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Lelu [443]
3 years ago
9

Quitting tobacco use is usually very easy. true false

Physics
2 answers:
VladimirAG [237]3 years ago
8 0
False, because of the nicotine in it. Nicotine is addictive
LiRa [457]3 years ago
8 0
This will be false :)
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2 Pont
Reptile [31]

Answer:

A. It makes astronauts weightless.

Explanation:

Gravity does not make astronauts feel weightless. Astronauts are weightless because they are orbiting at the same rate as their shuttle.

Although the force of gravity weakens as one moves away from the earth surface, it does not mean that this force is absent in orbit

  • Gravitational force has a constant acceleration value near the earth surface which is commonly known to be 9.8m/s².
  • It is a force of attraction tending to hold and bind bodies together so far they have mass.
  • This force keeps every thing from escaping space-ward from the earth surface.
6 0
3 years ago
It takes 185 kj of work to accelerate a car from 23.0 m/s to 28.0 m/s. what is the car's mass?
Pachacha [2.7K]
<span>This question is based on conservation of energy as the work done would lead to change in kinetice energy of car change in KE = 1/2 mv(f)^2 - 1/2mv(i)^2 = 1/2m(v(f)^2-v(i)^2) where v(f) and v(i) are the final and initial speeds change in KE = 185kJ = 185,000J = 1/2 m((28m/s)^2-(23m/s)^2) 185,000=1/2 m(255m^2/s^2) solving for m m=1451kg</span>
6 0
3 years ago
A compound pulley is a type of system. How does a system work?
Nastasia [14]
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4 0
3 years ago
Why are male genitals outside the body?
BigorU [14]
Because if they were in the inside, then they would not be able to sexually reproduct and could not have offspring. 
7 0
3 years ago
A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe
Illusion [34]

Answer:

t = 5.59x10⁴ y

Explanation:

To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

A_{t} = A_{0}\cdot e^{- \lambda t}    (1)

<em>where A_{t}: is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.</em>

To find A₀ we can use the following equation:  

A_{0} = N_{0} \lambda   (2)

<em>where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon </em>

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:        

N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}

<em>where m_{T}: is the tree's carbon mass, N_{A}: is the Avogadro's number and m_{^{12}C}: is the ¹²C mass.  </em>

N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C    

Similarly, from equation (2) λ is:

\lambda = \frac{Ln(2)}{t_{1/2}}

<em>where t 1/2: is the half-life of ¹⁴C= 5700 years </em>

\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}

So, the initial activity A₀ is:  

A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y    

Finally, we can calculate the time from equation (1):

t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y              

I hope it helps you!

4 0
3 years ago
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