What is the smallest possible piece an element in broken down into my normal laboratory processes

If a reaction is exothermic and produces large quantities of heat reaching equilibrium, its equilibrium constant will be

natima [27]

The equilibrium constant will be lowered and the equilibrium will shift to the left if the heat being produced is not removed.

5 0

3 years ago

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1- ¿Cuál sería la energía de un objeto de 50 newton de peso que se encuentra sobre una estantería de 3 metros de altura? ¿Qué ti

Liula [17]

Responder:

<h3>150 Nm </h3><h3>Energía potencial </h3>

Explicación:

El tipo de energía que posee el objeto se conoce como energía potencial. <u>La energía potencial es la energía que posee un objeto, mi virtud de su posición. </u>

Energía potencial = masa * aceleración debido a la gravedad * altura

Dado que Force = masa * aceleración debido a la gravedad

Energía potencial = Fuerza * altura

Fuerza dada = 50N y altura = 3 m

Energía potencial = 50 * 3

Energía potencial = 150 Nm

6 0

3 years ago

Displacement vectors of 4 km north, 2 km south, 5 km north, and 5 km south combine to a total displacement of

valentinak56 [21]

south = -(north)

Displacement = (4 km north) + (2 km south) + (5 km north) + (5 km south)

Displacement = (4 km north) - (2 km north) + (5 km north) - (5 km north)

Displacement = (4 - 2 + 5 - 5) km north

<u>Displacement = 2 km north </u>

6 0

3 years ago

1-A car moves toward east 12km is represented as A and it turns towards south 16km is represented as B. What is the resultant ve

hjlf

1. A-20 km south east

The car's displacement consists of two components into two different directions. Using a system of coordinates in which x represents the east direction and y represents the south direction, the two displacements are:

$d_x = 12 km$ east

$d_y = 16 km$ south

Since the two components are orthogonal to each other, we can find the resultant displacement by using Pythagorean's theorem:

$d=\sqrt{d_x^2+d_y^2}=\sqrt{(12 km)^2+(16 km)^2}=\sqrt{400}=20 km$

and the direction is between the two original directions, so south-east.

2. D. 10 m/s

First of all, we need to calculate the total time the stone took to hit the ground. Since the vertical distance covered is S = 78.4 m, and since the motion is an accelerated motion with constant acceleration g=9.8 m/s^2, we have

$S=\frac{1}{2}gt^2$

From which we find the total time of the fall, t:

$t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(78.4 m)}{9.8 m/s^2}}=4 s$

Now we can consider the horizontal motion of the stone: we know that the stone travels for d = 40 m in a time of t = 4 s, therefore the horizontal velocity of the stone is

$v=\frac{d}{t}=\frac{40 m}{4 s}=10 m/s$

3. B=32.32 m

As in the previous problem, we have to calculate the total time it takes for the stone to reach the river first. Since the vertical distance covered is S = 20 m, we have

$t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(20 m)}{9.8 m/s^2}}=2.0 s$

And since the stone is traveling horizontally at v = 16 m/s, the horizontal distance covered is

$d=vt=(16 m/s)(2 s)=32 m$

So, the closest answer is B.

5 0

3 years ago

When did galileo discover projectile motion?

Tcecarenko [31]

Between 1589-1592 when he discovered projecctile motion

4 0

3 years ago