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Marina86 [1]
3 years ago
14

Michael has a substance that he puts in Container 1. The substance has a volume of 5 cubic meters. He then puts the substance in

Container 2, and it now has a volume of 10 cubic meters. In which phase(s) is the substance
Physics
1 answer:
Svetradugi [14.3K]3 years ago
7 0

Solids and liquids don't suddenly change their volumes.

But gases do ... they expand to fill whatever container you
put them in.

Michael's mystery substance is in the gaseous phase.

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18. Un avión de rescate de animales que vuela hacia el este a 36.0 m/s deja caer una paca de
Alona [7]

Answer:

Definimos momento como el producto entre la masa y la velocidad

P = m*v

(tener en cuenta que la velocidad es un vector, por lo que el momento también será un vector)

Sabemos que el peso de la paca de heno es 175N, y el peso es masa por aceleración gravitatoria, entonces.

Peso = m*9.8m/s^2 = 175N

m = (175N)/(9.8m/s^2) = 17.9 kg

Ahora debemos calcular la velocidad de la paca justo antes de tocar el suelo.

Sabemos que la velocidad horizontal será la misma que tenía el avión, que es:

Vx = 36m/s

Mientras que para la velocidad vertical, usamos la conservación de la energía:

E = U + K

Apenas se suelta la caja, esta tiene velocidad cero, entonces su energía cinética será cero y la caja solo tendrá energía potencial (Si bien la caja tiene velocidad horizontal en este punto, por la superposición lineal podemos separar el problema en un caso horizontal y en un caso vertical, y en el caso vertical no hay velocidad inicial)

Entonces al principio solo hay energía potencial:

U = m*g*h

donde:

m = masa

g = aceleración gravitatoria

h = altura  

Sabemos que la altura inicial es 60m, entonces la energía potencial es:

U = 175N*60m = 10,500 N

Cuando la paca esta próxima a golpear el suelo, la altura h tiende a cero, por lo que la energía potencial se hace cero, y en este punto solo tendremos energía cinética, entonces:

10,500N = (m/2)*v^2

De acá podemos despejar la velocidad vertical justo antes de golpear el suelo.

√(10,500N*(2/ 17.9 kg)) = 34.25 m/s

La velocidad vertical es 34.25 m/s

Entonces el vector velocidad se podrá escribir como:

V = (36 m/s, -34.25 m/s)

Donde el signo menos en la velocidad vertical es porque la velocidad vertical es hacia abajo.

Reemplazando esto en la ecuación del momento obtenemos:

P = 17.9kg*(36 m/s, -34.25 m/s)  

P = (644.4 N, -613.075 N)

6 0
3 years ago
You and a friend each carry identical boxes from the first floor of a building to a room located on the second floor, farther do
melamori03 [73]

Answer:

Work done in both the cases will be same

Explanation:

As we know that the work done against gravity is given as

W = F_g .d

here we know that gravitational force is a conservative force and the work done against gravitational force is independent of the path

So here the work done by person to move the object between two different heights will be independent of the path they choose

So for the first person and second person will be same in both the cases because the height through which the boxes are transferred will be same in both the cases

8 0
3 years ago
10. Josie and Trey were working on their physics project and both built catapults. Trey's catapult shot a ball
SVEN [57.7K]

Answer:

Josie's ball faster than T

3 0
2 years ago
Which diagram is the best representation of gas molecules in a closed container? A. Diagram A B. Diagram B C. Diagram C D. Diagr
WITCHER [35]
Gas "floats" so if there are examples or pictures it would be the one with the most evenly spread out "dots". 
8 0
3 years ago
In a physics lab, Asha is given a 10.7 kg uniform rectangular plate with edge lengths 67.3 cm by 53.5 cm . Her lab instructor re
Leya [2.2K]

Answer:

I=2.6363\ kg.m^2

Explanation:

Given:

dimension of uniform plate, (0.673\times 0.535)\ m^2

mass of plate, m=10.7\ kg

Now we find the moment of inertia about the center of mass of the rectangular plate is given as:

I_{cm}=\frac{1}{12} \times m(L^2+B^2)

where:

L= length of the plate

B= breadth of the plate

I_{cm}=\frac{1}{12} \times 10.7\times(0.673^2+0.535^2)

I_{cm}=0.6591\ kg.m^2

We know that the center of mass of the rectangular plane is at its geometric center which is parallel to the desired axis XX' .

Now we find the distance between the center of mass and the corner:

s=\frac{\sqrt{ (0.673^2+0.535^2)}}{2}

s=0.4299\ m

Now using parallel axis theorem:

I=I_{cm}+m.s^2

I=0.6591+10.7\times 0.4299^2

I=2.6363\ kg.m^2

6 0
3 years ago
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