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FromTheMoon [43]
3 years ago
15

A 75 g, 30 cm long rod hangs vertically on a friction less, horizontal axel passing through the center. A 10 g ball of clay trav

elling horizontally at 2.5 m/s hits and stick to the very bottom tip of the rod. To what maximum angle, measured from vertical, does the rod rotates when the ball of clay is attached to the rod?
Physics
1 answer:
kotegsom [21]3 years ago
7 0
Angular momentum is conserved, just before the clay hits and just after; 
<span>mv(L/2) = Iw </span>

<span>I is the combined moment of inertia of the rod, (1/12)ML^2 , and the clay at the tip, m(L/2)^2 ; </span>
<span>I = [(1/12)ML^2 + m(L/2)^2] </span>

<span>Immediately after the collision the kinetic energy of rod + clay swings the rod up so the clay rises to a height "h" above its lowest point, giving it potential energy, mgh. From energy conservation in this phase of the problem; </span>
<span>(1/2)Iw^2 = mgh </span>

<span>Use the "w" found in the conservation of momentum above; and solve for "h" </span>
<span>h = mv^2L^2/8gI </span>

<span>Next, get the angle by noting it is related to "h" as; </span>
<span>h = (L/2) - (L/2)Cos() </span>

<span>So finally </span>
<span>Cos() = 1- 2h/L = 1 - mv^2L/4gI </span>

<span>m=mass of clay </span>
<span>M=mass of rod </span>
<span>L=length of rod </span>
<span>v=velocity of clay</span>
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A large reflecting telescope has an objective mirror with a 14.0 m radius of curvature. What angular magnification in multiples
goldenfox [79]

Answer:

The magnification is  m =  -2.15

Explanation:

From the question we are told that

   The  radius is  r =  14.0 \ m

    The  focal length eyepiece is  f_e  =  3.25 \ m

Generally the objective focal length is mathematically represented as

        f_o  =  \frac{r}{2}

=>     f_o  =  \frac{14}{2}

=>     f_o  =  7 \ m

The  magnification is mathematically represented as

      m =  - \frac{f_o }{f_e }

=>    m =  - \frac{7 }{ 3.25 }

=>   m =  -2.15

5 0
4 years ago
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FromTheMoon [43]

Answer:

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Explanation:

6 0
2 years ago
The net external force on the propeller of a 3.0 kg model airplane is 6.8 N forward.What is the acceleration of the airplane? An
solong [7]

The correct answer to the question is 2.27 m/s^2 i.e the acceleration of the body is 2.27 m/s^2 along the forward direction.

CALCULATION:

As per the question, the net external force on the propeller of model airplane F =  6.8 N.

The mass of the model air plane m = 3.0 kg

We are asked to calculate the acceleration of the air plane.

From Newton's second law of motion, we know that the net external force acting on a body is equal to the product of mass with acceleration of that body.

Mathematically force F = m × a

                                  ⇒  a=\ \frac{F}{m}

                                          =\ \frac{6.8\ N}{3.0\ kg}

                                          =\ 2.27\ m/s^2                  [ans]

The direction of acceleration is along the direction of force. Hence, the acceleration of the propeller is 2.27 m/s^2 along forward direction.

8 0
4 years ago
At noon, ship A is 140 km west of ship B. Ship A is sailing east at 30 km/h and ship B is sailing north at 25 km/h. How fast (in
Luba_88 [7]

Answer:

The rate of change of distance between the two ships is 18.63 km/h

Explanation:

Given;

distance between the two ships, d = 140 km

speed of ship A = 30 km/h

speed of ship B = 25 km/h

between noon (12 pm) to 4 pm = 4 hours

The displacement of ship A at 4pm = 140 km - (30 km/h x 4h) =

140 km - 120 km = 20 km

(the subtraction is because A is moving away from the initial position and the distance between the two ships is decreasing)

The displacement of ship B at 4pm = 25 km/h x 4h = 100 km

Using Pythagoras theorem, the resultant displacement of the two ships at 4pm is calculated as;

r² = a²   +  b²

r² = 20²  +  100²

r = √10,400

r = 101.98 km

The rate of change of this distance is calculated as;

r² = a²   +  b²

r = 101.98 km, a = 20 km, b = 100 km

2r(\frac{dr}{dt} ) = 2a(\frac{da}{dt} )  + 2b(\frac{db}{dt} )\\\\r(\frac{dr}{dt} ) = a(\frac{da}{dt} )  + b(\frac{db}{dt} )\\\\101.98(\frac{dr}{dt} ) = 20(-30 )  + 100(25 )\\\\101.98(\frac{dr}{dt} ) = -600 + 2,500\\\\101.98(\frac{dr}{dt} ) = 1900\\\\\frac{dr}{dt}  = \frac{1900}{101.98} = 18.63 \ km/h

5 0
3 years ago
A copper transmission cable 180 km long and 11.0 cm in diameter carries a current of 135 A.
zheka24 [161]

Answer:

a) 43.98 V

b) E = 21.37 MJ

Explanation:

Parameters given:

Length of cable = 180 km = 180000 m

Diameter of cable = 11 cm = 0.11 m

Radius = 0.11 / 2 = 0.055 m

Current, I = 135 A

a) To find the potential drop, we have to find the voltage across the wire:

V = IR

=> V = IρL / A

where R = resistance

L = length of cable

A = cross-sectional area

ρ = resistivity of the copper wire = 1.72 * 10^(-8) Ωm

Therefore:

V = (135 * 1.72 * 10^(-8) * 180000) / (π * 0.055^2)

V = 43.98 V

The potential drop across the cable is 43.98 V

b) Electrical energy is given as:

E = IVt

where t = time taken = 1 hour = 3600 s

Therefore, the energy dissipated per hour is:

E = 135 * 43.98 * 3600

E = 21.37 MJ (mega joules, 10^6)

7 0
3 years ago
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