Answer:
Decreases
Explanation:
Ideal gas law:
PV = nRT
where P is absolute pressure,
V is volume,
n is number of moles,
R is gas constant,
and T is absolute temperature.
If V is constant and T decreases, then P must decrease.
Answer:
a) a = 0.477 m/s^2
b) u = 0.04862
Explanation:
Given:-
- The rotational speed of the turntable N = 33 rev/min
- The watermelon seed is r = 4.0 cm away from axis of rotation.
Find:-
(a) Calculate the acceleration of the seed, assuming that it does not slip. (b) What is the minimum value of the coefficient of static friction between the seed and the turntable if the seed is not to slip
Solution:-
- First determine the angular speed (w) of the turntable.
w = 2π*N / 60
w = 2π*33 / 60
w = 3.456 rad/s
- The watermelon seed undergoes a centripetal acceleration ( α ) defined by:
α = w^2 * r
α = 3.456^2 * 0.04
α = 0.477 m / s^2
- The minimum friction force (Ff) is proportional to the contact force of the seed.
- The weight (W) of the seed with mass m acts downwards. The contact force (N) can be determined from static condition of seed in vertical direction.
N - W = 0
N = W = m*g
- The friction force of the (Ff) is directed towards the center of axis of rotation, while the centripetal force acts in opposite direction. The frictional force Ff = u*N = u*m*g must be enough to match the centripetal force exerted by the turntable on the seed.
Ff = m*a
u*m*g = m*a
u = a / g
u = 0.477 / 9.81
u = 0.04862
Answer:
Doubled
Explanation:
F = (kq1q2) / r^2
F and q (Either q1 or q2) are directly proportional, so double the charge would also double the electruc force between the charges.
Answer:
![\boxed{F_{net} = 28.7 \ N}](https://tex.z-dn.net/?f=%5Cboxed%7BF_%7Bnet%7D%20%3D%2028.7%20%5C%20N%7D)
![\boxed{a = 2.1 \ m/s^2}](https://tex.z-dn.net/?f=%5Cboxed%7Ba%20%3D%202.1%20%5C%20m%2Fs%5E2%7D)
Explanation:
<u><em>Finding the net force:</em></u>
<u><em>Firstly , we'll find force of Friction:</em></u>
![F_{k} = (micro)_{k}mg](https://tex.z-dn.net/?f=F_%7Bk%7D%20%3D%20%28micro%29_%7Bk%7Dmg)
Where
is the coefficient of friction and m = 13.6 kg
![F_{k} = (0.16)(13.6)(9.8)\\](https://tex.z-dn.net/?f=F_%7Bk%7D%20%3D%20%280.16%29%2813.6%29%289.8%29%5C%5C)
![F_{k} = 21.32 \ N](https://tex.z-dn.net/?f=F_%7Bk%7D%20%3D%2021.32%20%5C%20N)
<u><em>Now, Finding the net force:</em></u>
![F_{net} = F - F_{k}\\F_{net} = 50 - 21.32\\](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%20F%20-%20F_%7Bk%7D%5C%5CF_%7Bnet%7D%20%3D%2050%20-%2021.32%5C%5C)
![F_{net} = 28.7 \ N](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%2028.7%20%5C%20N)
<u><em>Finding Acceleration:</em></u>
![a = \frac{F_{net}}{m}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7BF_%7Bnet%7D%7D%7Bm%7D)
![a = \frac{28.7}{13.6}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B28.7%7D%7B13.6%7D)
![a = 2.1 \ m/s^2](https://tex.z-dn.net/?f=a%20%3D%202.1%20%5C%20m%2Fs%5E2)