Newton's first and magnitude of thrust?

Question

If you're riding your bike at constant velocity, then the magnitude of the force of thrust -a- the magnitude of force of friction, but the two forces are in -b- directions.

a) is less than, is greater than, is equal to
b) opposite, the same, perpendicular

What would the answer be and why?

Answer 1

forces are opposite and equal, otherwise you would change speed/direction.

Answer 2

one million)tension is comparable to the mass and its acceleration(i think of this is pronounced interior the 2nd regulation of action which grew to become into made by way of Newton) in formulation f=ma substitute the given 14.6N=(.350kg)(a) a=14.6N/.350kg a=40-one.seventy one m/s^2 this is the cost(because of the fact it purely contains the relatively volume or quantity and there is not any direction, if it has this is going to be a vector already) 2)a=(v2-v1)/t you already be conscious of the acceleration so in simple terms substitute the values 40-one.71m/s^2=(v2-0)/2.5s (0 because of the fact the rocket all started from relax) v2=(2.5s)(40-one.71m/s^2) v2=104.28m/s(the cost you're seeking for) desire this facilitates....