Answer:
C
Explanation:
Newtons second law explains this the most because for every action their is an equal and opposite reaction. The reaction of you turning, causes the reaction of your whole body to turn with the bike.
Answer:
![\boxed{\sf Viscosity \ of \ glycerine \ (\eta) = 14.382 \ poise}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%5Csf%20Viscosity%20%5C%20of%20%5C%20glycerine%20%5C%20%28%5Ceta%29%20%3D%2014.382%20%5C%20poise%7D%20)
Given:
Radius of ball bearing (r) = 1.5 mm = 0.15 cm
Density of iron (ρ) = 7.85 g/cm³
Density of glycerine (σ) = 1.25 g/cm³
Terminal velocity (v) = 2.25 cm/s
Acceleration due to gravity (g) = 980.6 cm/s²
To Find:
Viscosity of glycerine (
)
Explanation:
![\boxed{ \bold{v = \frac{2}{9} \frac{( {r}^{2} ( \rho - \sigma)g)}{ \eta} }}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%20%5Cbold%7Bv%20%3D%20%20%5Cfrac%7B2%7D%7B9%7D%20%20%5Cfrac%7B%28%20%7Br%7D%5E%7B2%7D%20%28%20%5Crho%20-%20%20%5Csigma%29g%29%7D%7B%20%5Ceta%7D%20%7D%7D)
![\sf \implies \eta = \frac{2}{9} \frac{( {r}^{2}( \rho - \sigma)g )}{v}](https://tex.z-dn.net/?f=%20%5Csf%20%5Cimplies%20%5Ceta%20%3D%20%20%5Cfrac%7B2%7D%7B9%7D%20%20%5Cfrac%7B%28%20%7Br%7D%5E%7B2%7D%28%20%5Crho%20-%20%20%5Csigma%29g%20%29%7D%7Bv%7D%20)
Substituting values of r, ρ, σ, v & g in the equation:
![\sf \implies \eta = \frac{2}{9} \frac{( {(0.15)}^{2} \times (7.85 - 1.25) \times 980.6)}{2.25}](https://tex.z-dn.net/?f=%20%5Csf%20%5Cimplies%20%5Ceta%20%3D%20%20%5Cfrac%7B2%7D%7B9%7D%20%20%5Cfrac%7B%28%20%7B%280.15%29%7D%5E%7B2%7D%20%20%5Ctimes%20%20%287.85%20-%201.25%29%20%5Ctimes%20980.6%29%7D%7B2.25%7D%20)
![\sf \implies \eta = \frac{2}{9} \frac{(0.0225 \times 6.6 \times 980.6)}{2.25}](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20%5Ceta%20%3D%20%20%5Cfrac%7B2%7D%7B9%7D%20%20%5Cfrac%7B%280.0225%20%5Ctimes%206.6%20%5Ctimes%20980.6%29%7D%7B2.25%7D%20)
![\sf \implies \eta = \frac{2}{9} \times \frac{145.6191}{2.25}](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20%5Ceta%20%3D%20%20%5Cfrac%7B2%7D%7B9%7D%20%20%5Ctimes%20%20%5Cfrac%7B145.6191%7D%7B2.25%7D%20)
![\sf \implies \eta = \frac{2}{9} \times 64.7196](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20%5Ceta%20%3D%20%20%5Cfrac%7B2%7D%7B9%7D%20%20%5Ctimes%2064.7196)
![\sf \implies \eta = 2 \times 7.191](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20%5Ceta%20%3D%20%202%20%5Ctimes%207.191)
![\sf \implies \eta = 14.382 \: poise](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20%5Ceta%20%3D%20%2014.382%20%5C%3A%20poise)
<span>Mechanical association learning used by an actor to memorize his lines</span>
The spring is initially stretched, and the mass released from rest (v=0). The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.100 s. This corresponds to half oscillation of the system. Therefore, the period of a full oscillation of the system is
![T=2 t = 2 \cdot 0.100 s = 0.200 s](https://tex.z-dn.net/?f=T%3D2%20t%20%3D%202%20%5Ccdot%200.100%20s%20%3D%200.200%20s)
Which means that the frequency is
![f= \frac{1}{T}= \frac{1}{0.200 s}=5 Hz](https://tex.z-dn.net/?f=f%3D%20%5Cfrac%7B1%7D%7BT%7D%3D%20%5Cfrac%7B1%7D%7B0.200%20s%7D%3D5%20Hz%20%20)
and the angular frequency is
![\omega=2 \pi f = 2 \pi (5 Hz)=31.4 rad/s](https://tex.z-dn.net/?f=%5Comega%3D2%20%5Cpi%20f%20%3D%202%20%5Cpi%20%285%20Hz%29%3D31.4%20rad%2Fs)
In a spring-mass system, the maximum velocity of the object is given by
![v_{max} = A \omega](https://tex.z-dn.net/?f=v_%7Bmax%7D%20%3D%20A%20%5Comega)
where A is the amplitude of the oscillation. In our problem, the amplitude of the motion corresponds to the initial displacement of the object (A=0.500 m), therefore the maximum velocity is
Answer:
A. -30 N
Explanation:
not sure but if the box isn't moving then the force opposite of Polly would be equal to the force she's exerting.