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2 years ago

Good morning! can someone please answer this, ill give you brainliest and your earning 50 points.

Physics

1 answer:

tatyana61 [14]2 years ago

7 0

Answer:

With the exception of natural gas fossil fuel energies have seen an overall decrease since year 2000.

Explanation:

Just observe the graph

The Fossil fuels include-Petroleum ,coal and natural gases.

The graph of these goes fastly up during 20th century but ,then it stabled at the peak then has not shown any solid increase .
Observe other energy sources graph ,nuclear energy etc have shown quite speedy increase .

Option C is correct

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What is the gravitational potential energy of a 63.5kg skateboarder at the top of a 0.61m high ramp

Lady bird [3.3K]

Answer:

1.1 m/s

Explanation:

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4 years ago

Help!! A. 0.5 s B. 1.0 s C. 1.5 s D. 2.0 s

algol13

Answer:

c. 1.5 s

Explanation:

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3 years ago

Does the distance a person kicks a soccer ball

3241004551 [841]

Answer:

Independent variable: how far the soccer ball is kicked.

Dependent variable: how well the person does on their math test

Explanation:

The distance the ball is kicked is what the scientist can change. On the other hand, in this experiment, how well the person does on their test supposedly relies on how far the soccer ball is kicked.

5 0

3 years ago

Un depósito de gran superficie se llena de agua hasta una altura de 0,3 m. En el fondo del depósito hay un orificio de 5 cm2 de

ahrayia [7]

Answer:

a) El caudal de salida del chorro es $1.213\times 10^{-3}\,\frac{m^{3}}{s}$ .

Explanation:

a) Asúmase que el tanque se encuentra a presión atmósferica y que la sima del tanque tiene una altura de 0 metros. La rapidez de salida del chorro del depósito se determined a partir del Principio de Bernoulli, cuya línea de corriente entre la cima y la sima del tanque queda descrita por la siguiente ecuación:

$\Delta z = \frac{v_{out}^{2}}{2\cdot g}$

Donde:

$\Delta z$ - Diferencia de altura, medida en metros.

$g$ - Constante gravitacional, medida en metros por segundo al cuadrado.

$v_{out}$ - Rapidez de salida del chorro, medida en metros por segundo.

Se despeja la rapidez de salida del chorro:

$v_{out} = \sqrt{2\cdot g \cdot \Delta z}$

Si $g = 9.807\,\frac{m}{s^{2}}$ y $\Delta z = 0.3\,m$ , entonces la rapidez de salida del chorro es:

$v_{out} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.3\,m)}$

$v_{out} \approx 2.426\,\frac{m}{s}$

Ahora, la cantidad de líquido que sale del depósito por unidad de tiempo se obtiene al multiplicar la rapidez de salida del chorro por el área transversal del orificio. Esto es:

$\dot V_{out} = v_{out}\cdot A_{t}$

Donde:

$v_{out}$ - Rapidez de salida del chorro, medida en metros por segundo.

$A_{t}$ - Área transversal del orificio, medido en metros cuadrados.

$\dot V_{out}$ - Caudal de salida del chorro, medido en metros cúbicos por segundo.

Dado que $v_{out} = 2.426\,\frac{m}{s}$ y $A_{t} = 5\,cm^{2}$ , el caudal de salida del chorro es:

$\dot V_{out} = \left(2.426\,\frac{m}{s} \right)\cdot (5\,cm^{2})\cdot \left(\frac{1}{10000}\,\frac{m^{2}}{cm^{2}} \right)$

$\dot V_{out} = 1.213\times 10^{-3}\,\frac{m^{3}}{s}$

El caudal de salida del chorro es $1.213\times 10^{-3}\,\frac{m^{3}}{s}$ .

5 0

3 years ago

Two masses 1.2kg and 1.8kg are connected to the ends of a rod of length 2m. Find the moment of inertia about the axes, 1)going t

frutty [35]

Answers: 1) 3 kg m²

2) 2.88 kg m²

Explanation: Question 1

I = m(r)²+ M(r)²

I = 1.2 kg × (1 m )² +1.8 kg ×(1 m )²

∴ I = 3 kg m²

Question 2

ACCORDING TO THE DIAGRAM DRAWN FOR QUESTION 2

we have to decide where the center of gravity (G) lies and obviously it should lie somewhere near to the greater mass. (which is 1.8 kg). Since we don't know the distance from center of gravity(G) to the mass (1.8 kg) we'll take it as 'x' and solve!!

moments around 'G'

F₁ d ₁ = F₂ d ₂

12 (2-X) = 18 (X)

24 -12 X =18 X

∴ X = 0.8 m

∴ ( 2 - x ) = 1.2 m

∴ Moment of inertia (I) going through the center of mass of two masses,

⇒ I = m (r)² +M (r)²

⇒ I = 1.2 × (1.2)² + 1.8 × (0.8)²

⇒ I = 1.2 × 1.44 + 1.8 × 0.64

⇒ I = 1.728 + 1.152

⇒ ∴ I = 2.88 kg m²

∴ THE QUESTION IS SOLVED !!!

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8 0

3 years ago