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3 years ago

1. Si comprimes un globo hasta reducir su volumen a un tercio de su valor original, ¿cuánto aumenta la presión en su interior?

Physics

1 answer:

harina [27]3 years ago

4 0

Answer:

See step by step sexplanation

Explanation:

1.-Sabemos que la relación:

P₁ * V₁ = P₂ * V₂

Para una temperatura constante debe mantenerse entonces si el globo se comprime hasta llevarlo a 1/3 de su valor inicial, entonces necesariamente para cumplir con la relación mencionada, la presión aumenta tres veces su valor original

2.-La definición de presión es fuerza por unidad de superficie, entonces la fuerza es determinada por la altura de la columna de liquido en el recipiente y no por la cantidad total de liquido, de acuerdo a esto habrá más presión en la base del florero, ya que la columna de agua tiene más altura.

3.-No se puede estar de acuerdo con el criterio del plomero. En su solución no plantea el aumento de la altura del tanque, para el logro del aumento de la presión que es realmente lo que hay que hacer

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A(n) 636 kg elevator starts from rest. It moves upward for 4.5 s with a constant acceleration until it reaches its cruising spee

zmey [24]

Answer:

The average power delivered by the elevator motor during this period is 6.686 kW.

Explanation:

Given;

mass of the elevator, m = 636 kg

initial speed of the elevator, u = 0

time of motion, t = 4.5 s

final speed of the elevator, v = 2.05 m/s

The upward force of the elevator is calculated as;

F = m(a + g)

where;

m is mass of the elevator

a is the constant acceleration of the elevator

g is acceleration due to gravity = 9.8 m/s²

$a = \frac{v-u}{t} \\\\a = \frac{2.05 -0}{4.5} \\\\a = 0.456 \ m/s^2$

F = (636)(0.456 + 9.8)

F = (636)(10.256)

F = 6522.816 N

The average power delivered by the elevator is calculated as;

$P_{avg} = \frac{1}{2} (Fv)\\\\P_{avg} = \frac{1}{2} (6522.816 \ \times \ 2.05)\\\\P_{avg} = 6685.89 \ W\\\\P_{avg} = 6.68589 \ kW\\\\P_{avg} = 6.686 \ k W$

Therefore, the average power delivered by the elevator motor during this period is 6.686 kW.

3 0

3 years ago

First extinguish a match or candle by blasting it violently. Why?

Andreas93 [3]

Answer:

The wax vapor on burning candles could reignite the flame.

Explanation:

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3 years ago

As you are trying to move a heavy box of mass m, you realize that it is too heavy for you to lift by yourself.There is no one ar

Ira Lisetskai [31]

Answer: magnitude of applied force is FA = mg + F

Where F is the resultant force downward that the rope moves with

Explanation:

Force downwards F is,

F = FA - T

T is the upwards tension force on the rope

FA is the actual applied force in pulling the rope down.

Therefore, T = FA - F .....equ. (1)

For the box to move up with force ma ( it's mass times its acceleration upwards) upwards tension on the roap must exceed its own weight mg ( it's mass times acceleration due to gravity 9.8m/s^2)

Therefore, ma = T - mg

T = ma + mg ..... equ. (2)

Equating equ. 1 and 2

T = FA - F = ma + mg

Therefore FA = ma + mg + F

But at constant velocity a = 0

Magnitude of applied force becomes

FA = mg + F

See image below

5 0

3 years ago

The 68-kg crate is stationary when the force P is applied. Determine the resulting acceleration of the crate if (a) P = 0, (b) P

bogdanovich [222]

Explanation:

Mass of the crate, m = 68 kg

We need to find the resulting acceleration if :

(a) Force, P = 0

P = m a

⇒ a = 0

(b) P = 181 N

$a=\dfrac{P}{m}$

$a=\dfrac{181\ N}{68\ kg}$

$a=2.67\ m/s^2$

(c) P = 352 N

$a=\dfrac{P}{m}$

$a=\dfrac{352\ N}{68\ kg}$

$a=5.17\ m/s^2$

Hence, this is the required solution.

5 0

3 years ago

The position of a particle moving along the x axis varies in time according to the expression x = 4t 2, where x is in meters and

Pie

Answer:

a) X = 17.64 m

b) X = 17.64 + 4∆t^2 + 16.8∆t

c) Velocity = lim(∆t→0)⁡〖∆X/∆t〗 = 16.8 m/s

Explanation:

a) The position at t = 2.10s is:

X = 4t^2

X = 4(2.10)^2

X = 17.64 m

b) The position at t = 2.10 + ∆t s will be:

X = 4(2.10 + ∆t)^2

X = 17.64 + 4∆t^2 + 16.8∆t m

c) ∆X is the difference between position at t = 2.10s and t = 2.10 + ∆t so,

∆X= 4∆t^2 + 16.8∆t

Divide by ∆t on both sides:

∆X/∆t = 4∆t + 16.8

Taking the limit as ∆t approaches to zero we get:

Velocity =lim(∆t→0)⁡〖∆X/∆t〗 = 4(0) + 16.8

Velocity = 16.8 m/s

8 0

3 years ago

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