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goldfiish [28.3K]
3 years ago
10

A type of cuckoo clock keeps time by having a mass bouncing on a spring, usually something cute like a cherub in a chair. What f

orce constant is needed to produce a period of 0.320 s for a 0.0190-kg mass?
Physics
1 answer:
Paul [167]3 years ago
8 0

Answer:

k = 7.32 N/m

Explanation:

We know that the time period of a spring mass system is given as follows:

T = 2π√(m/k)

T² = 4π²m/k

k = 4π²m/T²

where,

k = force constant = ?

m = mass = 0.019 kg

T = Time Period = 0.32 s

Therefore,

k = 4π²(0.019 kg)/(0.32 s)²

<u>k = 7.32 N/m</u>

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kifflom [539]
In an atom of hydrogen the orbit radius is given by the formula:
r = n² · α₀

where:
n = number of orbit = 15
α₀ = Bohr radius (innermost radius) = 0.529 Â

Since d = 2 · r, we can write:
d = n² · d₀
   = 15² · 1.06
   =  238.5 Â

Hence, the <span>diameter of the fifteenth orbit of the hydrogen atom is 238.5 </span>Â.
8 0
3 years ago
Series circuit when you had one bulb and battery voltage was at 9 volts, what was current into battery?
KengaRu [80]

Answer:

incomplete question, resistor must be there

Explanation:

7 0
3 years ago
Approximating the eye as a single thin lens 2.70 cmcm from the retina, find the focal length of the eye when it is focused on an
Lena [83]

Answer:

0.37 cm

Explanation:

The image is formed on the retina which is at a constant distance of 2.70 cm to the lens. Therefore, image distance = 2.70 cm.

The object is at a distant of 265 cm to the lens of the eye.

From lens formula,

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

where: f is the focal length, u is the object distance and v is the image distance.

Thus, u = 265.00 cm and v = 2.70 cm.

\frac{1}{f} = \frac{1}{265} + \frac{27}{10}

  = \frac{10+7155}{2650}

\frac{1}{f}  = \frac{7165}{2650}

⇒ f = \frac{2650}{7165}

      = 0.37

The focal length of the eye is 0.37 cm.

8 0
3 years ago
Which of the following is group 1? a. alkali metals b. alkaline earth metals c. halogens d. transition metals
Elanso [62]
The answer is A. Alkali metals
4 0
4 years ago
Read 2 more answers
1.33 m^3 of fluid flows out of a pipe in 24.5 s. the fluid leaves the pipe at 3.55 m/s. what is the area of the pipe. (unit=m^2)
Margaret [11]

Answer:

0.015meter^{2}

Explanation:

Total volume of water coming out = 1.33meter^{3}

Also volume = Cross sectional area*Length covered

Length covered = Velocity *time

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                           =86.97 meter

Let the cross sectional area be A.

1.33 = 86.97*A

A =0.015meter^{2}

3 0
4 years ago
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