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Alexxx [7]
4 years ago
5

2. A mass M is suspended from a frictionless hinge on a massless stick of length b. A massless rod is held perpendicular to the

first rod (making ski-lift ``T-bar'') and another mass m slides frictionlessly along the second rod. Both rods pivot in the plane of the T-bar. Write equations of motion for both masses under the hilarious assumption that the masses can pass through each other without interacting in any way. What are the equations for small oscillations of m and M
Physics
1 answer:
stealth61 [152]4 years ago
6 0

Answer: f=Mg for mass M

And f=mg for m

Equation for small oscillation of m is T=2Π√m/k

For M it will be

T=2Π√M/k

Explanation: the force acting on mass M will be f=Mg in a downward direction f=mg in a downward direction.

Net force acting downward will be f=g(M+m).

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How far will an arrow travel if it shot horizontally at 85.3m/s and it is 1.5m above the ground ?
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3 years ago
A plane flies 446 km east from city A to city B in 43.0 min and then 939 km south from city B to city C in 1.10 h. For the total
NeTakaya

Answer:

(a) Magnitude is 1039 km

(b) Direction of the displacement is 64.59^{\circ} South of East

(c) Average velocity magnitude is 570.88 km

(d) The direction of average velocity is 64.59^{\circ} South of East

(e) Average speed is 759.34 km/h

Solution:

Distance moved from A to B in East direction, \vec{AB} = 446 km

Distance moved from B to C in South direction, \vec{BC} = - 939 km

Time taken to move from A to B, t = 43.0 min = 0.72 h

Time taken to move from B to C, t' = 1.10 h

Now,

(a) The magnitude of displacement of the plane is provided by AC as shown in fig 1 and can be given as:

AC = \sqrt{(AB)^{2} + (BC)^{2}}

AC = \sqrt{(446)^{2} + (- 939)^{2}} = 1039 km

(b) Direction of the displacement is given by:

tan\theta = \frac{\vec{BC}}{\vec{AB}}

\theta = tan^{- 1}(\frac{- 939}{\vec{446}}) = - 64.59^{\circ}

64.59^{\circ} South of East

(c) Magnitude of the average speed is given by:

v_{avg} = \frac{AC}{t + t'}

v_{avg} = \frac{1039}{1.82} = 570.88 km/h

(d) The direction of the average velocity is the same as that of the displacement, i.e., 64.59^{\circ} South of East.

(e) The average speed of the [plane is given by:

v'_{avg} = \frac{Total\ Distance\ Traveled}{Total\ Time}

v'_{avg} = \frac{446 + 939}{1.10 + 0.72} = 759.34 km/h

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