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3 years ago

Which contributes most to our average dose of background radiation?

Physics

1 answer:

Anuta_ua [19.1K]3 years ago

6 0

C) Natural and Artificial sources

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A balloon filled with helium gas at 20°C occupies 4.91 L at 1.00 atm. The balloon is immersed in liquid nitrogen at -196°C, whil

Marizza181 [45]

Answer:

V = 0.248 L

Explanation:

To do this, use the following equation:

P1*V1/T1 = P2*V2/T2

This equation is used to find a relation between two differents conditions of a same gas, which is this case. From this equation we can solve for V2.

Solving for V2:

V2 = P1*V1*T2/T1*P2

Temperature must be at Kelvin, so, we have to sum the temperature 273 to convert it in K.

Replacing the data we have:

V2 = 1 * 4.91 * (-196+273) / 5.2 * (20+273)

V2 = 378.07 / 1523.6

V2 = 0.248 L

7 0

4 years ago

Para el siguiente conjunto de medidas, calcule EL ERROR RELATIVO PORCENTUAL: 1.34 m, 1.35 m, 1.37 m y 1.36 m

ivanzaharov [21]

Answer:

Ver explicacion abajo

Explanation:

En este caso para poder calcular el error relativo porcentual, es necesario calcular primero el error absoluto, que se calcula de la siguiente forma:

Error absoluto = Resultado exacto - aproximación

Sin embargo, no tenemos el resultado exacto de las medidas, pero podriamos conocerlo tomando el promedio de estas medidas y este es el que tomaremos como el verdadero resultado de las medidas:

Promedio de medidas = 1.34 + 1.35 + 1.37 + 1.36 / 4

Promedio de medidas = 1.355 m

Ya que tenemos el promedio, podemos calcular el error absoluto de cada medida y luego el error relativo porcentual:

Ea1 = 1.355 - 1.34 = 0.015

Ea2 = 1.355 - 1.35 = 0.005

Ea3 = 1.37 - 1.355 = 0.015

Ea4 = 1.36 - 1.355 = 0.005

Ya que tenemos los 4 errores absolutos, es posible calcular el porcentual:

%error relativo = (Error absoluto / resultado exacto) * 100

Aplicando la expresión con cada uno de los valores tenemos:

%Er1 = (0.015/1.34) * 100 = 1.12%

%Er2 = (0.005/1.35) * 100 = 0.37%

%Er3 = (0.015/1.37) * 100 = 1.09%

%Er4 = (0.005/1.36) * 100 = 0.37%

Espero que te sirva.

4 0

3 years ago

The vitreous humor, a transparent, gelatinous fluid that fills most of the eyeball, has an index of refraction of 1.34. Visible

Leto [7]

Answer:

a) 298.5 nm , 522.4 nm and b) radiation frequency does not change

Explanation:

When electromagnetic radiation reaches a medium with a different index of refraction, the medium vibrates the molecules, as if it were a resonance process, whereby the medium vibrates at the same frequency as the incident light.

On the other hand, when the light reaches another medium its average speed within the medium changes, it is now less than the speed of light in a vacuum (c) for this to happen as we saw that the frequency is constant there must be a change in the wavelength of the radiation that is characterized by the ratio

λₙ = λ₀ / n

λₙ = 400 nm in the void

λₙ = 400 / 1.34

λₙ= 298.5 nm

λ₀ = 700 nm

λₙ = 700 / 1.34

λₙ = 522.4 nm

The radiation frequency does not change

4 0

3 years ago

A playground merry-go-round has radius 2.40 m and moment of inertia 2100 kg⋅m2 about a vertical axle through its center, and it

daser333 [38]

Answer:

a) 0.31 rad/s

b) 100 J

c) 6.67 W

Explanation:

(a) the force would generate a torque of:

$T = FR = 18 * 2.4 = 43.2 Nm$

According to Newton 2nd law, the angular acceleration would be

$\alpha = \frac{T}{I} = \frac{43.2}{2100} = 0.021 rad/s^2$

It starts from rest, then after 15s it would achieve a speed of

$\omega = \alpha t = 0.021 * 15 = 0.31 rad/s$

(b) The distance angle swept by it is:

