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3 years ago

What are the standard units of measurement for scientific experiments

2 answers:

8 0

Metric units would be the best, this is the most popular, or most famous measurements the US uses.
Hope that this helps you! =)

Ivanshal [37]3 years ago

4 0

In science, the metric units are most often the units used for experiments. These units are not often used in America otherwise.

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In a physics lab experiment for the determination of moment of inertia, a team weighs an object and finds a mass of 2.15 kg. The

Maksim231197 [3]

Answer:

0.339 kgm²

Explanation:

We know the period of this pendulum, T = 2π√(I/mgh) where I = moment of inertia of the object about the pivot axis, m = mass of object = 2.15 kg, g = acceleration due to gravity = 9.8 m/s² and h = distance of center of mass of object from pivot point = 0.163 m.

Since T = 2π√(I/mgh), making I subject of the formula, we have

I = mghT²/4π²

Now since it takes 241 s to complete 113 cycles, then it takes 241 s/113 cycles to complete one cycle.

So, T = 241 s/113 = 2.133 s

So, Substituting the values of the variables into I, we have

I = mghT²/4π²

I = 2.15 kg × 9.8 m/s² × 0.163 m × (2.133 s)²/4π²

I = 15.63/4π² kgm²

I = 0.396 kgm²

Now from the parallel axis theorem, I = I' + mh² where I' = moment of inertia of object with respect to its center of mass about an axis parallel to the pivot axis

I' = I - mh²

I' = 0.396 kgm² - 2.15 kg × (0.163 m)²

I' = 0.396 kgm² - 0.057 kgm²

I' = 0.339 kgm²

7 0

3 years ago

What is the energy released in this B- nuclear reaction 2K-> 2Ca0,e? (The atomic mass of 42 K is 41.962403 u and that of 42Ca

Katyanochek1 [597]

<u>Answer:</u> The energy released in the given nuclear reaction is 3.526 MeV.

<u>Explanation:</u>

For the given nuclear reaction:

$_{19}^{42}\textrm{K}\rightarrow _{20}^{42}\textrm{Ca}+_{-1}^{0}\textrm{e}$

We are given:

Mass of $_{19}^{42}\textrm{K}$ = 41.962403 u

Mass of $_{20}^{42}\textrm{Ca}$ = 41.958618 u

To calculate the mass defect, we use the equation:

$\Delta m=\text{Mass of reactants}-\text{Mass of products}$

Putting values in above equation, we get:

$\Delta m=(41.962403-41.958618)=0.003785u$

To calculate the energy released, we use the equation:

$E=\Delta mc^2\\E=(0.003785u)\times c^2$

$E=(0.003785u)\times (931.5MeV)$ (Conversion factor: $1u=931.5MeV/c^2$ )

$E=3.526MeV$

Hence, the energy released in the given nuclear reaction is 3.526 MeV.

7 0

3 years ago

A bullet whose mass is 30.2 g leaves a rifle with a muzzle velocity of 1,000 m/s. It strikes a block of wood (mass 5 kg), initia

AveGali [126]

This problem here is an example of inelastic collision where kinetic energy is not conserved but momentum is. We calculate as follows:

m1v1 + m2v2 = (m1 + m2)v3
v3 = m1v1 + m2v2 / m1 + m2
v3 = (30.2)(1000) + (5000)(0) / (30.2 + 5000)
v3 = 6.00 m/s

5 0

3 years ago

When two forces or more forces act on an object, the forces are

sashaice [31]

The forces are net forces. I think this is correct.

5 0

3 years ago

What it mean head on collision?

murzikaleks [220]

I don’t understand the question

6 0

3 years ago

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