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olganol [36]
3 years ago
8

A student thinks that any real vibration must be damped. Is the student correct? If so, give convincing reasoning. If not, give

an example of a real vibration that keeps constant amplitude forever if the system is isolated.
Physics
1 answer:
Stels [109]3 years ago
6 0

Answer:

Yes the student is correct

Explanation:

The first law of thermodynamics states that energy can neither be created nor destroyed

The second law of thermodynamics states that the entropy (disorderliness) of an isolated system always increases

Therefore, whereby energy is not supplied to maintain the orderly oscillatory motion with constant amplitude, the amplitude of the system is bound to reduce with time that is the vibration of the system must be damped

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Four charges 7 × 10−9 C at (0 m, 0 m), −9 × 10−9 C at (3 m, 3 m), 7 × 10−9 C at (1 m, 3 m), and −8 × 10−9 C at (−3 m, 2 m), are
Ivanshal [37]

Answer:

Magnitude of the resulting force on the 7 nC charge at the origin:

Fn₁= 23.95*10⁻⁹ N

Explanation:

Look at the attached graphic:

Charges of positive signs exert repulsive forces on q₁ + and charges of negative signs exert attractive forces on q₁ +.

q₁ experiences three forces (F₂₁,F₃₁,F₄₁) and we calculate them with Coulomb's law:

F = (k*q₁*q)/(d)²

d_{12} = \sqrt{3^{2}+3^{2}  }  = \sqrt{18} m : distance from q₁ to q₂

(d₁₂)² = 18 m²

d_{13} =\sqrt{1^{2}+3^{2}  } = \sqrt{10} m  : distance from q₁ to q₃

(d₁₃)² = 10 m²

d_{14} =\sqrt{3^{2}+2^{2}  } = \sqrt{13} m  : distance from q₁ to q₄

(d₁₄)² = 13 m²

K=  8.98755 × 10⁹ N *m²/C²

q₁=  7*10⁻⁹C

k*q₁=8.98755*10⁹ *7*10⁻⁹= 62.9

F₂₁= (62.9)*(9* 10⁻⁹) /(18) = 31.45*10⁻⁹ C

F₃₁= (62.9)*(7* 10⁻⁹) /(10) = 44*10⁻⁹ C

F₄₁= (62.9)*(8* 10⁻⁹) /(13) = 38.7*10⁻⁹ C

x-y components of the net force on q₁ (Fn₁):

α= tan⁻¹(3/3)= 45°  ,  β= tan⁻¹(3/1)= 71.56° , θ= tan⁻¹(2/3)= 33.69°

Fn₁x = F₂₁x+ F₃₁x+F₄₁x

F₂₁x =+ F₂₁*cosα =+ (31.45*10⁻⁹)* (cos 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*cosβ = - ( 44*10⁻⁹)* (cos 71.56°) = -13.91 *10⁻⁹ N

F₄₁x= -F₄₁*cosθ = -(38.7*10⁻⁹)* (cos 33.69°) = -32.2*10⁻⁹ N

Fn₁x = (+22.24 - 13.91 - 32.2)*10⁻⁹ N

Fn₁x = -23.87 *10⁻⁹ N

Fn₁y = F₂₁y+ F₃₁y+F₄₁y

F₂₁x =+ F₂₁*sinα =+ (31.45*10⁻⁹)* (sin 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*sinβ = - ( 44*10⁻⁹)* (sin 71.56°) = -41.74 *10⁻⁹ N

F₄₁x= +F₄₁*sinθ = +(38.7*10⁻⁹)* (sin 33.69°) =+21.47*10⁻⁹ N

Fn₁y = (22.24 -41.74+21.47)*10⁻⁹ N  

Fn₁y = 1.97*10⁻⁹ N

Magnitude of the resulting force on the 7 nC charge at the origin (q₁):

F_{n1} =\sqrt{(Fn_{1x} )^{2}+(Fn_{1y} )^{2} }

F_{n1} =\sqrt{(23.87 )^{2}+(1.97 )^{2} }

Fn₁= 23.95*10⁻⁹ N

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3 years ago
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hodyreva [135]
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A 0.48 kg circus monkey is about to be shot from a cannon as part of his thrilling circus act. Draw a free body diagram labeling
Alenkinab [10]
What is it that you need help on?
6 0
2 years ago
William WangHomework #33 Regents Review (3)0989Assignment Mode : Open (Time On Task) 9 of 25 ListenDuring a collision, an 84-kil
Lemur [1.5K]

Answer:

1.7\cdot 10^3 N

Explanation:

The impulse theorem states that the product between the force and the time interval of the collision is equal to the change in momentum:

F \Delta t = m \Delta v

where

F is the force

\Delta t is the time interval

m is the mass

\Delta v is the change in velocity

Here we have

m = 84 kg

\Delta t = 1.2 s

\Delta v = 24 m/s

So we can solve the equation to find the force:

F= \frac{m \Delta v}{\Delta t }=\frac{(84 kg)(24 m/s)}{1.2 s}=1680 N \sim 1.7\cdot 10^3 N

4 0
3 years ago
ASAP please 25 points if right
garik1379 [7]

Answer:

1. The resistance of any physical object to any velocity

2. It continues in it's existing state of rest or uniform motion

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5 0
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