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3 years ago

Why aren’t descriptive investigations repeatable

1 answer:

8 0

It involves questions but it doesn't involve hypothesis. So we are still trying to find info by asking the question, and it's also a waste of time asking the question over and over, because the hypothesis wasn't formulated yet so that's why they are not repeatable.

Hope this helps!

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What is normal for a spring that obeys hook's law ?

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A spring that obeys Hooke's law has a spring force constant of 272 N/m. This spring is then stretched by 28.6 cm

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3 years ago

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Centripetal is a word used in science to mean

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: the force that is necessary to keep an object moving in a curved path and that is directed inward toward the center of rotation a string on the end of which a stone is whirled about exerts centripetal force on the stone — compare centrifugal force.

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3 years ago

What are used to transfer electrical energy from a power plant all the way to a home? generators turbines power transformers pow

defon

Answer:

Power lines

Explanation:

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Turbines also induce currents from water waves that is transmitted.

Transformers change A.C to D.C or vice versa.

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3 years ago

Exposure to a sufficient quantity of ultraviolet will redden the skin, producing erythema - a sunburn. The amount of exposure ne

ivann1987 [24]

Answer:

Energy = 7.83 x 10⁻¹⁹ J

Energy = 6.63 x 10⁻¹⁹ J

Explanation:

The energy of a photon in terms of wavelength can be calculated by the following formula:

$Energy = \frac{hc}{\lambda}\\$

where,

h = Plank's Constant = 6.63 x 10⁻³⁴ Js

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of light

Now, for λ = 254 nm = 2.54 x 10⁻⁷ m:

$Energy = \frac{(6.63\ x\ 10^{-34}\ Js)(3\ x\ 10^8\ m/s)}{2.54\ x\ 10^{-7}\ m}\\$

Energy = 7.83 x 10⁻¹⁹ J

Now, for λ = 300 nm = 3 x 10⁻⁷ m:

$Energy = \frac{(6.63\ x\ 10^{-34}\ Js)(3\ x\ 10^8\ m/s)}{3\ x\ 10^{-7}\ m}\\$

Energy = 6.63 x 10⁻¹⁹ J

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3 years ago

A 0.0450 kg bullet is accelerated from rest to a speed of 425 m/s in a 2.25 kg rifle (which is inititally at rest). The pain of

Mrac [35]

Answer:

If the rifle is held loosely away from the shoulder, the recoil velocity will be of -8.5 m/s, and the kinetic energy the rifle gains will be 81.28 J.

Explanation:

By momentum conservation, and given the bullit and the recoil are in a straight line, the momentum analysis will be unidimentional. As the initial momentum is equal to zero (the masses are at rest), we have that the final momentum equals zero, so

$0=P_{f}=m_{b} *v_{b}+m_{r}*v_{r}$

now we clear $v_{r}$ and use the given data to get that

$v_{r}=-8.5\frac{m}{s}$

But we have to keep in mind that the bullit accelerate from rest to a speed of 425 m/s, then if the rifle were against the shoulder, the recoil velocity would be a fraction of the result obtained, but, as the gun is a few centimeters away from the shoulder, it is assumed that the bullit get to its final velocity, so the kick of the gun, gets to its final velocity $\bold{v_{r}}$ too.

Finally, using $v_{r}$ we calculate the kinetic energy as

$K=\frac{1}{2}m_{r}v_{r}^{2}=81.28J$

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3 years ago