Answer:
La tensión es 85.3 N.
Explanation:
Cuando el objeto gira en dirección horizontal, la sumatoria de fuerzas se puede calcular usando la segunda ley de Newton:
![\Sigma F_{x} =ma_{c}](https://tex.z-dn.net/?f=%20%5CSigma%20F_%7Bx%7D%20%3Dma_%7Bc%7D%20)
Dado que el movimiento es horizontal, el peso (que está en el eje y) no contribuye en la sumatoria de fuerzas en el eje x. Por lo que la única fuerza actuando sobre el objeto en la dirección del movimiento es la tensión.
En donde:
m: es la masa del objeto = 200 g = 0.200 kg
: es la aceleración centrípeta
La aceleración centrípeta viene dada por:
![a_{c} = \omega^{2} r](https://tex.z-dn.net/?f=a_%7Bc%7D%20%3D%20%5Comega%5E%7B2%7D%20r)
En donde:
ω: es la velocidad angular del objeto = 3 rev/s
r: es el radio = 1.20 m
Entonces, la tensión es:
![T = m\omega^{2} r = 0.200 kg(3\frac{rev}{s}*\frac{2\pi rad}{1 rev})^{2}*1.20 m = 85.3 N](https://tex.z-dn.net/?f=T%20%3D%20m%5Comega%5E%7B2%7D%20r%20%3D%200.200%20kg%283%5Cfrac%7Brev%7D%7Bs%7D%2A%5Cfrac%7B2%5Cpi%20rad%7D%7B1%20rev%7D%29%5E%7B2%7D%2A1.20%20m%20%3D%2085.3%20N)
Por lo tanto, la tensión es 85.3 N.
Espero que te sea de utilidad!
u have to divide it and u will get the answer
Answer:
82780.42123 m/s
14.45 days
Explanation:
m = Mass of the planet
M = Mass of the star = ![0.85\times 1.989\times 10^{30}\ kg=1.69065\times 10^{30}\ kg](https://tex.z-dn.net/?f=0.85%5Ctimes%201.989%5Ctimes%2010%5E%7B30%7D%5C%20kg%3D1.69065%5Ctimes%2010%5E%7B30%7D%5C%20kg)
r = Radius of orbit of planet = ![0.11\times 149.6\times 10^{9}\ m=16.456\times 10^{9}\ m](https://tex.z-dn.net/?f=0.11%5Ctimes%20149.6%5Ctimes%2010%5E%7B9%7D%5C%20m%3D16.456%5Ctimes%2010%5E%7B9%7D%5C%20m)
v = Orbital speed
The kinetic and potential energy balance of the planet and star system is given by
![\frac{GMm}{r^2}=\dfrac{mv^2}{r}\\\Rightarrow v=\sqrt{\dfrac{Gm}{r}}\\\Rightarrow v=\sqrt{\dfrac{6.67\times 10^{-11}\times 1.69065\times 10^{30}}{16.456\times 10^{9}}}\\\Rightarrow v=82780.42123\ m/s](https://tex.z-dn.net/?f=%5Cfrac%7BGMm%7D%7Br%5E2%7D%3D%5Cdfrac%7Bmv%5E2%7D%7Br%7D%5C%5C%5CRightarrow%20v%3D%5Csqrt%7B%5Cdfrac%7BGm%7D%7Br%7D%7D%5C%5C%5CRightarrow%20v%3D%5Csqrt%7B%5Cdfrac%7B6.67%5Ctimes%2010%5E%7B-11%7D%5Ctimes%201.69065%5Ctimes%2010%5E%7B30%7D%7D%7B16.456%5Ctimes%2010%5E%7B9%7D%7D%7D%5C%5C%5CRightarrow%20v%3D82780.42123%5C%20m%2Fs)
The orbital speed is 82780.42123 m/s
The orbital period is given by
![t=\dfrac{2\pi r}{v}\\\Rightarrow t=\dfrac{2\pi \times 16.456\times 10^{9}}{82780.42123}\\\Rightarrow t=1249040.48419\ seconds](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7B2%5Cpi%20r%7D%7Bv%7D%5C%5C%5CRightarrow%20t%3D%5Cdfrac%7B2%5Cpi%20%5Ctimes%2016.456%5Ctimes%2010%5E%7B9%7D%7D%7B82780.42123%7D%5C%5C%5CRightarrow%20t%3D1249040.48419%5C%20seconds)
Converting to days
![\dfrac{1249040.48419}{24\times 60\times 60}=14.45\ days](https://tex.z-dn.net/?f=%5Cdfrac%7B1249040.48419%7D%7B24%5Ctimes%2060%5Ctimes%2060%7D%3D14.45%5C%20days)
The orbital period is 14.45 days
<h2>
Answer: The volume will triple (increase by a factor of three).</h2>
Explanation:
The expression for an Ideal Gas is:
(1)
Where:
is the pressure of the gas
is the volume of the gas
the number of moles of gas
is the gas constant
is the absolute temperature of the gas
Finding
:
(2)
If we are told the the amount of gas
and pressure
remain constant, but we increase the temperature
by a factor of three; we will have to rewrite (2) with the new temperature
:
![T_{N}=3T](https://tex.z-dn.net/?f=T_%7BN%7D%3D3T)
(3)
(4)
Now, if we compare (2) with (4), it is clearly noticeable the volume of the gas has increased by a factor of 3.
Answer: 40 N
Explanation: Use the formula
F = m a
to solve this.
F = (8 kg) (5 m/s²)
F = 40 kgm/s²
F = 40 N