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2 years ago

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Anna applies a force of 19.5 newtons to push a book placed on a table. If the normal force of the book is 51.7 newtons, what is

the coefficient of kinetic friction?

2 answers:

andrey2020 [161]2 years ago

5 0

Considering that the book is moving with constant speed, the force applied by Anna must be the same that the friction force:

$F = F_R = F\cdot \mu_k\cdot N$

If we clear the previous equation:

$\mu_k = \frac{F}{F_R} = \frac{19.5\ N}{51.7\ N} = \bf 0.38$

barxatty [35]2 years ago

5 0

The force applied to the book is equal to the normal force of the book multiplied by the coefficient of kinetic friction. <span>The normal force is equal to the weight of the book. </span>In this case, the formula with the given data is equal to 19.5N = 51.7N * coeff. of friction. The coefficient of friction is equal to 0.38.

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Yes, a sled has inertia while sitting still.

Explanation:

From Newton's law of inertia, an object at rest will remain at rest unless it is acted upon by an external force. The reason the object will remain at rest unless an external force acts is because of inertia. Inertia means the resistance of an object to motion.

Thus, a sled hammer at rest will remain at rest unless it is acted upon by an external force. So we can conclude that it has Inertia.

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A rocket accelerates by burning its onboard fuel, so its mass decreases with time. Suppose the initial mass of the rocket at lif

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Height of the rocket be one minute after liftoff is 40.1382 km.

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v = velocity of rocket at time t

g = Acceleration due to gravity = $9.8 m/s^2$

$v_e$ = Constant velocity relative to the rocket = 2,900m/s.

m = Initial mass of the rocket at liftoff = 29000 kg

r = Rate at which fuel is consumed = 170 kg/s

Velocity of the rocket after 1 minute of the liftoff =v

t = 1 minute = 60 seconds'

Substituting all the given values in in the given equation:

$v(60)=-9.8 m/s^2\times 60 s-2,900m/s\times \ln (\frac{29,000 kg-170 kg/s\times 60 s}{2,9000 kg})$

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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:

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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:

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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!

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