Answer:
<em>The period of the motion will still be equal to T.</em>
<em></em>
Explanation:
for a system with mass = M
attached to a massless spring.
If the system is set in motion with an amplitude (distance from equilibrium position) A
and has period T
The equation for the period T is given as
where k is the spring constant
If the amplitude is doubled, the distance from equilibrium position to the displacement is doubled.
Increasing the amplitude also increases the restoring force. An increase in the restoring force means the mass is now accelerated to cover more distance in the same period, so the restoring force cancels the effect of the increase in amplitude. Hence, <em>increasing the amplitude has no effect on the period of the mass and spring system.</em>
Answer:
<em><u>M</u></em><em><u>a</u></em><em><u>t</u></em><em><u>h</u></em><em><u>e</u></em><em><u>m</u></em><em><u>a</u></em><em><u>t</u></em><em><u>i</u></em><em><u>c</u></em><em><u>a</u></em><em><u>l</u></em><em><u>l</u></em><em><u>y</u></em><em><u>:</u></em>
That will be
<em>=</em><em> </em><em>1</em><em>5</em><em>0</em><em>0</em><em> </em><em>x</em><em> </em><em>1</em><em>5</em><em> </em><em>x</em><em> </em><em>4</em><em>5</em><em>0</em><em>0</em>
<em>=</em><em> </em><em><u>1</u></em><em><u>0</u></em><em><u>1</u></em><em><u>,</u></em><em><u>2</u></em><em><u>5</u></em><em><u>0</u></em><em><u>,</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em>
Answer:
you absolute buffoon Use Ohms' Law: V = RI
V = (1x10^3)(5x10^-3) = 5 volts
Yes, this is in the range of normal household voltages.
Explanation:
Answer:
v =2.02
Explanation:
v^2=0.05-4.9
v^2=-4.85
square root both side
v=2.02
^^^^this is a not a perfect square
Answer:
D.None of these
Explanation:
The derivation of acceleration formula:
Let us call the 5kg mass and the 4kg mass . If the tension in the string is then for the mass
(1). <em>(the negative sign on the right side indicates that acceleration is downwards)</em>
And for the mass
(2). <em> (the acceleration is upwards, hence the positive sign)</em>
Solving for in the 2nd equation we get:
,
and putting this into the 1st equation we get:
Back to the question:
Using the formula for the acceleration we find
which is the acceleration that none of the given choices offer. Also, the acceleration of the two blocks is the same, because if it weren't, the difference in the instantaneous velocities of the objects would cause the string to break. Therefore, these two reasons make us decide that none of the choices are correct.