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3 years ago

Why is it important to have only one set of chemical symbols in the world ?

Physics

1 answer:

Pavlova-9 [17]3 years ago

3 0

Answer:

As you learned, scientists standardized the short-hand way we represent elements, by their chemical symbol. One of the main reasons this was developed was because using letters was the easiest way to represent the elements. Another reason that we use chemical symbols is to allow us to write chemical formulas easily

One set of chemical symbols means that every science student and scientist -- no matter what languages they speak -- can identify these chemicals.

I hope this helps

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The components of a 10.8-meters-per-second velocity at an angle of 34.° above the horizontal are

Kisachek [45]

Answer:

6.0 m/s vertical and 9.0 m/s horizontal

Explanation:

For the vertical component, we use the formula:

Sin(34°) = y / 10.8

Then we solve for y:

0.559 = y / 10.8

y = 6.0

And for the horizontal component, we use the formula:

Cos(34°) = x / 10.8

Then we solve for x:

0.829 = x / 10.8

y = 9.0

So the answer is " 6.0 m/s vertical and 9.0 m/s horizontal".

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2 years ago

Two identical silver spheres of mass m and radius r are placed a distance R (sphere 1) and 2R (sphere 2) from the Sun, respectiv

lys-0071 [83]

Answer:

The ratio of T2 to T1 is 1.0

Explanation:

The gravitational force exerted on each sphere by the sun is inversely proporational to the square of the distance between the sun and each of the spheres.

Provided that the two spheres have the same radius r, the pressure of solar radiation too, is inversely proportional to the square of the distance of each sphere from the sun.

Let F₁ and F₂ = gravitational force of the sun on the first and second sphere respectively

P₁ and P₂ = Pressure of solar radiation on the first and second sphere respectively

M = mass of the Sun

m = mass of the spheres, equal masses.

For the first sphere that is distance R from the sun.

F₁ = (GmM/R²)

P₁ = (k/R²)

T₁ = (F₁/P₁) = (GmM/k)

For the second sphere that is at a distance 2R from the sun

F₂ = [GmM/(2R)²] = (GmM/4R²)

P₂ = [k/(2R)²] = (k/4R²)

T₂ = (F₂/P₂) = (GmM/k)

(T₁/T₂) = (GmM/k) ÷ (GmM/k) = 1.0

Hope this Helps!!!

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3 years ago

A truck of mass 200kg rests on an inclined plane hindered from rolling down the surface by a storing sprint whose force constant

mixas84 [53]

Answer:

1.92 J

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 200 Kg

Spring constant (K) = 10⁶ N/m

Workdone =?

Next, we shall determine the force exerted on the spring. This can be obtained as follow:

Mass (m) = 200 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = m × g

F = 200 × 9.8

F = 1960 N

Next we shall determine the extent to which the spring stretches. This can be obtained as follow:

Spring constant (K) = 10⁶ N/m

Force (F) = 1960 N

Extention (e) =?

F = Ke

1960 = 10⁶ × e

Divide both side by 10⁶

e = 1960 / 10⁶

e = 0.00196 m

Finally, we shall determine energy (Workdone) on the spring as follow:

Spring constant (K) = 10⁶ N/m

Extention (e) = 0.00196 m

Energy (E) =?

E = ½Ke²

E = ½ × 10⁶ × (0.00196)²

E = 1.92 J

Therefore, the Workdone on the spring is 1.92 J

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2 years ago

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Anvisha [2.4K]

Explanation:

sksksksmsmamamamamamamamama

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2 years ago

the idling engines of a landing turbojet produce forward thrust when operating in a normal manner, but they can produce reverse

Finger [1]

Forward thrust has positive values and reverse thrust has negative values.

Thrust is a sudden push or pull in a certain direction.

Flight speed u = 150 km/h

1 km/h = $\frac{1}{3.6}$ km/s

therefore, 150 km/h = 41.67 km / s

The thrust force represents the horizontal or x-component of momentum equation:

$T = m_{exhaust} * U_{exhaust} - U_{flight}$

T = 50 * (150 - 41.67)

T = 5416.67 N

Therefore, the value of forward thrust is 5416.67 N.

Now the exhaust velocity is now vertical due to reverse thrust application, then it has a zero horizontal component,

thus thrust equation is:

$T = m_{exhaust} * U_{exhaust} - U_{flight}$

T = 50 * (0 - 41.67)

T = -2083.5 N

Therefore, the thrust force T is -2083.5 N in the reverse direction.

Now the exhaust velocity and flight velocity is zero, then it has a zero horizontal component, thus thrust is also zero because $U_{exhaust} = U_{flight} = 0\\$

T = 0

Therefore, there is no difference in two velocities in x direction.

The given question is incomplete, the complete question is,

"The idling engines of a landing turbojet produce forward thrust when operating in a normal manner, but they can produce reverse thrust if the jet is properly deflected. Suppose that while the aircraft rolls down the runway at 150 km/h the idling engine consumes air at 50 kg/s and produces an exhaust velocity of 150 m/s.

a. What is the forward thrust of this engine?

b. What are the magnitude and direction (i.e., forward or reverse) if the exhaust is deflected 90 degree without affecting the mass flow?

c. What are the magnitude and direction of the thrust (forward or reverse) after the plane has come to a stop, with 90 degree exhaust deflection and an airflow of 40 kg/s?"

To know more about thrust,

brainly.com/question/14552836

#SPJ1

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1 year ago