Can you give the data please?
Answer:
Rolling case achieves greater height than sliding case
Step-by-step explanation:
For sliding ball:
- When balls slides up the ramp the kinetic energy is converted to gravitational potential energy.
- We have frictionless ramp, hence no loss due to friction.So the entire kinetic energy is converted into potential energy.
- The ball slides it only has translational kinetic energy as follows:
ΔK.E = ΔP.E
0.5*m*v^2 = m*g*h
h = 0.5v^2 / g
For rolling ball:
- Its the same as the previous case but only difference is that there are two forms of kinetic energy translational and rotational. Thus the energy balance is:
ΔK.E = ΔP.E
0.5*m*v^2 + 0.5*I*w^2 = m*g*h
- Where I: moment of inertia of spherical ball = 2/5 *m*r^2
w: Angular speed = v / r
0.5*m*v^2 + 0.2*m*v^2 = m*g*h
0.7v^2 = g*h
h = 0.7v^2 / g
- From both results we see that 0.7v^2/g for rolling case is greater than 0.5v^2/g sliding case.
Answer should be -4
since there is no other context, we can assume that x is -4.
Answer:
Only option d is not true
Step-by-step explanation:
Given are four statements about standard errors and we have to find which is not true.
A. The standard error measures, roughly, the average difference between the statistic and the population parameter.
-- True because population parameter is mean and the statistic are the items. Hence the differences average would be std error.
B. The standard error is the estimated standard deviation of the sampling distribution for the statistic.
-- True the sample statistic follows a distribution with standard error as std deviation
C. The standard error can never be a negative number. -- True because we consider only positive square root of variance as std error
D. The standard error increases as the sample size(s) increases
-- False. Std error is inversely proportional to square root of n. So when n decreases std error increases
Answer:
negitive
Step-by-step explanation:
good luck on finals
