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3 years ago

How can you determine the type of radiation an atom emits as its nucleus decays

Physics

1 answer:

Pie3 years ago

7 0

Gamma raidiation and beta and alpha particles

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Which of the following statements describes an interaction between the geosphere and atmosphere?

german

B is the answer, I’m really good at this subject

8 0

3 years ago

Read 2 more answers

Sound enters the ear, travels through the auditory canal, and reaches the eardrum. The auditory canal is approximately a tube op

Juliette [100K]

Answer:

The fundamental frequency of can is 2.7 kHz.

Explanation:

Given that,

A typical length for the auditory canal in an adult is about 3.1 cm, l = 3.1 cm

The speed of sound is, v = 336 m/s

We need to find the fundamental frequency of the canal. For a tube open at only one end, the fundamental frequency is given by :

$f=\dfrac{v}{4l}\\\\f=\dfrac{336}{4\times 3.1\times 10^{-2}}\\\\f=2709.67\ Hz\\\\f=2.7\ kHz$

So, the fundamental frequency of can is 2.7 kHz. Hence, this is the required solution.

7 0

3 years ago

4. Una cuerda de acero de piano mide 1.60 m de longitud y 0.20 cm de diámetro. ¿Cuál es la tensión en la cuerda si se estira 0.2

bazaltina [42]

Answer:

$1030.83\ \text{N}$

Explanation:

$\Delta L$ = Cambio en la longitud de la cuerda = 0.25 cm

T = tensión en cuerda

A = Área de la cadena = $\dfrac{\pi}{4}d^2$

d = Diámetro de la cuerda = 0.2 cm

L = Longitud original de la cuerda = 1.6 m

El cambio de longitud de una cuerda viene dado por

$\Delta L=\dfrac{TL}{AE}\\\Rightarrow T=\dfrac{\Delta LAE}{L}\\\Rightarrow T=\dfrac{0.25\times 10^{-2}\times \dfrac{\pi}{4}(0.2\times 10^{-2})^2\times 210\times 10^9}{1.6}\\\Rightarrow T=1030.84\ \text{N}$

La tensión en la cuerda es $1030.84\ \text{N}$ .

8 0

3 years ago

Two persons manage to push a motorcar of mass 1200 kg at uniform velocity along a level road, The same motorcar can be pushed by

densk [106]

Answer:

40N by each person.

Explanation:

8 0

3 years ago

In a milkans apparatus,an oil drop of weight 2.0×10^-15kg accquires two surplus electrons. when a Potential difference of 620 vo

andriy [413]

Answer:

r = 9.92 mm

Explanation:

Given that,

Mass of oil drop, $m=2\times 10^{-15}\ kg$

It acquires 2 surplus electrons, q = +2e $=3.2\times 10^{-19}\ C$

Potential difference, V = 620 V

Thie potential difference is applied between the pair of horizontal metal plates the drop is in equilibrium.

We need to find the distance between the plates.

At equilibrium,

mg = qE

Since, E = V/r (r is distance between plates)

$mg=\dfrac{qV}{r}\\\\r=\dfrac{qV}{mg}\\\\r=\dfrac{3.2\times 10^{-19}\times 620}{2\times 10^{-15}\times 10}\\\\=0.00992\ m\\\\=9.92\ mm$

So, the distance between the plates is 9.92 mm.

6 0

3 years ago