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Aleksandr [31]
2 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]2 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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What will be the average velocity of a body falling in free fall on Earth for 3 s?
SpyIntel [72]

Answer:

29.4m/s

Explanation:

Given parameters:

Time  = 3s

Unknown:

Average velocity  = ?

Solution:

To solve this problem, we use the expression below:

      v  = u + gt

v is the average velocity

u is the initial velocity  = 0m/s

g is the acceleration due to gravity  = 9.8m/s²

t is the time

So;

        v  = 0 + (9.8 x 3)  = 29.4m/s

6 0
2 years ago
To test the quality of water sources in her area, Shameka collects five samples from each water source and tests the quality of
xenn [34]

The missing part of the incomplete question is given below:

Which important step of scientific design is Shameka conducting?

repetition

replication

verification of results

using controlled variables

Answer:

Verification of results

Explanation:

The way toward gathering five examples of water from various sources is conveyed to confirm the outcome. By gathering water from five distinct areas of a similar source the analyst can genuinely find out the nature of the water in her region of remain.  

On the off chance that after examples are tried it is found the water isn't sound, the outcomes would be acknowledged as it has been appropriately checked and a  proper move would be made.

Thus, the correct answer is - verification of results

6 0
3 years ago
A current of 4.00 mA flows through a copper wire. The wire has an initial diameter of 4.00 mm which gradually tapers to a diamet
lesya692 [45]

The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

  • <em>Current flowing in the wire, I = 4.00 mA</em>
  • <em>Initial diameter of the wire, d₁ = 4 mm = 0.004 m</em>
  • <em>Final diameter of the wire, d₂ = 1 mm = 0.001 m</em>
  • <em>Length of wire, L = 2.00 m</em>
  • <em>Density of electron in the copper, n = 8.5 x 10²⁸ /m³</em>

<em />

The initial area of the copper wire;

A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2

The final area of the copper wire;

A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2

The initial drift velocity of the electrons is calculated as;

v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s

The final drift velocity of the electrons is calculated as;

v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7}  \ m/s

The change in the mean drift velocity is calculated as;

\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s

The time of motion of electrons for the initial wire diameter is calculated as;

t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s

The time of motion of electrons for the final wire diameter is calculated as;

t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s

The average acceleration of the electrons is calculated as;

a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

Learn more here: brainly.com/question/22406248

7 0
2 years ago
1. The mass defect of iron-56 is 0.52875 amu. What is the energy equivalent of this mass?
Delvig [45]

Answer:

Hello! Your answer is BELOW

Explanation:

1.About 91.754% of all iron is iron-56. Of all nuclides, iron-56 has the lowest mass per nucleon. With 8.8 MeV binding energy per nucleon, iron-56 is one of the most tightly bound nuclei.

2.The atomic weight of lead is quite variable in nature because the three heaviest isotopes are the stable end-products of the radioactive decay of uranium (238U to 206Pb and 235U to 207Pb) and thorium (232Th to 208Pb).

3.Mass defect for uranium-238 is 3.983 × 10-25 kg.

4.Energy and Mass Are Relative

The equation E = mc^2 states that the amount of energy possessed by an object is equal to its mass multiplied by the square of the speed of light.

Hope I helped! Ask me anything if you have any questions. Brainiest plz!♥ Hope you make a 100%. Have a nice morning! -Amelia♥

7 0
3 years ago
What would a series circuit be used for?
igor_vitrenko [27]

Answer:

C

Explanation:

a series circuit would be an odd choice to power a battery or light a lamp when a direct would be much more efficient, and it's not converting types of energy, so C is the best possible answer

6 0
2 years ago
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