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nadezda [96]
3 years ago
9

You are asked to construct a mobile with four equal m = 141 kg masses, and three light rods of negligible mass and equal lengths

. The rods are of length 55 cm. (a) At what location on the level 1 rod should the free end of rod 2 be attached?
Physics
1 answer:
GarryVolchara [31]3 years ago
8 0

Answer:

The free end must be attached at a distance of 27.5 cm

Solution:

Mass, m = 141 kg

Length of the rods, L= 55 cm

Now,

As clear from fig. 1:

The free end of the rod 2 must be attached at:

F = 2 W

WL = W(55 - L)

2L = 55

L = 27.5 cm

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Answer:

The average acceleration of the ball during the collision with the wall is a=2,800m/s^{2}

Explanation:

<u>Known Data</u>

We will asume initial speed has a negative direction, v_{i}=-30m/s, final speed has a positive direction, v_{f}=26m/s, \Delta t=20ms=0.020s and mass m_{b}.

<u>Initial momentum</u>

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<u>final momentum</u>

p_{f}=mv_{f}=(26m/s)(m_{b})=26m_{b}\ m/s

<u>Impulse</u>

I=\Delta p=p_{f}-p_{i}=26m_{b}\ m/s-(-30m_{b}\ m/s)=56m_{b}\ m/s

<u>Average Force</u>

F=\frac{\Delta p}{\Delta t} =\frac{56m_{b}\ m/s}{0.020s} =2800m_{b} \ m/s^{2}

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Therefore, a=\frac{2800m_{b} \ m/s^{2}}{m_{b}} =2800m/s^{2}

8 0
3 years ago
Which descriptions best fit the labels? X: kinetic energy Y: potential energy X: potential energy Y: kinetic energy X: mechanica
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A photon with an energy E = 2.12 GeV creates a proton-antiproton pair in which the proton has a kinetic energy of 96.0 MeV. What
Oksanka [162]

Answer:

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Explanation:

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We need to calculate the kinetic energy of the anti proton

Using formula of energy

E_{photon}=m_{p}c^2+m_{np}c^2+K.E_{p}+K.E_{np}

We know that,

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Put the value into the formula

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Hence, The kinetic energy of the anti proton is 147.4 MeV.

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