-b}-\frac{a}{v(v+b)^{\ 1/2}}$

Differentiating the equation of state with respect to temperature at constant volume,

$\left(\frac{\ d P}{\ d T}\right)_{v}=\frac{R}{v-b}-\frac{1}{2}- \frac{a}{v(v+b)^} T^{\frac{-1}{2}}$

$\begin{aligned}&\left(\frac{\ dP}{\ dT}\right)_{V}=\frac{R}{v-b}+\frac{a}{2 v(v+b) T^{3 / 2}}\end{aligned}$

Differentiating the equation of the state with respect to volume at constant temperature.

$\left(\frac{\ dP}{\ dv}\right)_{\gamma}=+(-1) \times R T(v-b)^{-1-1}+\frac{a}{b T^{1 / 2}}\left(\frac{1}{v^{2}}-\frac{1}{(v+b)^{2}}\right)\)\\\(\left(\frac{\ dP}{\ dv}\right)_{r}=-\frac{R T}{(v-b)^{2}}+\frac{a}{T^{1 / 2}}\left(\frac{2 v+b}{v^{2}(v+b)^{2}}\right)$

Substituting both eq (3) and eq (4) in eq (2)

We get,

${cp{} - } c_{v}=\frac{T\left(\frac{R}{v-b}+\frac{a}{2 v(v+b) T^{3 / 2}}\right)^{2}}{\left(\frac{R T}{(v-b)^{2}}-\frac{a(2 v+b)}{T^{1 / 2} v^{2}(v+b)^{2}}\right)}$

Specific heat difference equation,

$c_{p} -c_{v}}=\frac{T\left(\frac{R}{v-b}+\frac{a}{2 v(v+b)^{T}^{3 / 2}}\right)^{2}}{\left(\frac{R T}{(v-b)^{2}}-\frac{a(2 v+b)}{T^{1 / 2} v^{2}(v+b)^{2}}\right)}$

7 0

2 years ago