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Sophie [7]
3 years ago
5

What is a real life example of something impermeable? (please help now!!) (due tomorrow!!)

Physics
1 answer:
Nutka1998 [239]3 years ago
6 0
Impermeable is a substance that is water proofed e.g glass, aluminium e.t.c <span />
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3. What are the challenges of looking for Dyson spheres?
S_A_V [24]

1. it is difficult to search for it . Because infrared rays will never penetrate through earth atmosphere.

2. we are unaware of how it looks like and we only know it is red and will glow . A damaged star also looks like this.

3. Dust also makes is hard to detect Dyson spheres . So we will get confused between Dyson sphere and a star surrounded by dust.

5 0
3 years ago
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The current theory of the structure of the
Mariana [72]

Answers:

a) 2.82(10)^{21} kg

b) 1410 J

c) 36.62 m/s

Explanation:

<h3>a) Mass of the continent</h3>

Density \rho  is defined as a relation between mass m and volume V:

\rho=\frac{m}{V} (1)

Where:

\rho=2720 kg/m^{3} is the average density of the continent

m is the mass of the continent

V is the volume of the continent, which can be estimated is we assume it as a a slab of rock 5300 km on a side and 37 km deep:

V=(length)(width)(depth)=(5300 km)(5300 km)(37 km)=1,030,330,000 km^{3} \frac{(1000 m)^{3}}{1 km^{3}}=1.03933(10)^{18} m^{3}

Finding the mass:

m=\rho V (2)

m=(2720 kg/m^{3})(1.03933(10)^{18} m^{3}) (3)

m=2.82(10)^{21} kg (4) This is the mass of the continent

<h3>b) Kinetic energy of the continent</h3>

Kinetic energy K is given by the following equation:

K=\frac{1}{2}mv^{2} (5)

Where:

m=2.82(10)^{21} kg is the mass of the continent

v=4.8 \frac{cm}{year} \frac{1 m}{100 cm} \frac{1 year}{365 days} \frac{1 day}{24 hours} \frac{1 hour}{3600 s}=1(10)^{-9} m/s is the velocity of the continent

K=\frac{1}{2}(2.82(10)^{21} kg)(1(10)^{-9} m/s)^{2} (6)

K=1410 J (7) This is the kinetic energy of the continent

<h3>c) Speed of the jogger</h3>

If we have a jogger with mass m=77 kg and the same kinetic energy as that of the continent 1413 J, we can find its velocity by isolating v from (5):

v=\sqrt{\frac{2 K}{m}} (6)

v=\sqrt{\frac{2 (1413 J)}{77 kg}}

Finally:

v=36.62 m/s This is the speed of the jogger

5 0
3 years ago
What is your initial speed if you accelerate at 5.8 m/s/s for 3.0 seconds and achieve a final speed of 45 m/s?
Natali5045456 [20]

Answer:

27.6 m/s

Explanation:

hopefully it makes sense and is visible

:)

8 0
2 years ago
When Jennifer is out for a
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The acceleration is the principal subordinate of the speed if the speed is steady the subsidiary is invalid if the speed is diminishing the subsidiary is negative. When discussing so much stuff we consider the momentary esteem.

<span>Note that when you back off, you back off by and large yet can locally in time quicken a tiny bit, suppose amid 1/tenth of a sec since you achieved a segment of the street which was slanting. In any case, this does not change the way that when the speed diminishes, the quickening is negative.</span>
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Several types of radiation may be emitted during radioactive decay. The equation represents
Fittoniya [83]
I think the answer is B
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3 years ago
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