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3 years ago

What qualifications do you need to be a research scientist?

Physics

1 answer:

Vedmedyk [2.9K]3 years ago

5 0

The qualifications boil down to: College education.

In most university or industrial research organizations, you might be able to work there as a member of the team who doesn't get much pay or much respect, with research going on all around you directed by other people, after you've gotten you Master's degree.

But you really don't have a shot at leading anything, or having much to say about what's being researched or how, until you have a PhD degree in the field where you'd like to do the research.

(Did I mention how proud I was to be present about 6 weeks ago, in a land far away, when my daughter was awarded a PhD degree in Molecular Biology ? I didn't want to let you get away without hearing about that.)

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How many volts would it take to push 1 amp through a resistance of 1 ohm?

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Are the two expressions equvalent 7(8x+5) and 48x + 35

Evgesh-ka [11]

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3 years ago

All objects emit ______ radiation?<br> A-electromagnetic<br> B-kinetic<br> D-solar

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2 years ago

The barometric pressure at sea level is 30 in of mercury when that on a mountain top is 29 in. If the specific weight of air is

stealth61 [152]

To solve this problem we will apply the concepts related to pressure, depending on the product between the density of the fluid, the gravity and the depth / height at which it is located.

For mercury, density, gravity and height are defined as

$\rho_m = 846lb/ft^3$

$g = 32.17405ft/s^2$

$h_1 = 1in = \frac{1}{12} ft$

For the air the defined properties would be

$\rho_a = 0.0075lb/ft^3$

$g = 32.17405ft/s^2$

$h_2 = ?$

We have for equilibrium that

$\text{Pressure change in Air}=\text{Pressure change in Mercury}$

$\rho_m g h_1 = \rho_a g h_2$

Replacing,

$(846)(32.17405)(\frac{1}{12}) = (0.0075)(32.17405)(h_2)$

Rearranging to find $h_2$

$h_2 = \frac{(846)(32.17405)(\frac{1}{12}) }{(0.0075)(32.17405)}$

$h = 9400ft$

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7 0

2 years ago

outward from a wall just above floor level. A 1.5 kg box sliding across a frictionless floor hits the end of the spring and comp

sweet [91]

Answer:

v = 0.489 m/s

Explanation:

It is given that,

Mass of a box, m = 1.5 kg

The compression in the spring, x = 6.5 cm = 0.065 m

Let the spring constant of the spring is 85 N/m

We need to find the velocity of the box (v) when it hit the spring. It is based on the conservation of energy. The kinetic energy of spring before collision is equal to the spring energy after compression i.e.

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2$

$v=\sqrt{\dfrac{kx^2}{m}} \\\\v=\sqrt{\dfrac{85\times (0.065)^2}{1.5}} \\\\v=0.489\ m/s$

So, the speed of the box is 0.489 m/s.

3 0

3 years ago