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3 years ago

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Un globo de aire caliente tiene un volumen de 560 mL a la presión atmosférica normal y una temperatura del aire de 400 ºC. Cuand

o está en ascensión, y con el quemador de gas aumentamos la temperatura hasta los 700 ºC. ¿cuál será el nuevo volumen

1 answer:

dalvyx [7]3 years ago

4 0

Explanation:

charles law V1/T1 =V2/T2

560 x 673 =V2/973

376880 = V2/973

V2 = 376880 x 973 = 366704240mL

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A woman falls to the ground while wearing a parachute. The air resistance on the parachute of the parachute is 500N. If the woma

scoundrel [369]

The gravitational force on the woman is A) 500 N

Explanation:

There are two forces acting on the woman during her fall:

The force of gravity, $F_G$ , acting downward
The air resistance, $F_D$ , acting upward

According to Newton's second law, the net force acting on the woman is equal to the product between the woman's mass and her acceleration:

$\sum F=ma$

where m is the mass of the woman and a her acceleration.

The net force can be written as

$\sum F = F_G - F_D$

Also, we know that the woman falls at a constant velocity (5 m/s), this means that her acceleration is zero:

$a=0$

Combining the equations together, we get:

$F_G-F_D = 0$

which means that the magnitude of the gravitational force is equal to the magnitude of the air resistance:

$F_G=F_D=500 N$

Learn more about forces and Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

5 0

3 years ago

Read 2 more answers

Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore f

tatiyna

Answer:

a) h=3.16 m, b) v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

Em₀ = U = mg h

final point. Lowest point

$Em_{f}$ = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

K = ½ m $v_{cm }^{2}$ + ½ $I_{cm}$ w²

angular and linear speed are related

v = w r

w = v / r

K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

Em₀ = Em_{f}

mg h = ½ v_{cm }^{2} (m + I_{cm} / r²) (1)

h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

I_{cm} = ⅔ m r²

we substitute

h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

h = ½ v_{cm }^{2}/g 5/3

h = 5/6 v_{cm }^{2} / g

let's calculate

h = 5/6 6.1 2 / 9.8

h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

I_{cm} = ½ m r²

we substitute

v_{cm } = √ [2gh / (1 + ½)]

v_{cm } = √(4/3 gh)

let's calculate

v_{cm } = √ (4/3 9.8 3.16)

v_{cm }^ = 6.43 m / s

4 0

3 years ago

Sunitha can type 1800 words in half an hour. What is her typing speed in words per minute?

Andre45 [30]

Answer:

60words/minute

Explanation:

If Sunitha can type 1800 words in half an hour, this can be expressed as;

1800 words = 30 minutes

To get her typing speed per minute, we will use the formula

Speed = Number of words/Time used

Typing speed = 1800/30

Typing speed = 60words/minute

Hence her typing speed in words per minute is 60words/minute

6 0

2 years ago

Which component measures the potential difference across a branch in a circuit? A. switch B. resistor C. ammeter D. voltmeter

NNADVOKAT [17]

Answer:

Voltmeter

Explanation:

7 0

3 years ago

Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2

Bess [88]

Answer:

The acceleration of $M_2$ is $a = 0.7156 m/s^2$

Explanation:

From the question we are told that

The mass of first block is $M_1 = 2.25 \ kg$

The angle of inclination of first block is $\theta _1 = 43.5^o$

The coefficient of kinetic friction of the first block is $\mu_1 = 0.205$

The mass of the second block is $M_2 = 5.45 \ kg$

The angle of inclination of the second block is $\theta _2 = 32.5^o$

The coefficient of kinetic friction of the second block is $\mu _2 = 0.105$

The acceleration of $M_1 \ and\ M_2$ are same

The force acting on the mass $M_1$ is mathematically represented as

$F_1 = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

=> $M_1 a = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

Where T is the tension on the rope

The force acting on the mass $M_2$ is mathematically represented as

$F_2 = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

$M_2 a = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

At equilibrium

$F_1 = F_2$

So

$T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

making a the subject of the formula

$a = \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}$

substituting values $a = \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}$

=> $a = 0.7156 m/s^2$

3 0

3 years ago