M)³ / 6 = 4.2e9 m³
<span>so its mass is </span>
<span>M = 3300kg/m³ * 4.2e9m³ = 1.4e13 kg </span>
<span>and so its KE at 16 km/s = 16000 m/s is </span>
<span>KE = ½ * 1.4e13kg * (16000m/s)² = 1.8e21 J
</span># of bombs N = 1.8e21J / 4.0e16J/bomb = 44 234 bombs
<span>give or take.
</span>
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Answer:
C) A ratio based on the mass of a carbon-12 atom
Explanation:
For each element, there is a characteristic number of protons, this number represent a type of atom and its called the atomic number
On the other hand we have the number of neutrons, and for one element which difers only on the number of neutrons, we have different isotopes from one element.
Together, protons and neutrons define the mass number
One property linked with the mass number, is the atomic mass, which is formerly expressed as "uma" (atomic mass unit)
By definition, one atom of carbon-12 (with 6 neutrons) has an atomic mass of 12 umas
Answer:Inertial mass
Explanation:When we measure gravitational mass we find the strength of an object's interaction with a gravitational field.
When we measure inertial mass we find an object's resistance to being accelerated by a force.
An object's gravitational mass and inertial mass are the same.
We apply a force and measure the resulting acceleration, so we can use Newton’s second law to find the inertial mass.
Answer:
0.54454
104.00902 N
Explanation:
m = Mass of wheel = 100 kg
r = Radius = 0.52 m
t = Time taken = 6 seconds
= Final angular velocity
= Initial angular velocity
= Angular acceleration
Mass of inertia is given by
![I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{100\times 0.52^2}{2}\\\Rightarrow I=13.52\ kgm^2](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7Bmr%5E2%7D%7B2%7D%5C%5C%5CRightarrow%20I%3D%5Cdfrac%7B100%5Ctimes%200.52%5E2%7D%7B2%7D%5C%5C%5CRightarrow%20I%3D13.52%5C%20kgm%5E2)
Angular acceleration is given by
![\alpha=\dfrac{\tau}{I}\\\Rightarrow \alpha=\dfrac{\mu fr}{I}\\\Rightarrow \alpha=\dfrac{\mu 50\times 0.52}{13.52}](https://tex.z-dn.net/?f=%5Calpha%3D%5Cdfrac%7B%5Ctau%7D%7BI%7D%5C%5C%5CRightarrow%20%5Calpha%3D%5Cdfrac%7B%5Cmu%20fr%7D%7BI%7D%5C%5C%5CRightarrow%20%5Calpha%3D%5Cdfrac%7B%5Cmu%2050%5Ctimes%200.52%7D%7B13.52%7D)
Equation of rotational motion
![\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=\omega_i+\dfrac{\mu (-50)\times 0.52}{13.52}t\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{\mu (-50)\times 0.52}{13.52}\times 6\\\Rightarrow 0=6.28318-11.53846\mu\\\Rightarrow \mu=\dfrac{6.28318}{11.53846}\\\Rightarrow \mu=0.54454](https://tex.z-dn.net/?f=%5Comega_f%3D%5Comega_i%2B%5Calpha%20t%5C%5C%5CRightarrow%20%5Comega_f%3D%5Comega_i%2B%5Cdfrac%7B%5Cmu%20%28-50%29%5Ctimes%200.52%7D%7B13.52%7Dt%5C%5C%5CRightarrow%200%3D60%5Ctimes%20%5Cdfrac%7B2%5Cpi%7D%7B60%7D%2B%5Cdfrac%7B%5Cmu%20%28-50%29%5Ctimes%200.52%7D%7B13.52%7D%5Ctimes%206%5C%5C%5CRightarrow%200%3D6.28318-11.53846%5Cmu%5C%5C%5CRightarrow%20%5Cmu%3D%5Cdfrac%7B6.28318%7D%7B11.53846%7D%5C%5C%5CRightarrow%20%5Cmu%3D0.54454)
The coefficient of friction is 0.54454
At r = 0.25 m
![\omega_f=\omega_i+\dfrac{0.54454 (-50)\times 0.52}{13.52}6\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{0.54454 f\times 0.25}{13.52}6\\\Rightarrow 2\pi=0.06041f\\\Rightarrow f=\dfrac{2\pi}{0.06041}\\\Rightarrow f=104.00902\ N](https://tex.z-dn.net/?f=%5Comega_f%3D%5Comega_i%2B%5Cdfrac%7B0.54454%20%28-50%29%5Ctimes%200.52%7D%7B13.52%7D6%5C%5C%5CRightarrow%200%3D60%5Ctimes%20%5Cdfrac%7B2%5Cpi%7D%7B60%7D%2B%5Cdfrac%7B0.54454%20f%5Ctimes%200.25%7D%7B13.52%7D6%5C%5C%5CRightarrow%202%5Cpi%3D0.06041f%5C%5C%5CRightarrow%20f%3D%5Cdfrac%7B2%5Cpi%7D%7B0.06041%7D%5C%5C%5CRightarrow%20f%3D104.00902%5C%20N)
The force needed to stop the wheel is 104.00902 N