Need help ASAP please helppp

A(n) 69.8 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 60.4 m away from the shuttle

alukav5142 [94]

Answer:

The time taken is 6.7 min

Explanation:

Using the linear momentum conservation theorem, we have:

$m_1*v_{o1}+m_2*v_{o2}=m_1*v_{f1}+m_2*v_{f2}$

when she was 60.4m from the shuttle, she has zero speed, so the initial velocity is zero.

$m_1*0+m_2*0=m_1*v_{f1}+m_2*v_{f2}\\m_1*_{f1}=-m_2*v_{f2}\\v_{f1}=-\frac{m_2*v_{f2}}{m_1}\\\\v_{f1}=-\frac{0.886kg*12m/s}{69.8kg}\\\\V_{f1}=-0.15m/s$

That is 0.15m/s in the opposite direction of the camera.

the time taken to get to the shuttle is given by:

$t=\frac{d}{v_{f1}}\\\\t=\frac{60.4m}{0.15m/s}\\\\t=403s\\t_{min}=403s*\frac{1min}{60s}=6.7min$

6 0

3 years ago

Calculate the amount of energy produced in joules by 100- watt light bulb lit for 2.5 hours.

9966 [12]

100/2.5 is 40.
40 is the energy that is being produced

4 0

4 years ago

Need help with this!!! <br> 15 points!!!!

irinina [24]

Answer:

// EDU HACK REDDIT EDITON //

if (running == null) {

var running = null; console.log("Program Loading...");

var iframe = document.getElementById("stageFrame");

for (i = 0; i < 1; i++) {

setInterval(function() {

if (running == null) { running = true; console.log('Program Loaded! Enjoy!') }

if (iframe.contentWindow.document.getElementsByClassName('FrameRight') != null) { iframe.contentWindow.document.getElementsByClassName('FrameRight')[0].onclick(); }

if (iframe.contentWindow.document.getElementById("invis-o-div") != null) { iframe.contentWindow.document.getElementById("invis-o-div").remove(); }

},1000);

}

} else {

console.log("Program already running!")

}

Explanation:

5 0

3 years ago

A 22.0 kg bucket of concrete is connected over a very light frictionless pulley to a 375 N box on the roof of a building as show

Veronika [31]

Answer:

vf = 3.27[m/s]

Explanation:

In order to solve this problem we must analyze each body individually and find the respective equations. The free body diagram of each body (box and bucket) should be made, in the attached image we can see the free body diagrams and the respective equations.

With the first free body diagram, we determine that the tension T should be equal to the product of the mass of the box by the acceleration of this.

With the second free body diagram we determine another equation that relates the tension to the acceleration of the bucket and the mass of the bucket.

Then we equalize the two stress equations and we can clear the acceleration.

a = 3.58 [m/s^2]

As we know that the bucket descends 1.5 [m], this same distance is traveled by the box, as they are connected by the same rope.

$x = \frac{1}{2} *a*t^{2}\\1.5 = \frac{1}{2}*(3.58) *t^{2} \\t = 0.91 [s]$

And the speed can be calculated as follows:

$v_{f}=v_{o}+a*t\\v_{f}=0+(3.58*0.915)\\v_{f}= 3.27[m/s]$

7 0

3 years ago

Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms f

vampirchik [111]

Answer:

$\large \boxed{\text{30 rev/s}}$

Explanation:

This question is based on the Law of Conservation of Angular Momentum.

Angular momentum (L) equals the moment of inertia (I) times the angular speed (ω).

L = Iω

If momentum is conserved,

I₁ω₁ = I₂ω₂

Data:

I₁ = 3.5 kg·m²s⁻¹

ω₁ = 6.0 rev·s⁻¹

I₂ = 0.70 kg·m²s⁻¹

Calculation:

$\begin{array}{rcl}I_{1}\omega_{1} &= &I_{2}\omega_{2}\\\text{3.5 kg$\cdot$m$^{2}$}\times \text{6.0 rev/s} &= &\text{0.70 kg$\cdot$m$^{2}$}\times\omega_{2}\\\text{21 rev/s} &= &0.70\omega_{2}\\\omega_{2} & = & \dfrac{\text{21 rev/s}}{0.70}\\\\&=&\textbf{30 rev/s}\\\end{array}\\\text{The skater's final rotational speed is $\large \boxed{\textbf{30 rev/s}}$}$

8 0

4 years ago