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9966 [12]
2 years ago
15

; (b) A uniform beam 150cm long weighs 3.5kg and

Physics
1 answer:
noname [10]2 years ago
4 0

Explanation:

(b) A uniform beam 150cm long weighs 3.5kg and

supported on knife-edges at its ends. The beam

supports a weight 7kg at a distance 30cm from

one end. Find the reactions of the supports.

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An air-track cart with mass m1=0.28kg and initial speed v0=0.75m/s collides with and sticks to a second cart that is at rest ini
arsen [322]
Kinetic energy is calculated through the equation,

   KE = 0.5mv²

At initial conditions,

  m₁:  KE = 0.5(0.28 kg)(0.75 m/s)² = 0.07875 J

  m₂ : KE = 0.5(0.45 kg)(0 m/s)² = 0 J

Due to the momentum balance,

   m₁v₁ + m₂v₂ = (m₁ + m₂)(V)

Substituting the known values,

   (0.29 kg)(0.75 m/s) + (0.43 kg)(0 m/s) = (0.28 kg + 0.43 kg)(V)

   V = 0.2977 m/s

The kinetic energy is,
   KE = (0.5)(0.28 kg + 0.43 kg)(0.2977 m/s)²
   KE = 0.03146 J

The difference between the kinetic energies is 0.0473 J. 
7 0
2 years ago
A truck covered 2/7 of a journey at an average speed of 40
slavikrds [6]

Answer:8h

Explanation:

8 0
2 years ago
Read 2 more answers
Lorenzo is making a prediction. “I learned that nonmetals increase in reactivity when moving from left to right. So I predict th
nadezda [96]
That prediction is not correct because Xenon is extremely stable; column 18 of the periodic table contains the noble gasses, which are stable because their outer-most energy levels are completely filled. Having the octet (8) of valence electrons means that the element no longer needs to lose or gain electrons to gain stability.

The column 17 elements are unstable because they only have one valence electron short of the stable octet configuration of the noble gasses.
6 0
2 years ago
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A civil engineer wishes to redesign the curved roadway in the example What is the Maximum Speed of the Car? in such a way that a
vlabodo [156]

Answer:

24.3 degrees

Explanation:

A car traveling in circular motion at linear speed v = 12.8 m/s around a circle of radius r = 37 m is subjected to a centripetal acceleration:

a_c = \frac{v^2}{r} = \frac{12.8^2}{37} = 4.43 m/s^2

Let α be the banked angle, as α > 0, the outward centripetal acceleration vector is split into 2 components, 1 parallel and the other perpendicular to the road. The one that is parallel has a magnitude of 4.43cosα and is the one that would make the car slip.

Similarly, gravitational acceleration g is split into 2 component, one parallel and the other perpendicular to the road surface. The one that is parallel has a magnitude of gsinα and is the one that keeps the car from slipping outward.

So gsin\alpha = 4.43cos\alpha

\frac{sin\alpha}{cos\alpha} = \frac{4.43}{g} = \frac{4.43}{9.81} = 0.451

tan\alpha = 0.451

\alpha = tan^{-1}0.451 = 0.424 rad = 0.424*180/\pi \approx 24.3^0

3 0
2 years ago
A 4.0-mF capacitor initially charged to 50 V and a 6.0-mF capacitor charged to 30 V are connected to each other with the positiv
Juli2301 [7.4K]

Answer:

<em>The final charge on the 6.0 mF capacitor would be 12 mC</em>

Explanation:

The initial charge on 4 mF capacitor  = 4 mf  x 50 V = 200 mC

The initial Charge on 6 mF capacitor  = 6 mf x 30 V =180 mC

Since the negative ends are joined together  the total charge on both capacity would be;

q = q_{1} -q_{2}

q = 200 - 180

q = 20 mC

In order to find the final charge on the 6.0 mF capacitor we have to find the combined voltage

q = (4 x V) + (6 x V)

20 = 10 V

V = 2 V

For the final charge on 6.0 mF;

q = CV

q = 6.0 mF x 2 V

q =  12 mC

Therefore the final charge on the 6.0 mF capacitor would be 12 mC

5 0
2 years ago
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