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4 years ago

Determine how many Altuves you could travel in a day if you were moving at 45.000 mph all day. Altuve = 1.65 m

Physics

1 answer:

dimaraw [331]4 years ago

8 0

Answer:

The answer is below

Explanation:

Speed is the ratio of total distance traveled to the total time taken. It is given by the formula:

Speed = Distance / time taken

Given that the speed = The polynomial 45.000 mph.

In a day there are 24 hours, hence, the distance traveled is:

Speed = Distance / time taken

45 mph = Distance / 24 hours

Distance = 45 mph × 24 h = 1080 miles

But 1 mile = 1609.34 m

1080 miles = $1080\ miles*\frac{1609.34 \ m}{1\ miles}=1738092\ m$

Given that Altuve = 1.65 m

Hence, the number of Altuve traveled = 1738092 m / 1.65 m = 1053389

1053389 Altuves was traveled

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A flywheel with a radius of 0.200m starts from rest and accelerates with a constant angular acceleration of 0.900rad/s2.

kobusy [5.1K]

Answer:

The magnitude of the radial acceleration is 0.754 rad/s²

Explanation:

Given;

radius of the flywheel, r = 0.2 m

initial angular velocity of the flywheel, $\omega_i = 0$

angular acceleration of the flywheel, a = 0.900 rad/s².

angular distance, θ = 120⁰

the angular distance in radian = $\frac{120}{180} \pi = \frac{2 \pi}{3} \ rad$

Apply the following kinematic equation to determine the final angular velocity;

$\omega _f^2 = \omega _i ^2 + 2\alpha \theta\\\\\omega _f^2 = 0 + 2(0.9)(\frac{2\pi}{3} )\\\\\omega _f^2 = 3.7699\\\\\omega _f = \sqrt{ 3.7699} \\\\ \omega _f = 1.942 \ rad/s$

The magnitude of the radial acceleration is calculated as;

$\alpha _r = \omega ^2r\\\\\alpha _r = (1.942)^2 (0.2)\\\\\alpha _r =0.754 \ rad/s^2$

Therefore, the magnitude of the radial acceleration is 0.754 rad/s²

4 0

3 years ago

An airplane is flying at 300 mi/h at 4000 m standard altitude. As is typical, the air velocity relative to the upper surface of

posledela

Answer:

57330.17766 Pa

Explanation:

$P_1$ = Pressure at 4000 m = 61660 Pa

$\rho$ = Density at 4000 m = 0.8194 kg/m³

(Values taken from table of properties of air)

$V_1$ = Velocity of jet = $300\times \frac{1609.34}{3600}=134.11\ m/s$

$V_2$ = Velocity at max thickness

$V_2=1.26V_1=1.26\times 134.11=168.9786\ m/s$

From Bernoulli's equation

$P_1+\frac{1}{2}\rho V_1^2=P_2+\frac{1}{2}\rho V_2^2\\\Rightarrow P_2=P_1+\frac{1}{2}\rho V_1^2-\frac{1}{2}\rho V_2^2\\\Rightarrow P_2=61660+\frac{1}{2}\times 0.8194\times 134.11^2-\frac{1}{2}\times 0.8194\times 168.9786^2\\\Rightarrow P_2=57330.17766\ Pa$