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3 years ago

6

If you begin with 40 grams of a radioactive isotope and end with 10 grams, how many half-lifes of the radioactive isotope have p

assed ?

1 answer:

Rom4ik [11]3 years ago

8 0

30 grams of radioactive isotope have passed.

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How are eras and periods of the geologic time scale named?

Natalka [10]

Answer:

mark me brainliest plz

Explanation:

In the early 1800's a system for naming geologic time periods was devised using four periods of geologic time. They were named using Latin root words, Primary, Secondary, Tertiary and Quaternary. ... Keep in mind that this chart is focused on geologic time periods. There are also geologic Eons, Eras, and epochs.

3 0

3 years ago

Read 2 more answers

Un proyectil es lanzado horizontalmente desde una altura de 12 metro con una velocidad de 80 m/sg. a.Calcular el tiempo de vuelo

Naily [24]

Answer:

t= 1,56 s , x= 124,8 m , v = (80 i^ - 15,288 j ) m/s

Explanation:

Este es un ejercicio de lanzamiento proyectiles, comencemos por encontrar el tiempo que tarda en llegar al piso

y = y₀ + $v_{oy}$ t – ½ g t²

en este caso la altura inicial es y₀= 12 m y llega a y=0 , como es lanzado horizontalmente la velocidad vertical es cero (v_{oy}=0)

0 = y₀ – ½ g t²

t= √ (2 y₀/g)

calculemos

t= √ ( 2 12 / 9,8)

t= 1,56 s

El alcance del proyectil es la distancia horizontal recorrida

x = v₀ₓ t

x = 80 1,56

x= 124,8 m

La velocidad de impacto cuando toca el suelo

vx = v₀ₓ = 80 ms

$v_{y}$ = v_{oy} – gt

v_{y} = - 9,8 1,56

v_{y} = - 15,288 m/s

la velocidad es

v = (80 i^ - 15,288 j ) m/s

Traducttion

This is a projectile launching exercise, let's start by finding the time it takes to reach the ground

y = y₀ + v_{oy} t - ½ g t²

in this case the initial height is i = 12 m and it reaches y = 0, as it is thrown horizontally the vertical speed is zero ( $v_{oy}$ = 0)

0 =y₀I - ½ g t²

t = √ (2y₀ / g)

let's calculate

t = √ (2 12 / 9.8)

t = 1.56 s

Projectile range is the horizontal distance traveled

x = v₀ₓ t

x = 80 1.56

x = 124.8 m

Impact speed when it hits the ground

vₓ = v₀ₓ = 80 ms

v_{y} = v_{oy} - gt

v_{y} = - 9.8 1.56

v_{y} = - 15,288 m / s

the speed is

v = (80 i ^ - 15,288 j) m / s

7 0

3 years ago

a glass bottle of soda is sealed with a screw cap the absolute pressure of the gas inside the bottle is 1.80*10^5 pa. assuming t

eduard

The force exerted by a pressure of any gas over a surface its given by the formula P=F/S (where P is pressure, F force and S surface).

We can multiply both sides of the formula by S to obtain the force.

P*S=(F*S)/S

P*S=F

Solve for P=1.80*10^5 Pa and S=4.10*10^-4 m^2 ([Pa] =[N/m^s])

(1.80*10^5 N/m^s) * (4.10*10^-4 m^2) =F

73.8 N =F

8 0

3 years ago

A 0.311 kg tennis racket moving 30.3 m/s east makes an elastic collision with a 0.0570 kg ball moving 19.2 m/s west. Find the ve

Troyanec [42]

The velocity of tennis racket after collision is 14.96m/s

<u>Explanation:</u>

Given-

Mass, m = 0.311kg

u1 = 30.3m/s

m2 = 0.057kg

u2 = 19.2m/s

Since m2 is moving in opposite direction, u2 = -19.2m/s

Velocity of m1 after collision = ?

Let the velocity of m1 after collision be v

After collision the momentum is conserved.

Therefore,

m1u1 - m2u2 = m1v1 + m2v2

$v1 = (\frac{m1-m2}{m1+m2})u1 + (\frac{2m2}{m1+m2})u2$

$v1 = (\frac{0.311-0.057}{0.311+0.057})30.3 + (\frac{2 X 0.057}{0.311 + 0.057}) X-19.2\\\\v1 = (\frac{0.254}{0.368} )30.3 + (\frac{0.114}{0.368}) X -19.2\\ \\v1 = 20.91 - 5.95\\\\v1 = 14.96$

Therefore, the velocity of tennis racket after collision is 14.96m/s

7 0

3 years ago

Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d

Tasya [4]

Answer:

I(x) = 1444×k × ${\pi}$

I(y) = 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$

Explanation:

Given data

function = x^2 + y^2 ≤ 36

function = x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to $\pi /2$ for θ

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ y²p dA

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ (a sinθ)²(k × a²) adθda

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (sin²θ)dθ

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (1-cos2θ)/2 dθ

I(x) = k $({r}^{6}/6)^(5)_0$ × ${θ/2 - sin2θ/4}^{\pi /2}_0$

I(x) = k × $({6}^{6}/6)$ × ( ${\pi /4} - sin\pi /4)$

I(x) = k × $({6}^{5})$ × ${\pi /4}$

I(x) = 1444×k × ${\pi}$ .....................1

and we can say I(x) = I(y) by the symmetry rule

and here I(o) will be I(x) + I(y) i.e

I(o) = 2 × 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$ ......................2

3 0

3 years ago