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Aleksandr [31]
3 years ago
14

A 4.5-kg object oscillates on a horizontal spring with an amplitude of 3.8 cm. Its maximum acceleration is 26 m/s2 . Find (a) th

e force constant k, (b) the frequency, and (c) the period of the motion.
Physics
1 answer:
zimovet [89]3 years ago
3 0

Answer:

F = - K X     force constant for spring

a = F / m      maximum acceleration

F = 4.5 kg * 26 m/s^2 = 117 Newtons

(A)  K = 117 N / .038 m = 3079 N/m

ω = (K/M)^1/2 = (117/5)^1/2 = 4.84 / sec

(B) f = ω / 2 pi = 4.84 / 6.28 = .77 /sec

(C) P = 1 / f = 1/.77   = 1.30 sec

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5 0
3 years ago
Two high-current transmission lines carry currents of 29.0 A and 78.0 A in the same direction and are suspended parallel to each
jarptica [38.1K]

Answer with Explanation:

We are given that

I_1=29 A

I_2=78 A

d=38 cm=\frac{38}{100}=0.38 m

1 m=100 cm

a.Length of segment,l=20 m

Magnetic force ,F=\frac{2\mu_0I_1I_2 l}{4\pi d}

\frac{\mu_0}{4\pi}=10^{-7}

Substitute the values

F=\frac{10^{-7}\times 29\times 78\times 20}{0.38}=0.0119 N

Hence, the magnetic force exert by each segment on the other=0.0119 N

b.We know that when current carrying in the wires are in same direction then the force will attract to each other.

Hence, the force will be attractive.

4 0
3 years ago
Read 2 more answers
An RL circuit contains a resistor with R = 6800 Ω and an inductor with L = 2300 µH. If the impedance of this circuit is 160,000
Rainbow [258]

| Impedance | = √ [R² +(ωL)²]

R² = 6800² = 4.624 x 10⁷
 
(ωL)² = (2 · π · f · 2.3 · 10⁻³)²

          = 2.0884 x 10⁻⁴  f²

| Z | =  √[ (4.624 x 10⁷) + (2.0884 x 10⁻⁴ f²) ]  =  1.6 x 10⁵

     (1.6 x 10⁵)²  =  (4.624 x 10⁷) + (2.0884 x 10⁻⁴ f²)

     (2.56 x 10¹⁰) - (4.624 x 10⁷)  =  2.0884 x 10⁻⁴ f²


Frequency² =   (2.56 x 10¹⁰ - 4.624 x 10⁷)  /  2.0884 x 10⁻⁴

                    =       2.555 x 10¹⁰ / 2.0884 x 10⁻⁴

                    =          1.224 x 10¹⁴ 

                    =          122,400 GHz          <== my calculation

                                      11.1 MHz           <== online impedance calculator

Obviously, I must have picked up some rounding errors
in the course of my calculation. 
  











7 0
3 years ago
Sometimes balance point may not be obtained on the potentiometer wire why​
scoundrel [369]
Solution
Let a cell of emf E be connected across the entire length L of a potentiometer wire . Now , if the balance point is obtained at a length l during measurement of an unknown voltage

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The balance point is not on the potentiometer wire - this statement means that

. In that case ,
l > L
V > E
8 0
3 years ago
Select the correct answer.
VladimirAG [237]

Answer:

It's effective temperature.

Explanation:

8 0
3 years ago
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