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3 years ago

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On a drive from one city to? another, victor averaged 39 mph. if he had been able to average 72 ?mph, he would have reached his

destination 11 hrs earlier. what is the driving distance between one city and the? other?

1 answer:

PilotLPTM [1.2K]3 years ago

5 0

Let the unknown distance be xmiles
x/39-x/72=11hr
72x-39x/2808=11hr
33x/2808=11
33x= 30888
x=936miles
U can substitue back to check
at speed of 72mph, he would need 936/72=13hrs
at speed of 39mph, he would need 936/39=24hr
the difference is 24-13=11

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$\rho$ = Density

$v_i =$ Velocity

Re-organizing the expression we can get that

$P_1 - P_2 = \frac{1}{2}\rho (v_2^2 - v_1^2)$

Our values are given as

$v_1 = 40m/s$

$v_2 = 55m/s$

$\rho_{water} = 1.2kg/m^3 \rightarrow$ Normal Conditions

Replacing we have,

$P_1 -P_2 = \frac{1}{2}*1.2*(55^2-40^2)$

$P_1 - P_2 = 855Pa$

If we consider that there is a balance between the two states, the Force provided by gravity is equivalent to the Support Force, therefore

$F_l = F_g$

Here the lift force is the product between the pressure difference previously found by the effective area of the aircraft, while the Force of gravity represents the weight. There,

$F_g = W$

$F_l = (P_2-P_1)A$

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$(P_1 - P_2)*A = W$

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3 years ago

Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L

antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

$\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}$

Similarly, for rod 2

$\alpha_2 = \dfrac{\tau_2}{I_2}.$

Now, the moment of inertia for rod 1 is

$I_1 = \dfrac{1}{3}ML^2$ ,

and the torque acting on it is (about the center of mass)

$\tau_1 = Mg\dfrac{L}{2};$

therefore, the angular acceleration of rod 1 is

$\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},$

$\boxed{\alpha_1 = \dfrac{3g}{2L} }$

Now, for rod 2 the moment of inertia is

$I_2 = \dfrac{1}{3}(2M)(2L)^2$

$I_2 = \dfrac{8}{3} ML^2,$

and the torque acting is (about the center of mass)

$\tau _2 = (2M)g \dfrac{(2L)}{2}$

$\tau _2 = 2MgL;$

therefore, the angular acceleration $\alpha_2$ is

$\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.$

$\boxed{\alpha_2 = \dfrac{3g}{4L}}$

We see here that

$\dfrac{3g}{2L} > \dfrac{3g}{4L}$

therefore

$\boxed{\alpha_1 > \alpha_2.}$

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

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