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3 years ago

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Ondrea could drive a Jetson's flying car at a constant speed of 540.0 km/hr across oceans and space, approximately how long woul

d she take to drive from the Sun to Pluto in years? (Assume Pluto is at its average distance of 5.9 × 109 km from the Sun)

1 answer:

Tcecarenko [31]3 years ago

7 0

Answer:

The time taken in years is $x = 125 \ years$

Explanation:

From the question we are told that

The speed is $v = 540.00 \ km /hr = \frac{540 *1000}{3600} = 150\ m/s$

The distance from the sun to Pluto is $d = 5.9*10^{9} \ k m = 5.9*10^9 * 1000 = 5.9*10^{12} \ m$

Generally the time taken is mathematically represented as

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

=> $t = \frac{5.9*10^{11}}{150}$

=> $t = 3.933*10^{9}$

Converting to years

$1 year \to 3.154*10^7 \ s$

$x \ years \to 3.933*10^{9}$

=> $x = \frac{ 3.933*10^{9} * 1 }{ 3.154 *10^7}$

=> $x = 125 \ years$

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3 years ago

An object is thrown 16m/s straight up from a 7m tall cliff. How much time does it take to hit the ground below

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0,54 sec

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3 years ago

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2 years ago

Read 2 more answers

When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius

Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

$\tau = I\alpha$

where

I = Moment of inertia $\rightarrow \frac{1}{2}mr^2$ For a disk

$\alpha =$ Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

$2 \alpha \theta = \omega_f^2-\omega_i^2$

Where

$\omega_{f,i} =$ Final and Initial Angular velocity

$\alpha =$ Angular acceleration

$\theta =$ Angular displacement

Our values are given as

$\omega_i = 0 rad/s$

$\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega_f = 47.12rad/s$

$\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad$

$r = 7cm = 7*10^{-2}m$

$m = 17g = 17*10^{-3}kg$

Using the expression of angular acceleration we can find the to then find the torque, that is,

$2\alpha\theta=\omega_f^2-\omega_i^2$

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}$

$\alpha = \frac{47.12^2-0^2}{2*6\pi}$

$\alpha = 58.89rad/s^2$

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

$\tau = I\alpha$

$\tau = (\frac{1}{2}mr^2)\alpha$

$\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)$

$\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m$

Therefore the torque exerted on it is $2.45*10^{-3}N\cdot m$

3 0

2 years ago

Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.0 kg gibbon

aleksklad [387]

Answer:

230 N

Explanation:

At the lowest position , the velocity is maximum hence at this point, maximum support force T is given by the branch.

The swinging motion of the ape on a vertical circular path , will require

a centripetal force in upward direction . This is related to weight as follows

T - mg = m v² / R

R is radius of circular path . m is mass of the ape and velocity is 3.2 m/s

T = mg - mv² / R

T = 8.5 X 9.8 + 8.5 X 3.2² / .60 { R is length of hand of ape. }

T = 83.3 + 145.06

= 228.36

= 230 N ( approximately )

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2 years ago