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3 years ago

Como se da la transferencia de energia entre los objetos?

1 answer:

6 0

El calor es la transferencia de energía de un objeto más cálido a un objeto más fresco. El calor puede transferirse de tres maneras: por conducción, por convección y por radiación. La conducción es la transferencia de energía de una molécula a otra por contacto directo.

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If the capacity of the lungs of an adult were 3.8 L, how many moles of air would the lungs contain at body temperature and atmos

Ahat [919]

Answer:

0.14955 mol

Explanation:

We can solve this problem using the ideal gas law

$P \ V = \ n \ R \ T$

where P is the pressure, V the volume, n the number of moles, R the ideal gas constant and T the temperature.

We can use the atmospheric pressure as 1 atm, and the body temperature as 36.5 °C, in Kelvin this is:

$T_{body} = 36.5 \° C = (36.5 + 273.15) K = 309.65 \ K$

The ideal gas constant is:

$R = 0.082057 \frac{L \ atm}{ K \ mol}$

taking all this in consideration, the number of moles will be:

$n = \frac{P \ V}{ R \ T }$

$n = \frac{1 \ atm * 3.8 \ L }{ 0.082057 \frac{L \ atm}{ K \ mol}$ * 309.65 \ K } [/tex]

$n = 0.14955 \ mol$

3 0

3 years ago

a baseball is hit straight up into the air. If the initial velocity was 20 m/s, how high will the ball go?How long will it be un

AleksAgata [21]

1) The motion of the ball is an uniformly accelerated motion, with constant acceleration equal to $g=-10 m/s^2$ (the negative sign means it is directed towards the ground).

For an uniformly accelerated motion, we can use the following relationship:
$2gh=v_f^2-v_i^2$
where h is the maximum height reached by the ball, $v_i = 20 m/s$ is its initial velocity and $v_f$ the velocity of the ball when it reaches the maximum height. But $v_f=0$ (when the ball reaches the maximum height, it stops before going down, so its velocity at that moment is zero), so we can use the relationship to calculate h, the maximum height:
$h=- \frac{v_i^2}{2g} =- \frac{20 m/s)^2}{2 \cdot (-10 m/s^2)} =20 m$

2) We can find the time the ball takes to return to the ground by requiring that the space covered by the ball returns to zero:
$S(t)=0$
where for an uniformly accelerated motion,
$S(t)=v_i t + \frac{1}{2} gt^2 =0$
By solving this, we have two solutions: one is t=0, which corresponds to the moment the player hits the ball, the second one is
$t=- \frac{2 v_i}{g}=- \frac{2 \cdot 20 m/s}{-10 m/s^2}=4 s$
so, the ball returns to the ground after 4 s.

3) The velocity of the ball when it returns to the ground is given by:
$v(t) = v_i +gt=20 m/s + (-10 m/s^2)(4 s)=-20 m/s$
where the negative sign means the ball is going downwards.

8 0

3 years ago

The speed of the bus is 40km/hr. what does it mean?

kiruha [24]

Answer:

The speed of the bus is 40 km/hr so this means the bus is travelling at a speed of 40 km per hour.

5 0

4 years ago

Question 8 a-e plz

N76 [4]

Answer:

(a) t = 0 s

(b) t = 0 s, 30 s, 55 s

(d) t = 10 s to t = 15 s

(e) a = 6 m/s^2

Explanation:

(a) The car is at starting position at t = 0 s and v = 0 m/s.

(b) The velocity of car is zero when the time is t = 0 s, 30 s and 55 s.

(d) The fastest speed is 60m/s from t = 10 s to t = 15 s.

(e) The slope of the velocity time graph gives acceleration.

a = (60 - 0) / (10 - 0) = 6 m/s^2

5 0

3 years ago

ListenA bicycle and its rider have a combined mass of 80. kilograms and a speed of 6.0 meters per second. What is the magnitude

Setler [38]

Answer:

a) $1.2\times 10^2\ N$

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

a = Acceleration

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{0-6}{4}\\\Rightarrow a=-1.5\ m/s^2$

The acceleration of the bicycle and rider is -1.5 m/s²

Force

$F=ma\\\Rightarrow F=80\times -1.5\\\Rightarrow F=-120\ N=-1.2\times 10^2\ N$

The magnitude of the average force needed to bring the bicycle and its rider to a stop is $1.2\times 10^2\ N$

3 0

3 years ago