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3 years ago

The temperature of an object is directly related to.

Physics

2 answers:

Alisiya [41]3 years ago

6 0

It is related to preasure

Colt1911 [192]3 years ago

6 0

<span>the speed of the particle's motion</span>

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Ground-based radio telescopes can collect data from distant objects in space

Butoxors [25]

Answer:

A. during the day or night and in any weather conditions.

Explanation:

Ground-based radio telescopes can be used to collect data from distant objects in space during the day or night in any weather condition.

They do not depend or are they affected by weather and they pass well through them.

Telescopes are devices used to obtain information about distant bodies usually astronomical in nature.
Optical telescopes use the visible range of light and they are overwhelmed by the sun during the day.
Bad weather conditions can also diminish the reception of light.
They work best at night.
Radio telescopes uses electromagnetic radiations and can work at any time and during any weather.

6 0

3 years ago

The amount of oxygen in the reactants is 4 atoms. In the products, the oxygen is distributed to water (H2O) and O2 gas. Which co

svp [43]

Answer:

2 in front of water and 1 in front of oxygen

Explanation:

3 0

3 years ago

Read 2 more answers

2.65 In a standard tensile test a steel rod of 22-mm diameter is subjected to a tension force of 75 kN. Knowing that n 5 0.30 an

Schach [20]

To solve this problem it is necessary to apply the concepts related to uniaxial deflection for which the training variable is applied, determined as

$\delta = \frac{Pl}{AE}$

Where,

P = Tensile Force

L= Length

A = Cross sectional Area

E = Young's modulus

PART A) The elongation of the bar in a length of 200 mm caliber, could be determined through the previous equation, then

$\delta = \frac{Pl}{AE}$

$\delta = \frac{(75*10^3)(200)}{(\pi/4*22^2)(200*10^3)}$

$\delta = 0.1973mm$

Therefore the elongaton of the rod in a 200mm gage length is $0.1973mm$

PART B) To know the change in the diameter, we apply the similar ratio of the change in length for which,

$\upsilon = \frac{l_{a}}{l_{s}}$

Where,

$\upsilon =$ Poission's ratio

$l_{a}$ = Lateral strain

$l_{s}$ = Linear strain

$\upsilon = \frac{\delta/d}{\delta/l}$

$0.3 = \frac{\delta/22}{0.1973/200}$

$0.3(\frac{0.1973}{200}) = \frac{\delta}{22}$

$\delta = 6.5109*10^{-3}mm$

Therefore the change in diameter of the rod is $6.5109*10^{-3}mm$

5 0

3 years ago

Which of the following layers in Uranus’ atmosphere is the densest part? Stratosphere Troposphere Thermosphere Uranus does not h

VLD [36.1K]

Answer: Troposphere

Explanation:

Fairly over the "surface" of Uranus lies the troposphere, where the atmosphere is the densest. The temperature ranges from a minus 243 degrees Fahrenheit (minus 153 degrees Celsius) to minus 370 F (minus 218 C) , with the upper regions being the coldest. This makes the atmosphere of Uranus the coldest within the solar system. Inside the troposphere are layers of clouds — water clouds at the least weights, with ammonium hydro sulfide clouds over them. Alkali and hydrogen sulfide clouds come following. At long last, lean methane clouds lay on the best. The troposphere amplifies 30 miles (50 kilometers) from the surface of the planet.

Using this information, we can determine that the Troposphere is the densest layer of Uranus's atmosphere.

7 0

3 years ago

When an ideal diatomic gas is heated at constant

ira [324]

Answer:

Q=∆U+W

Explanation:

work done+ change in internal energy = heat supplied to change the internal energy

(1st law of thermodynamics)

7 0

3 years ago