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murzikaleks [220]
2 years ago
5

Which of the following are examples of both projectile motion and 2-dimensional motion?

Physics
2 answers:
Elanso [62]2 years ago
5 0

Answer:

A

is answer

because of questions

Artyom0805 [142]2 years ago
4 0

Answer:

D. A football thrown upward at an angle.        

Explanation:

When an object is thrown in space, it is known as projectile and its motion is known as projectile motion. 1 dimensional motion refers to the motion either in horizontal or vertical direction. 2-dimensional motion refers to the motion in both horizontal and vertical direction. A football thrown at an angle will have constant horizontal velocity and accelerated motion in vertical direction. Thus, it is an example of projectile motion as well as two-dimensional motion.

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Is electricity a fuel
labwork [276]

YES, ELECTRICITY CONCERNS ENERGY WHICH IS USED AS A FUEL . IN MODERN DAY TECH, MOST MACHINES USE ELECTRICITY AS A FUEL SUCH AS THE ELECTRONIC TRAIN IN TOKYO, JAPAN.

8 0
2 years ago
Read 2 more answers
Cuanto cambia la entropía de 0.50 kg de vapor de mercurio [Lv: 2.7 x 10⁵ j/kg ] al calentarse en su punto de ebullición de 357°
lord [1]

Answer:

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

Explanation:

Por definición de entropía (S), medida en joules por Kelvin, tenemos la siguiente expresión:

dS = \frac{\delta Q}{T} (1)

Donde:

Q - Ganancia de calor, en joules.

T - Temperatura del sistema, en Kelvin.

Ampliamos (1) por la definición de calor latente:

dS = \frac{L_{v}}{T}\cdot dm (1b)

Donde:

m - Masa del sistema, en kilogramos.

L_{v} - Calor latente de vaporización, en joules

Puesto que no existe cambio en la temperatura durante el proceso de vaporización, transformamos la expresión diferencial en expresión de diferencia, es decir:

\Delta S = \frac{\Delta m \cdot L_{v}}{T}

Como vemos, el cambio de la entropía asociada al cambio de fase del mercurio es directamente proporcional a la masa del sistema. Si tenemos que m = 0.50\,kg,L_{v} = 2.7\times 10^{5}\,\frac{J}{kg} and T = 630.15\,K, entonces el cambio de entropía es:

\Delta S = \frac{(0.50\,kg)\cdot \left(2.7\times 10^{5}\,\frac{J}{kg} \right)}{630.15\,K}

\Delta S = 214.235 \,\frac{J}{K}

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

3 0
3 years ago
The earth's tectonic plates are directly propelled by the convection in the __________.
frozen [14]

The missing word here is <u>Asthenosphere.</u><u> </u>

The convection in the asthenosphere directly propels the tectonic plates of the earth.


Did you know that the asthenosphere is thought to remain malleable because of heat from deep within the Earth? It is thought to be lubricating the earth's tectonic plates' undersides and enabling movement.


The older, denser portions of the lithosphere that are dragged downward in subduction zones are stored in the asthenosphere, according to the theory of plate tectonics.

The lithosphere above is stressed by convection currents, and the cracking that frequently results manifests as earthquakes.

Magma is forced upward through volcanic vents and spreading centers by convection currents produced within the asthenosphere, which also results in the formation of new crust.

Learn why properties of the asthenosphere are important: brainly.com/question/11484043

#SPJ4

3 0
1 year ago
Which substance cannot be seperated physically or chemically?
scoray [572]

Answer:

What is a Pure Substance?

Explanation:

It is something which cannot be divided into parts by physical means, as it's all made up of the same thing. Pure substances are either elements or compounds. Elements can NOT be separated into other types of matter (physically or chemically).

4 0
3 years ago
Read 2 more answers
The rate (in cubic feet per hour) that a spherical snowball melts is proportional to the snowball's volume raised to the 2/3 pow
Darina [25.2K]

Answer:

A 3 feet radius snowball will melt in 54 hours.

Explanation:

As we can assume that the rate of  snowball takes to melt is proportional to the surface area, then the rate for a 3 feet radius will be:

T= A(3 ft)/A(1 ft) * 6 hr

A is the area of the snowballs. For a spherical geometry is computing as:

A=4.pi.R^2

Then dividing the areas:

A(3 feet)/A(1 foot) = (4 pi (3 ft)^2)/(4 pi (1 ft)^2) =  (36pi ft^2)/(4pi ft^2)= 9

Finally, the rate for the 3 feet radius snowball is:

T= 9 * 6 hr = 54 hr

6 0
3 years ago
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