$\theta = \frac{\alpha t^2}{2} = \frac{0.021 * 15^2}{2} = 2.314 rad$

Hence the work by the child

$W = T\theta = 43.2 *2.314 \approx 100 J$

c) Average power to work per time unit

$P = \frac{W}{t} = \frac{100}{15} = 6.67 W$

7 0

3 years ago

I need answers and solvings to these questions

den301095 [7]

1) The period of a simple pendulum depends on B) III. only (the length of the pendulum)

2) The angular acceleration is C) $15.7 rad/s^2$

3) The frequency of the oscillation is C) 1.6 Hz

4) The period of vibration is B) 0.6 s

5) The diameter of the nozzle is A) 5.0 mm

6) The force that must be applied is B) 266.7 N

Explanation:

The period of a simple pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

where

T is the period

L is the length of the pendulum

g is the acceleration of gravity

From the equation, we see that the period of the pendulum depends only on its length and on the acceleration of gravity, while there is no dependence on the mass of the pendulum or on the amplitude of oscillation. Therefore, the correct option is

B) III. only (the length of the pendulum)

The angular acceleration of the rotating disc is given by the equation

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f$ is the final angular velocity

$\omega_i$ is the initial angular velocity

t is the time elapsed

For the compact disc in this problem we have:

$\omega_i = 0$ (since it starts from rest)

$\omega_f = 300 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=31.4 rad/s$ is the final angular velocity

t = 2 s

Substituting, we find

$\alpha = \frac{31.4-0}{2}=15.7 rad/s^2$

For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation

$a=-\omega^2 x$

where

a is the acceleration

x is the displacement

$\omega$ is the angular frequency of the system

For the oscillator in this problem, we have the following relationship

$a=-100 x$

which implies that

$\omega^2 = 100$

And so

$\omega = \sqrt{100}=10 rad/s$

Also, the angular frequency is related to the frequency $f$ by

$f=\frac{\omega}{2\pi}$

Therefore, the frequency of this simple harmonic oscillator is

$f=\frac{10}{2\pi}=1.6 Hz$

When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write

$mg=kx$

where

m is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

k is the spring constant

x = 8.0 cm = 0.08 m is the stretching of the spring

We can re-arrange the equation as

$\frac{k}{m}=\frac{g}{x}=\frac{9.8}{0.08}=122.5$

The angular frequency of the spring is given by

$\omega=\sqrt{\frac{k}{m}}=\sqrt{122.5}=11.1 Hz$

And therefore, its period is

$T=\frac{2\pi}{\omega}=\frac{2\pi}{11.1}=0.6 s$

According to the equation of continuity, the volume flow rate must remain constant, so we can write

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the hose, with $r_1 = 5 mm$ being the radius of the hose

$v_1 = 4 m/s$ is the speed of the petrol in the hose

$A_2 = \pi r_2^2$ is the cross-sectional area of the nozzle, with $r_2$ being the radius of the nozzle

$v_2 = 16 m/s$ is the speed in the nozzle

Solving for $r_2$ , we find the radius of the nozzle:

$\pi r_1^2 v_1 = \pi r_2^2 v_2\\r_2 = r_1 \sqrt{\frac{v_1}{v_2}}=(5)\sqrt{\frac{4}{16}}=2.5 mm$

So, the diameter of the nozzle will be

$d_2 = 2r_2 = 2(2.5)=5.0 mm$

According to the Pascal principle, the pressure on the two pistons is the same, so we can write

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

where

$F_1$ is the force that must be applied to the small piston

$A_1 = \pi r_1^2$ is the area of the first piston, with $r_1= 2 cm$ being its radius

$F_2 = mg = (1500 kg)(9.8 m/s^2)=14700 N$ is the force applied on the bigger piston (the weight of the car)

$A_2 = \pi r_2^2$ is the area of the bigger piston, with $r_2= 15 cm$ being its radius

Solving for $F_1$ , we find

$F_1 = \frac{F_2A_1}{A_2}=\frac{F_2 \pi r_1^2}{\pi r_2^2}=\frac{(14700)(2)^2}{(15)^2}=261 N$

So, the closest answer is B) 266.7 N.

Learn more about pressure:

brainly.com/question/4868239

brainly.com/question/2438000

#LearnwithBrainly

5 0

3 years ago