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3 years ago

Explain why a balloon expands when taken to a higher elevation if the temperature remains constant?

Physics

2 answers:

gladu [14]3 years ago

8 0

B) The decrease in atmospheric pressure relative to pressure inside the balloon causes it to expand.

julia-pushkina [17]3 years ago

3 0

Boyle said that, if temperature of ideal gas is kept constant then pressure and volume of ideal gas are inversely proportional to each other.

Pα1/V
Pressure decreases with elevation therefore volume of balloon increases.

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La distancia entre dos ciudades es de 350km.Si un autocar emplea 4h en el viaje,determina su velocidad media y exprésala.En el s

Oxana [17]

Answer:

v = 87.5 km/h

Explanation:

To find the speed of the car you use the following formula:

$v=\frac{x}{t}$ (1)

x: distance traveled by the car = 350km

t: time = 4h

you replace the values of x and t in the equation (1):

$v=\frac{x}{t}=\frac{350km}{4h}=87.5\frac{km}{h}$

hence, the speed of the car is 87.5km/h

4 0

3 years ago

Which of the following is the smallest in mass?

s344n2d4d5 [400]

⭐Hola User._____________

⭐Here is Your Answer...!!!

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↪⭐〽Electrons are the smallest in the mass than all other particles .

5 0

3 years ago

PLEASE PLEASE HELP ME!

Vlad1618 [11]

B: heat is transferred as thermal energy by the interaction of moving particles

6 0

3 years ago

On an essentially frictionless, horizontal ice rink, a skater moving at 5.0 m/s encounters a rough patch that reduces her speed

madreJ [45]

Answer:

The length of the rough patch is 4.345 meters.

Explanation:

According to the Work-Energy Theorem, change in kinetic energy is equal to the dissipated work due to friction. That is:

$K_{1} = K_{2} + W_{loss}$

Where:

$K_{1}$ , $K_{2}$ - Initial and final kinetic energy, measured in joules.

$W_{loss}$ - Work losses due to friction.

By applying definitions of kinetic energy and work, the expression described above is expanded:

$\frac{1}{2}\cdot m \cdot v_{1}^{2} = \frac{1}{2}\cdot m \cdot v_{2}^{2} + f \cdot \Delta s$

Where:

$v_{1}$ , $v_{2}$ - Initial and final speed of the skater, measured in meters per second.

$m$ - Mass of the skater, measured in kilograms.

$\Delta s$ - Length of the rough patch, measured in meters.

$f$ - Friction force, measured in newtons.

According to the statement, friction force is represented by the following expression:

$f = r \cdot m \cdot g$

Where:

$r$ - Ratio of friction force to weight, dimensionless.

$g$ - Gravitational constant, measured in meters per square second.

Then,

$\frac{1}{2}\cdot m \cdot v_{1}^{2} = \frac{1}{2}\cdot m \cdot v_{2}^{2} + r \cdot m \cdot g \cdot \Delta s$

The equation is simplified algebraically and patch length is cleared afterwards:

$\frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2}) = r \cdot g \cdot \Delta s$

$\Delta s = \frac{v_{1}^{2}-v_{2}^{2}}{2 \cdot r \cdot g }$

Given that $v_{1} = 5\,\frac{m}{s}$ , $v_{2} = 2.5\,\frac{m}{s}$ , $r = 0.22$ and $g = 9.807 \,\frac{m}{s^{2}}$ , the length of the rough patch is:

$\Delta s = \frac{\left(5\,\frac{m}{s} \right)^{2}-\left(2.5\,\frac{m}{s} \right)^{2}}{2\cdot (0.22)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$\Delta s = 4.345\,m$

The length of the rough patch is 4.345 meters.

6 0

3 years ago

A 30 g particle is undergoing simple harmonic motion with an amplitude of 2.0 ✕ 10-3 m and a maximum acceleration of magnitude 8

KiRa [710]

Answer:

Explanation:

By using the Newton second law and the position equation for a simple harmonic motion we have

$F=ma\\a_{max}=\omega^{2}A\\x=Acos(\omega t+ \phi)\\$

where a is the acceleration, w is the angular frecuency and \phi is the phase constant. We can calculate w from the equation for the maximum acceleration

$\omega=\sqrt{\frac{a_{max}}{A}}=\sqrt{\frac{8*10^{3}m/s^{2}}{2*10^{-3}}}=2000rad/s$

(a).

$F=ma=m\omega^{2}Acos(\omega t + \phi)\\F=(30*10^{-3}kg)(2*10^{-3}m)(2000\frac{rad}{s})^{2}cos(2000\frac{rad}{s} t - \frac{\pi}{2})=240N*cos(2000\frac{rad}{s} t - \frac{\pi}{2})$

(b). $T=\frac{2\pi}{\omega}=\frac{2\pi}{2000}=3.14*10^{-13} s$

(c). $v_{max}=A\omega=(2*10^{-3}m)(2000\frac{rad}{s})=4\frac{m}{s}$

(d). The mecanical energy is the kinetic energy when the velocity is a maximum

$E_{m}=E_{k}(v_{max})=\frac{mv_{max}^{2}}{2}=\frac{30*10^{-3}kg(4\frac{m}{s})^{2}}{2}=0.024J$

3 0

4 years ago

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