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3 years ago

Which of the following would be an example of basic research

2 answers:

5 0

Basic research, also called pure research or fundamental research, is scientific research aimed to improve scientific theories for improved understanding or prediction of natural or other phenomena.

(source: wikipedia)

olga55 [171]3 years ago

4 0

Answer: Newton's discoveries of the laws of motion.

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You just calibrated a constant volume gas thermometer. The pressure of the gas inside the thermometer is 286.0 kPa when the ther

diamong [38]

Answer:

$T_{2} = 606.69 K$

Explanation:

In that the gas thermometer is a constant volume, it is satisfied that:

$\frac{P_{1} }{T_{1} } = \frac{P_{2} }{T_{2} }$

How the boiling water is under regular atmospheric pressure, then

$T_{1} = 373 .15 K$

Thus

$\frac{286000}{373.15} = \frac{465000}{T_{2} }$

$T_{2} = 606.69 K$

5 0

3 years ago

Each of two small non-conducting spheres is charged positively, the combined charge being 40 μC. When the two spheres are 50 cm

Phantasy [73]

Answer:

1.44 x 10⁻⁶ C

Explanation:

$q_{1}$ = charge on one sphere

$q_{2}$ = charge on other sphere

$q$ = Total charge on the two spheres = 40 μC

$q_{1}+ q_{2}$ = $q$

$q_{1}+ q_{2}$ = 40 x 10⁻⁶

$q_{1}$ = (40 x 10⁻⁶) - $q_{2}$ eq-1

$r$ = distance between the two spheres = 50 cm = 0.50 m

$F$ = magnitude of force between the two spheres = 2.0 N

Magnitude of force between the two spheres is given as

$F = \frac{k q_{1} q_{2}}{r^{2}}$

$2.0 = \frac{(9\times 10^{9}) ((40\times 10^{-6}) - q_{2}) q_{2}}{0.50^{2}}$

$q_{2}$ = 1.44 x 10⁻⁶ C

7 0

3 years ago

300,000,000m/s × 2 = —— m

nekit [7.7K]

Answer:

6 x 10⁸ m

Explanation:

300,000,000 m/s x 2 s

= 3 x 10⁸ x 2

= 6 x 10⁸ m

7 0

3 years ago

The box now rests on a frictionless ramp angled at 15◦ . The mover pulls up on a rope attached to the box to move it up the incl

Nataly [62]

Answer:

F = 84.61 N

Explanation:

As in the figure, since there is no friction so if component of Force applied along the incline is greater than the component of weight along the incline, then the object will move up the incline.

component of Force along the incline = F cos(23° - 15°) = F cos(8°)

component of weight along the incline = 33*g*sin(15°) = 33*9.81*sin(15°)

Equating the above two components of forces will give the minimum Force required.

F cos(8°) = 33*9.81*sin(15°)

F = 33*9.81*sin(15°) / cos(8°) (calculate the value using a scientific calculator)