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3 years ago

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(AKS 1a) In 1996 Eddie Cheever recorded the fastest lap ever at the Indianapolis 500. His car completed one lap (4023 meters) in

a record time of 38.1 seconds. What was Eddie's average velocity during this lap?

1 answer:

Nataliya [291]3 years ago

3 0

Answer:

hi

Explanation:

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If the archerfish spits its water 30 degrees from the horizontal aiming at an insect 1.2 m above the surface of the water, how f

Burka [1]

Answer:

The speed is $v = 9.8 \ m/s$

Explanation:

From the question w are told that

The angle made is $\theta = 30^o$

The distance above the surface of the water is $h_{max} = 1.2 \ m$

The value of $g = 10 \ m/s^2$

The maximum height attained by the fish is mathematically evaluate as

$h_{max} = \frac{v^2 sin ^2 \theta }{2g }$

Making v which is the speed of the fish the subject of the formula

$v = \sqrt{ \frac{2gh_{max}}{ sin^2 \theta } }$

substituting values

$v = \sqrt{ \frac{2*10 *1.2 }{ [sin (30)]^2 } }$

$v = 9.8 \ m/s$

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What percentage of energy consumed in the united states is nonrenewable??

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It was estimated, according to the United States Energy Information Administration, that the United States still consumes almost 91% of non-renewable resources as its energy source. In addition, 32 % of it are natural gas, 28 % for petroleum and crude oil, and 21 % for coal power.

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Do thermohaline currents flow vertically or horizontally<br><br> A. Vertically<br> B. Horizontally

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The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

b

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$

$\frac{m(a+b)^2}{3} \omega_3^2 + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0$

$\frac{1.53(0.6+0.6)^2}{3} \omega_3^2 + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0$

$0.7344 \omega_3^2 = 2.128$

$\omega _3 = \sqrt{\frac{2.128}{0.7344} }$

$\omega _3 =1.70 \ rad/s$

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

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