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borishaifa [10]
4 years ago
13

Which properties should an object, system, or process have in order for it to be a useful standard of measurement for a physical

quantity such as length or time? (Choose all that apply) -precise-restricted availability-defined in SI units-small-reproducible
Physics
1 answer:
Temka [501]4 years ago
3 0

Answer:

Precise and Reproducible

Explanation:

For a system or a process to be useful a measurement standard for any physical quantity, it must be:

  • Precise, Precision is important in any measurement.
  • The values of the measurement obtained should be reproducible
  • The measurement may be large or small
  • It may or may not be in SI units
  • It should be available easily.
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What is the energy of the spring-mass system when the mass first passes through the equilibrium position? (you may wish to inclu
Brrunno [24]
The spring-mass system moves by simple harmonic motion, where there is a continuous conversion from elastic potential energy to kinetic energy and viceversa.

The total mechanical energy of the system at any moment of the motion is
E=U+K= \frac{1}{2}kx^2 +  \frac{1}{2}mv^2
where the first term U is the elastic potential energy, with k being the spring constant and x the displacement of the spring with respect to its rest position, and the second term K is the kinetic energy, with m being the mass of the object attached to the spring and v its speed.

The total energy E is constant during the oscillation of the spring, but the values of U and K change. In fact, when the displacement of the spring is maximum (x is maximum), then all the energy is potential energy U, because the speed of the object is zero (it's the moment when the mass is changing direction). On the contrary, when the mass crosses the equilibrium position (rest position) of the spring, then the potential energy is zero (U=0) because the displacement is zero (x=0), and so all the energy is kinetic energy of the motion, and so K is maximum.
3 0
3 years ago
The mass of a sample of sodium bicarbonate is 2. 1 kilograms (kg). There are 1,000 grams (g) in 1 kg, and 1 Times. 109 nanograms
slega [8]

2.1 kg of sodium bicarbonate is equal to the 2.1 x 10¹² ng of sample. Option D is correct.

Mass is the quantity of the substance in the body or object. The SI unit of mass is Kilogram.

There are other units of measure,

  • Milligram: 1 g is equal to the \bold {10^3 \ mg}
  • Micro-gram: 1 g is equal to \bold {10^{6} \ \mu g}
  • Nano-gram: 1 g is  is equal to\bold {10^{9} \ ng}

First convert kg to gram,

Since, 1 Kg = 1000 g

2.1 kg = grams of sample

So,

Do the cross multiplication,

\rm mass\ of\ sample = \dfrac {2.1\ kg \times  1000\ g }{ 1 kg}\\\\\rm mass\ of\ sample =2100 g

Now, convert 2100 g to nano-grams

Since, 1 g = 1 x 10⁹ ng

2100 g = ng of sample

So,

Do the cross multiplication,

\rm mass\ of\ sample  = \dfrac {2100 g \times  1 \times  10^9 ng }{1\ g}\\\\\rm mass\ of\ sample  = 2.1 \times 10^1^2 ng

Therefore, 2.1 kg of  sodium bicarbonate is equal to the 2.1 x 10¹² ng of sample.

To know more about Mass units,

brainly.com/question/489186

3 0
3 years ago
The index of refraction of a transparent substance is 1.7, and the speed of light in air is 3.00 · 108 s. Calculate the speed of
Natalija [7]
Given the index of refraction, n and speed of light in the vacuum, c, we can solve for the speed of light in the transparent substance by the equation below.
                                                n = c / v
where v is our unknown.
Substituting the known values,
                                            1.7 = (3 x 10^8 m/s) / v
The value of v is equal to 1.76 x 10^8 m/s. 
5 0
4 years ago
Hi, I am having some difficulties in solving this question. Could someone please explain this question to me in detail. The thin
aniked [119]

Answer:

a) 3.0×10⁸ m

b) 0 m

Explanation:

Displacement is the distance from the starting position to the final position.

a) In half a year, the Earth travels from one point on the circle to the point on the exact opposite side of the circle (from 0° to 180°).  The distance between the points is the diameter of the circle.

x = 2r

x = 2 (1.5×10⁸ m)

x = 3.0×10⁸ m

b) In a full year, the Earth travels one full revolution, so it ends up back where it started.  The displacement is therefore 0 m.

8 0
4 years ago
Suppose that the period of a particular ideal mass-spring system is 5 s . What would be the period of the system if the mass wer
yan [13]

Answer: T2 = 7.07s

Explanation: The period of a loaded spring of spring constant k and mass m is given by

T= 2π √m/k

With 2π constant and k, it can be seen with little algebra that

T² is proportional to mass m

Hence (T1)²/m1 = (T2) ²/m2

Where T1 = 5, T2 =?, let m1 = m hence m2 = 2m.

By substituting, we have that

5²/m = (T2) ²/2m

25 / m = (T2) ²/2m

25 × 2m = (T2) ² × m

25 × 2 = (T2) ²

50 = (T2) ²

T2 = √50

T2 = 7.07s

5 0
4 years ago
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