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Anni [7]
3 years ago
15

A(n) 71.1 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 70.2 m away from the shuttle

and moving with zero speed relative to the shuttle. She has a(n) 0.67 kg camera in her hand and decides to get back to the shuttle by throwing the camera at a speed of 12 m/s in the direction away from the shuttle. How long will it take for her to reach the shuttle? Answer in minutes. Answer in units of min.
Physics
1 answer:
denpristay [2]3 years ago
4 0

Answer:

10.347 minutes.

Explanation:

According to F = ma, she exerts force on camera of the magnitude

F = 0.67Kg*12m/s^{2} = 8.04N, assuming it took her one second to accelerate camera to 12m/s, then by newtons third law, which says every action has equal and opposite reaction , the camera exerts the same amount of force on the astronaut which gives her acceleration of a = \frac{8.04N}{70.2Kg} = 0.1130801680m/s^2.

and velocity of V = 0.1130801680m/s.

at this velocity , the astronaut has to cover the distance of 70.2 meters, it will take her 620.7985075s = 10.347 min to get to the shuttle (using S = vt).

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Kamila [148]

Answer:

(a) \frac{m}{s^2}

(b) \frac{m}{s^3}

(c) 1 s

(d) 20 m

(e) 1 m

(f) 0\frac{m}{s}

(g) -12\frac{m}{s}

(h) -36\frac{m}{s}

(i) -72\frac{m}{s}

(j) -6\frac{m}{s^2}

(k) -18\frac{m}{s^2}

(l) -30\frac{m}{s^2}

(m) -42\frac{m}{s^2}

Explanation:

Since <em>x</em> is measured in meters and <em>t</em> in seconds, constants <em>a </em>and <em>b</em> must have units that gives meters when multiplied by square and cubic seconds respectivly, so that would mean \frac{m}{s^2} for <em>a </em>and \frac{m}{s^3} for <em>b</em>.

We can get the velocity <em>v </em>equation by deriving the position with respect to <em>t</em>, which gives:

v=6*t-6*t^2

And the acceleration <em>a</em> equation by deriving again:

a=6-12*t

Now for getting the maximun position between 0 and 4, we must find to points where the positions first derivate is equal to cero and evaluate those points. That is <em>v=0</em>, which gives

6*t-6*t^2=0\\6t*(1-t)=0\\t=0 or t=1

For <em>t = 0</em>,<em> x = 0</em> so the maximun position is archieved at 1 second, which gives <em>x = 1 meter</em>.

For obtaining it's displacement <em>r</em>, we can integrate the velocity from 0 seconds to 4 seconds, which gives the mean value of the position in that interval:

r=\frac{1}{4}*(2*4^3-3*4^2)m\\r= 20m

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8 0
4 years ago
What acceleration will you give to a 25kg box if you push it with a force of 85N
Grace [21]
<h3>Answer:</h3>

3.4 m/s²

<h3>Explanation:</h3>

We are given;

  • Mass of the box as 25 kg
  • Force is 85 N

We are required to determine the acceleration;

  • According to second newton's law of motion force is given by the product of mass and acceleration.
  • That is;

Force = ma

Rearranging the formula;

a = F ÷ m

Therefore;

acceleration = 85 N ÷ 25 kg

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Thus, the acceleration of the box will be 3.4 m/s²

5 0
3 years ago
In about 5 billion years, at the end of its lifetime, our sun will end up as a white dwarf, having about the same mass as it doe
masya89 [10]
The formula of density is given by

Density = Mass ÷ Volume

We have:
Mass = 1.989 × 10³⁰ kg
Volume = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (7500)^3 = 1.767*10^12 km³

Density = \frac{(1.989)(10^{30})}{(1.767)(10^{12})}=1.13×10¹⁸ kg/km³

Converting 1.13 × 10¹⁸ kg/km³ to g/cm³

1.13 × 10¹⁸ kg = 1.13 × 10¹⁸ × 10³ = 1.13 × 10²¹ grams
1 km³ = 1 × 10⁶ cm³ 

(1.13 × 10²¹) ÷ 10⁶ = 1.13 × 10¹⁵ gr/cm³

Answer: Density 1.13 × 10¹⁵ gr/cm³
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Answer:

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