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3 years ago

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1. How is nuclear notation for an electron different than we symbolize electrons in the rest of chemistry? Do your best, don’t s

tress.
2. Why is gamma radiation written with two zeroes?

3. How is gamma radiation different from alpha and beta radiation?

1 answer:

dexar [7]3 years ago

8 0

Answer:

i think During gamma decay the nucleus emits radiation without actually changing its composition: We start with a nucleus with 12 protons and 12 neutrons, and we end up with a nucleus with 12 protons and 12 neutrons… but somehow radiation gets released along the way!

Explanation:

hope this hels u

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A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it

Degger [83]

Answer:

a) $\omega_1=8.168\,rad.s^{-1}$

b) $n_1=7.735 \,rev$

c) $\alpha_1 =0.6864\,rad.s^{-2}$

d) $\alpha_2=4.1454\,rad.s^{-2}$

e) $t_2=1.061\,s$

Explanation:

Given that:

initial speed of turntable, $N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}$

full speed of rotation, $N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}$

time taken to reach full speed from rest, $t_1=11.9\,s$

final speed after the change, $N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}$

no. of revolutions made to reach the new final speed, $n_2=11\,rev$

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

$\omega=\frac{2\pi.N}{60}\, rad.s^{-1}$

where N = angular speed in rpm.

putting the respective values from case 1 we've

$\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}$

$\omega_1=8.168\,rad.s^{-1}$

(c)

using the equation of motion:

$\omega_1=\omega_0+\alpha . t_1$

here α is the angular acceleration

$78=0+\alpha_1\times 11.9$

$\alpha_1 = \frac{8.168 }{11.9}$

$\alpha_1 =0.6864\,rad.s^{-2}$

(b)

using the equation of motion:

$\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1$

$8.168^2=0^2+2\times 0.6864\times n_1$

$n_1=48.6003\,rad$

$n_1=\frac{48.6003}{2\pi}$

$n_1=7.735\, rev$

(d)

using equation of motion:

$\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2$

$12.5664^2=8.168^2+2\alpha_2\times 11$

$\alpha_2=4.1454\,rad.s^{-2}$

(e)

using the equation of motion:

$\omega_2=\omega_1+\alpha_2 . t_2$

$12.5664=8.168+4.1454\times t_2$

$t_2=1.061\,s$

4 0

3 years ago

Temperature is a measure of the average speed of the ____ of a thing.

Pie

Answer:

it is atoms and molecules

7 0

3 years ago

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An advantage of light microscopes compared to electron microscopes is that light microscopes _____.

ivolga24 [154]

<span>C is the correct answer. Electron microscopes require a vacuum to work, so living cells cannot be seen because they cannot respire. Light microscopes use a ray of visible light instead of a beam of electrons to magnify something so it can be seen by the naked eye. There are two different types of electron microscope: transmission (TEM) and scanning (SEM).</span>

5 0

2 years ago

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Calcular la energía cinética de un Coyote de 30 kg que corre tras su presa con una velocidad de 65 km/h

Mumz [18]

Respuesta:

4.9 × 10³ J

Explicación:

Paso 1: Información provista

Masa del coyote (m): 30 kg

Velocidad del coyote (v): 65 km/h

Paso 2: Convertir la velocidad del coyote de km/h a m/s

Vamos a usar los siguientes factores de conversion:

1 km = 1000 m

1 h = 3600 s

65 km/h × (1000 m/1 km) × (1 h/3600 s) = 18 m/s

Paso 3: Calcular la energía cinética del coyote (K)

Usarémos la siguiente fórmula.

K = 1/2 × m × v²

K = 1/2 × 30 kg × (18 m/s)² = 4.9 × 10³ J

6 0

3 years ago

(a) During an ice skating performance, an initially motionless 80.0-kg clown throws a fake barbell away. The clown’s ice skates

nlexa [21]

Answer:

a) Barbell mass = .2kg

b) Energy gained = 10 Joules

c)Newton's Third Law of Motion; explained more below

Explanation:

I'm always weary of Physics problems so I apologize ahead of time if this comes out wrong.

a) We know that in a system, the initial energy input must equal the final energy that is output. I'm going to assume part b) is regarding the energy of the clown's movement; meaning the clown is our "system". And since the system disregards friction we have a system where the initial side has the energy of the clown and the barbell, while the final side just has the final energy of the clown. This gives us:

(1/2)(Clown mass)(Initial clown velocity)² + (1/2)(Barbell mass)(Initial barbell velocity)² = (1/2)(Clown mass)(Final Clown Velocity)²

The clown is initially standing still on the ice, so the initial clown velocity is 0; which wipes out that part of the equation. Plugging in our knowns, we end up with:

(1/2)(Barbell mass)(10m/s)² = (1/2)(80kg)(.500m/s)²

Isolating the barbell mass and doing some calculating leaves us with .2kg.

b) Again, assuming the clown is our main focus, the energy gained by the throw simply equals (1/2)(Clown mass)(Final clown velocity)².

(1/2)(80)(.500)² = 10 Joules

c) The energy gained, as well as the movement of the clown, is explained by Newton's Third Law of Motion. "For every action there is an equal and opposite reaction" might sound familiar. In this case, the clown acted on the barbell and the barbell reacted on her. Had she not been on ice, the friction of the ground would've kept her in place; but then you couldn't get a Physics problem out of it eh?

4 0

3 years ago