Physics, 16.01.2020 20:31 prettygirl1053473

In order to be safe going around a loop the centripetal acceleration cannot exceed (see problems from yesterday).

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Answer from: Elliewop

in order to be safe the centripetal acceleration should be higher than 39.2 m/s²

Explanation:

Assuming no friction while travel around the loop then we can apply the conservation of energy

where

K = kinetic energy = 1/2*m*v²

and m= mass , v= velocity at the bottom

K = kinetic energy = m*g*h , where g= gravity and h= height from the bottom of the loop

then by conservation of energy

Energy in the bottom of the loop = K + V(=0 because of h=0) = K = (1/2)*m*v²

Energy in the top of the loop = K(=0 in the limiting case of v=0 , that is falling from the loop) + V= V = m*g* 2R

where R= radius of the loop

then

Energy in the bottom of the loop= Energy in the top of the loop

(1/2)*m*v²= m*g* 2R

(1/2)*v²= 2*g* R

v²/R = 4*g

the centripetal acceleration ac is defined as

ac= v²/R= 4*g

then since g=9.8 m/s²

ac= 4*9.8 m/s² = 39.2 m/s²

therefore in order to be safe the centripetal acceleration should be higher than 39.2 m/s²

Answer from: Quest

it's when something is going so fast it makes a loud noise, breaking the sound barrier, for example: a jet makes a loud noise, not only cause of the jet but because they reached a point where they went so fast that you can year the noise from various parts, kinda like a police car, lets say the car was going so fast that you hear the siren of the cop car right next to you, but the car is probably a mile a way, it's cos the siren was next to you a few seconds if you understand me

Answer from: Quest

Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge, $v = \dfrac{e_p}{q}$ then the following gives us the potential difference, $\delta v = v_b - v_a$ $\begin{aligned} \delta v & = \dfrac{e_p_b}{q} - \dfrac{e_p_a}{q} \\ & = \dfrac{\delta e_p}{q} \end{aligned}$ the formula $\delta v = \delta e_p / q$ may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get $\delta e_p = q\delta v$ using q = 8.00 × 10⁻¹⁹ c and δv = 2.00 v, we get $\begin{aligned} \delta e_p & = (8.00\times10^{-19} \text{ c})(2.00\text{ v}) \\ & = 1.60\times 10^{-18}\text{ j} \end{aligned}$ the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.

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In order to be safe going around a loop the centripetal acceleration cannot exceed (see problems fr...