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sdas [7]
3 years ago
9

Now that you are familiar with MRI's, nanotechnology and micro-bots, use your imagination to brainstorm other probable invention

s. Think of at least 5 more inventions that could be used to solve humanity's issues. Describe your inventions using a minimum of 2 sentences. Include its purpose and how it would work.
Physics
1 answer:
dlinn [17]3 years ago
8 0

Cancer research, solar panel production and agricultural innovation will all be key areas for nano tech, and so will clothing design, cosmetics manufacturing and many others are some of the new developments.

<u>Explanation:</u>

Nanotechnology is also being applied to or developed for application to a variety of industrial and purification processes. Purification and environmental cleanup applications include the desalination of water, water filtration, wastewater treatment, groundwater treatment, and other nano remediation.

Nanotechnology offers the potential for new and faster kinds of computers, more efficient power sources and life-saving medical treatments. Potential disadvantages include economic disruption and possible threats to security, privacy, health and the environment.

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A ring is attached at the center of the underside of a trampoline. A sneaky teenager crawls under the trampoline and uses the ri
Karo-lina-s [1.5K]

Answer:

Explanation:

Given

Mass of mother m=79\ kg

Mother is released with a speed of v=2.5\ m/s

Assuming we need to find elastic Potential energy stored in Trampoline

Kinetic energy acquired  by Mother comes from Elastic Potential Energy stored in the spring

I.e. Elastic Potential Energy=Kinetic Energy of mother

\frac{1}{2}kx^2=\frac{1}{2}mv^2

E=\frac{1}{2}\times 79\times 2.5^2=246.875\ J

So 246.875 J of Energy is stored in the trampoline as he pull the ring

5 0
3 years ago
A string fixed at both ends is 8.40 m long and has a mass of 0.120 kg. It is subjected to a tension of 96.0 N and set oscillatin
Luden [163]

Answer:

81.9756 m/s

16.8 m

4.8795 Hz

Explanation:

m = Mass of string = 0.12 kg

L = Length of string = 8.4 m

T = Tension on string = 96 N

Linear density is given by

\mu=\dfrac{m}{L}\\\Rightarrow \mu=\dfrac{0.12}{8.4}

Spee of the wave is given by

v=\sqrt{\dfrac{T}{\mu}}\\\Rightarrow v=\sqrt{\dfrac{96}{\dfrac{0.12}{8.4}}}\\\Rightarrow v=81.9756\ m/s

The speed of the waves on the string is 81.9756 m/s

Wavelength is given by

\lambda=2L\\\Rightarrow \lambda=2\times 8.4\\\Rightarrow \lambda=16.8\ m

The longest possible wavelength is 16.8 m

Frequency is given by

f=\dfrac{v}{\lambda}\\\Rightarrow f=\dfrac{81.9756}{16.8}\\\Rightarrow f=4.8795\ Hz

The frequency of the wave is 4.8795 Hz

3 0
4 years ago
An object is propelled straight up from ground level with an initial velocity of 48 feet per second. Its height at time t is mod
Ket [755]

Answer:

Explanation:

For a. its max height and when it occurs. First the max height. That's a y-dimension thing, and in the y-dimension we have this info:

v₀ = 48 ft/s

a = -32 ft/s/s

v = 0 (the max height of an object occurs when the final velocity of the object is 0). Use the following equation for this part of the problem:

v² = v₀² + 2aΔx and filling in:

0=48^2+2(-32)Δx and

0 = 2300 - 64Δx and

-2300 = -64Δs so

Δx = 36 feet.

Now for the time it takes to get to this max height. Final velocity is still 0 here, but the equation is a different one for this part of the problem. Use:

v = v₀ + at and filling in:

0 = 48 - 32t and

-48 = -32t so

t = 1.5 sec.  That's part a. Onto part b:

The object hits the ground when its displacement, Δx, is 0. Use this equation for this problem:

Δx = v₀t + \frac{1}{2}at^2 and filling in:

0=48t+\frac{1}{2}(-32)t^2 and

0=48t-16t^2 and

0 = 16t(3 - t) so

t = 0 and t = 3.  t = 0 is before the object is propelled, so it makes sense that at 0 seconds, the object was still on the ground, right? Then at 3 seconds, it's back on the ground. (Isn't math just perfectly, beautifully sensible!?) Now onto part c:

We are looking for the time interval when the object is >32 feet. So we use the same equation we just used, but with an inequality instead of an equals sign:

48t+\frac{1}{2}(-32)t^2 >32 and get everything on one side and factor it again:

-16t^2+48t-32>0 and we find that

1 < t < 2 so the time interval is between 1 and 2 seconds that the object is over 32 feet in the air.

8 0
3 years ago
The relationship between the volume and temperature of a gas was first put forward by the French scientist Jacques-Alexandre-Cés
stellarik [79]

Answer:

1 , 2 and , 4 on usa test prep

Explanation:

4 0
3 years ago
The mass of the sun is 1.99×1030kg and its distance to the Earth is 1.50×1011m.
Vladimir79 [104]
In order to calculate the gravitational force of the two bodies we use the formula which is expressed as:

F = GMm/R²

where <span>G = 6.67 x 10^-11 in SI unit, M and m are the mass of the two bodies and R is the distance between them. 

F = </span>6.67 x 10^-11 (1.99×10^30) (6×10^24) / (1.50×10^11)²
F = 3.53×10^22<span>N</span>
8 0
4 years ago
Read 2 more answers
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