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2 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]2 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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If you pull a resistant puppy with its leash in a horizontal direction, it takes 80 N to get it going. You can then keep it movi

netineya [11]

Answer:

The coefficient of static friction between the puppy and the floor is 0.7273.

Explanation:

The horizontal force applied to move the puppy from a steady state has to be greater than the force of static friction, after it is moving the force needs to be equal to be greater than the force of dynamic friction in order to maintain its movement. The force of static friction is given by:

$F_s = \mu_s*N$

Where $F_s$ is the static friction force, $\mu_s$ is the coefficient of static friction and $N$ is the normal force. Since there's no angle on the flor the normal force is equal to the weight of the puppy, therefore, $N = 110\text{ N}$ , to make the puppy moving we need to use a force of 80 N, therefore, $F_s = 80 \text{ N}$ , so we can solve for the coefficient as shown below:

$80 = \mu_s*110\\\mu_s = \frac{80}{110} = 0.7273\\$

The coefficient of static friction between the puppy and the floor is 0.7273.

5 0

2 years ago

Dane is standing on the moon holding an 8 kilogram brick 2 metres above the ground. How much energy is in the brick's gravitatio

Nadya [2.5K]

The gravitational potential energy of the brick is 25.6 J

Explanation:

The gravitational potential energy of an object is the energy possessed by the object due to its position in a gravitational field.

Near the surface of a planet, the gravitational potential energy is given by

$PE=mgh$

where

m is the mass of the object

g is the strength of the gravitational field

h is the height of the object relative to the ground

For the brick in this problem, we have:

m = 8 kg is its mass

g = 1.6 N/kg is the strenght of the gravitational field on the moon

h = 2 m is the height above the ground

Substituting, we find:

$PE=(8)(1.6)(2)=25.6 J$

Learn more about potential energy:

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2 years ago

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Step 8: Observe How Changes in the Speed of the Bottle Affect Beanbag Height

lina2011 [118]

Answer:

When the speed of the bottle is 2 m/s, the average maximum height of the beanbag is 0.10 m.

When the speed of the bottle is 3 m/s, the average maximum height of the beanbag is 0.43 m.

When the speed of the bottle is 4 m/s, the average maximum height of the beanbag is 0.87 m.

When the speed of the bottle is 5 m/s, the average maximum height of the beanbag is 1.25 m.

When the speed of the bottle is 6 m/s, the average maximum height of the beanbag is 1.86 m.

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2 years ago

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A battery that produces a potential difference V is connected to a 10-W light bulb. Later, the 10-W bulb is replaced with a 20-W

uysha [10]

Answer:

The battery supply the greatest current with the 20-W bulb

Explanation:

The electrical power at a certain time by a two-terminal device is the product of the potential difference between the terminals and the current intensity that passes through the device. For this reason the power is proportional to the current and voltage. Mathematically can be written as:

$P=VI$

Let:

$P_1=Electric\hspace{3}power\hspace{3}dissipated\hspace{3}by\hspace{3}the\hspace{3}10-W\hspace{3}bulb\\P_2=Electric\hspace{3}power\hspace{3}dissipated\hspace{3}by\hspace{3}the\hspace{3}20-W\hspace{3}bulb$

The voltage remains constant for every case so:

$I_1=Current\hspace{3}supply\hspace{3}by\hspace{3}the\hspace{3}battery\hspace{3}with\hspace{3}the\hspace{3}\hspace{3}10-W\hspace{3}bulb\\I_2=Current\hspace{3}supply\hspace{3}by\hspace{3}the\hspace{3}battery\hspace{3}with\hspace{3}the\hspace{3}\hspace{3}20-W\hspace{3}bulb$

Then:

$P_1=V*I_1=10W\\P_2=V*I_2=20W$

Analyzing the equations it makes sense that the battery supplies the highest current with the 20 W bulb, because the power only depends on the voltage and the current, if the voltage remains constant, the only reason for the power to increase is due to current.

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2 years ago