Answer:
A) continue to move to the right, with its speed increasing with time.
Explanation:
As long as force is positive , even when it is decreasing , it will create positive increase in velocity . Hence the body will keep moving with increasing velocity towards the right . The moment the force becomes zero on continuously decreasing , the increase in velocity stops and the body will be moving with the last velocity uniformly towards right . When the force acting on it becomes negative , even then the body will keep on going to the right till negative force makes its velocity zero . D uring this period , the body will keep moving towards right with decreasing velocity .
Hence in the present case A , is the right choice.
Explanation:
A. Walk the person around
Answer: vf1/vf2= 1/ sqrt(2)
Explanation :on the moon no drag force so we have only the force of gravity. aceleration is g(moon)= 1.62m/s2.the rest is basic kinematics
if the rock travels H to the bottom we can calculate velocity:
vo=0m/s (drops the rock) , yo=0
vf*vf= vo*vo+2g(y-yo)
when the rock is halfway y = H/2 so:
vf1*vf1=2*g*H/2 so vf1 = sqrt(gH)
when the rock reach the bottom y=H so:
vf2*vf2=2*g*H so vf2 = sqrt(2gH)
so vf1/vf2= 1/ sqrt(2)
good luck from colombia
Answer:
reliability
accuracy
Explanation:
If a reading of a measurement is consistently the same then the measurement is reliable.
If a reading of measurement is close the actual value of the measurement then the reading is accurate.
Here, a stationary tree shows reading 6 mph once and 0 mph another instant. So, neither the reading of a measurement is consistent not the reading of measurement is close the actual value.
Hence, the radar has problems in its reliability and accuracy
Answer:
speed of golf ball is 1.15 × m/s
and % of uncertainty in speed = 2.07 × %
Explanation:
given data
mass = 45.9 gram = 0.0459 kg
speed = 200 km/hr = 55.5 m/s
uncertainty position Δx = 1 mm = m
to find out
speed of the golf ball and % of speed of the golf ball
solution
we will apply here heisenberg uncertainty principle that is
uncertainty position ×uncertainty momentum ≥ ......1
Δx × ΔPx ≥
here uncertainty momentum ΔPx = mΔVx
and uncertainty velocity = ΔVx
and h = 6.626 × Js
so put here all these value in equation 1
× 0.0459 × ΔVx =
ΔVx = 1.15 × m/s
and
so % of uncertainty in speed = ΔV / m
% of uncertainty in speed = 1.15 × / 55.5
% of uncertainty in speed = 2.07 × %