The centripetal acceleration might better be expressed as −ω2r⃗ (t)−ω2r→(t) because it is a vector. The magnitude of the centripetal acceleration is v2radial/Rvradial2/R. The magnitude of the centripetal acceleration is v2tangential/Rvtangential2/R. A particle that is going along a path with local radius of curvature RRR at speed sss experiences a centripetal acceleration −s2/R−s2/R. If you are in a car turning left, the force you feel pushing you to the right is the force that causes the centripetal acceleration. In these statements vradialvradialv_radial refers to the component of the velocity of an object in the direction toward or away from the origin of the coordinate system or the rotation axis. Conversely, vtangentialvtangentialv_tangential refers to the component of the velocity perpendicular to vradialvradialv_radial. Identify the statement or statements that are false.

false  b) a = v²(radial) / r  and e)

Explanation:

Let's review given statement separately

a) centripetal acceleration

       a = v² / r

linear and angular velocity are related

     v = w r

we substitute

       a = w²r

this acceleration is directed to the center of the circle, so the vector must be negative

        a = - w r2

the bold are vector

True this statement

b) the magnitude is the scalar value

     a = v² / r

where v is the tangential velocity, not the radial velocity, so this statement is

False

c) This is true

      a = v²/ r

this speed is tangential

d) Newton's second law is

         F = m a

if the acceleration is centripetal

         F = m (- v² / r)

we substitute

         F = m (- s² / R)

the statement is true

e) when the car turns to the left, the objects have to follow in a straight line, which is why

you need a force toward the center of the circle to take the curve.

     Consequently there is no outward force

This statement is false

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

it about 600 mph from what i calculated