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LenKa [72]
2 years ago
15

A scientist is planning an experiment that will include the measurement of temperatures. She wishes to avoid using negative valu

es. The expected temperatures will be significantly colder than
the freezing point of water. Which temperature scale will the scientist most likely use?
O A. Kelvin
OB. Celsius
O C. Fahrenheit
OD. joule
Physics
1 answer:
cricket20 [7]2 years ago
8 0

Answer:

C.

Explanation:

Fahrenheit scale is a temperature scale that measures the boiling point of water at 212 degrees F and the freezing point at 32 degrees F. The Fahrenheit scale was developed by the German scientist Daniel Gabriel Fahrenheit in 1724.

In the given scenario, the scientist, who wishes to measure the temperature of her experiment which will expectantly be colder than the point of water, should use the Fahrenheit scale.

So, the correct answer is option C.

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3 years ago
what do we call a solar power plant that uses mirrors to focus the light of the sun on a central point
Mars2501 [29]

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Concentrating solar power (CSP) plants use mirrors to concentrate the sun's energy to drive traditional steam turbines or engines that create electricity.

Explanation:

4 0
3 years ago
A person observes a firework display for A safe distance of .750 km. Assuming that sound travels at 340 m/s in air what is the t
WINSTONCH [101]

Answer:

t = 2.2 s

Explanation:

Given that,

A person observes a firework display for A safe distance of 0.750 km.

d = 750 m

The speed of sound in air, v = 340 m/s

We need to find the between the person see and hear a firework explosion. let it is t. So, using the formula of speed.

v=\dfrac{d}{t}\\\\t=\dfrac{d}{v}\\\\t=\dfrac{750\ m}{340\ m/s}\\\\t=2.2\ s

So, the required time is 2.2 seconds.

3 0
2 years ago
An object with a mass M = 250 g is at rest on a plane that makes an angle θ = 30 o above the horizontal. The coefficient of kine
liubo4ka [24]

Answer:

v = 79.2 m/s

Solution:

As per the question:

Mass of the object, m = 250 g = 0.250 kg

Angle, \theta = 30^{\circ}

Coefficient of kinetic friction, \mu_{k} = 0.100

Mass attached to the string, m = 0.200 kg

Distance, d = 30 cm = 0.03 m

Now,

The tension in the string is given by:

Mgsin\theta + \mu_{k}Mgcos\theta + Ma = T        (1)

Also

T = m(g + a)

Thus eqn (1) can be written as:

Mgsin\theta + \mu_{k}Mgcos\theta + Ma = m(g - a)

Mgsin\theta + \mu_{k}Mgcos\theta + Ma = mg + ma

mg - Mgsin\theta - \mu_{k}Mgcos\theta = (M - m)a

a = \frac{0.2\times 9.8 - 0.250\times 9.8\times sin30^{\circ} - 0.1\times 0.250\times 9.8\times cos30^{\circ}}{0.250 - 0.200}

a = 10.45\ m/s^{2}

Now, the speed is given by the third eqn of motion with initial velocity being zero:

v^{2} = u^{2} + 2ad

where

u = initial velocity = 0

Thus

v = \sqrt{2ad}

v = \sqrt{2\times 10.45\times 0.03} = 0.792\ m/s

3 0
3 years ago
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