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natima [27]
3 years ago
10

Why must scientists be careful when studying nanotechnology?

Physics
1 answer:
JulijaS [17]3 years ago
6 0

Answer:

When studying nanotechnology, scientists must be aware that their ideas may not work out. Their work could be very time consuming and cost a lot of money. Finally, scientists do not yet know all of the effects of nanotechnology on human health.

Hope it helps u:)

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a 20 ft shipping container on a cargo ship has a mass of 24000 kg and a volume of 33.2m3. what is the density of the shipping co
Ira Lisetskai [31]

Answer:

722.89

Explanation:

mass=24000kg

volume=33•2

density=?

now,

density=mass/volume

=24000/33•2

=722•89

density=722•89 kg/m^3

3 0
3 years ago
Derivation 1.2 showed how to calculate the work of reversible, isothermal expansion of a perfect gas. Suppose that the expansion
stellarik [79]

Answer:

(a) The work done by the gas on the surroundings is, 17537.016 J

(b) The entropy change of the gas is, 73.0709 J/K

(c) The entropy change of the gas is equal to zero.

Explanation:

(a) The expression used for work done in reversible isothermal expansion will be,

where,

w = work done = ?

n = number of moles of gas  = 4 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas  = 240 K

= initial volume of gas  =  

= final volume of gas  =  

Now put all the given values in the above formula, we get:

The work done by the gas on the surroundings is, 17537.016 J

(b) Now we have to calculate the entropy change of the gas.

As per first law of thermodynamic,

where,

= internal energy

q = heat

w = work done

As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.

So, at constant temperature the internal energy is equal to zero.

Thus, w = q = 17537.016 J

Formula used for entropy change:

The entropy change of the gas is, 73.0709 J/K

(c) Now we have to calculate the entropy change of the gas when the expansion is reversible and adiabatic instead of isothermal.

As we know that, in adiabatic process there is no heat exchange between the system and surroundings. That means, q = constant = 0

So, from this we conclude that the entropy change of the gas must also be equal to zero.

Explanation:

6 0
3 years ago
A bowling ball has a mass of 6 kg. What happens to its momentum when its speed increases from 2m/s to 4 m/s?
Vitek1552 [10]
Here, Initial momentum = mu = 6*2 = 12 Kg m/s
Final momentum = mv = 6*4 = 24 Kg m/s

In short, Your Answer would be Option C

Hope this helps!
6 0
3 years ago
Read 2 more answers
A net force of 125 N accelerates a 25.0 kg mass. What is the resulting acceleration?
Neko [114]

Answer: a=5 m/s^2

Explanation:

The acceleration of an object can be calculated by using Newton's second law:

F=ma

where

F is the net force applied on the object

m is the mass of the object

a is its acceleration

In this problem, we have F=125 N and m=25.0 kg, so we can rearrange the equation to calculate the acceleration:

a=\frac{F}{m}=\frac{125 N}{25.0 kg}=5 m/s^2

5 0
3 years ago
Read 2 more answers
A 52.0-kg person, running horizontally with a velocity of +3.63 m/s, jumps onto a 15.2-kg sled that is initially at rest. (a) Ig
trasher [3.6K]

Answer:

The coefficient of kinetic friction between the sled and the snow is 0.0134

Explanation:

Given that:

M = mass of person = 52 kg

m = mass of sled = 15.2 kg

U = initial velocity of person = 3.63 m/s

u = initial velocity of sled = 0 m/s

After collision, the person and the sled would move with the same velocity V.

a) According to law of momentum conservation:

Total momentum before collision = Total momentum after collision

MU + mu = (M + m)V

V=\frac{MU+mu}{M+m}

Substituting values:

V=\frac{MU+mu}{M+m}=\frac{52(3.63)+15.2(0)}{52+15.2} =2.81m/s

The velocity of the sled and person as they move away is 2.81 m/s

b) acceleration due to gravity (g) = 9.8 m/s²

d = 30 m

Using the formula:

V^2=2\mu(gd)\\\mu=\frac{V^2}{2gd} \\\mu=\frac{2.81^2}{2*9.8*30} =0.0134

The coefficient of kinetic friction between the sled and the snow is 0.0134

3 0
3 years ago
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