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3 years ago

An archeologist on a dig finds a fragment of an ancient

Physics

1 answer:

marin [14]3 years ago

4 0

Answer:

19792.96488 years

Explanation:

$\tau_{\dfrac{1}{2}}$ = Half life of carbon-14 atom = 5730 years

Disintegration is given by

$\lambda=\dfrac{ln2}{\tau_{\dfrac{1}{2}}}\\\Rightarrow \lambda=\dfrac{ln2}{5730}\\\Rightarrow \lambda=0.00012\ /years$

The distintegration constant

$0.093N_0=N_0e^{-\lambda t}\\\Rightarrow 0.093N_0=N_0e^{-0.00012 t}\\\Rightarrow ln\dfrac{1}{0.093}=0.00012t\\\Rightarrow t=\dfrac{ ln\dfrac{1}{0.093}}{0.00012}\\\Rightarrow t=19792.96488\ years$

The age of the basket is 19792.96488 years

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A lunar module weighs 12 metric tons on the surface of the Earth. How much work is done in propelling the module from the surfac

zhuklara [117]

Answer:

W=76.55 miles.metric tons

Explanation:

Given that

Weight on the earth = 12 tons

So weight on the moon =12/6 = 2 tons

( because at moon g will become g/6)

As we know that

$F=\dfrac{K}{x^2}$

Here x= 1100 miles

F 2 tons

$2=\dfrac{K}{1100^2}$

$K=2.4\times 10^6$

We know that

Work = F. dx

$W=\int_{x_1}^{x_2}F.dx$

$W=\int_{1100}^{1140}\dfrac{2.4\times 10^6}{x^2}.dx$

$W=-2.4\times 10^6\left[\dfrac{1}{x}\right]_{1100}^{1140}$

$W=-2.4\times 10^6\left[\dfrac{1}{1140}-\dfrac{1}{1100}\right]$

W=76.55 miles.metric tons

6 0

3 years ago

A 5kg wheel rolls off a flat roof of a 50 m tall building at 12m/s.

Olegator [25]

Explanation:

a) Given in the y direction (taking down to be positive):

Δy = 50 m

v₀ = 0 m/s

a = 10 m/s²

Find: t

Δy = v₀ t + ½ at²

50 m = (0 m/s) t + ½ (10 m/s²) t²

t = 3.2 s

b) Given in the x direction:

v₀ = 12 m/s

a = 0 m/s²

t = 3.2 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (12 m/s) (3.2 s) + ½ (0 m/s²) (3.2 s)²

Δx = 38 m

6 0

3 years ago

Help!

Evgen [1.6K]

Answer:

factory:

-mechanical energy

-nuclear energy

-gravitational energy

Explanation:

8 0

3 years ago

Jane, looking for tarzan, is running at top speed 5.3 m/s and grabs a vine hanging vertically from a tall tree in the jungle. ho

lutik1710 [3]

v₀ = initial speed as tarzan grabs the vine = 5.3 m/s

v = final speed as the tarzan reach the maximum height = 0 m/s

h = maximum height gained by the tarzan

m = mass of tarzan

using conservation of energy

initial kinetic energy = final kinetic energy + potential energy

(0.5) m v²₀ = (0.5) m v² + m g h

(0.5) v²₀ = (0.5) v² + g h

(0.5) (5.3)² = (0.5) (0)² + (9.8) h

h = 1.43 m

3 0

3 years ago

A block, M1=10kg, slides down a smooth, curved incline of height 5m. It collides elastically with another block, M2=5kg, which i

erma4kov [3.2K]

Answer:

2.86 m

Explanation:

Given:

M₁ = 10 kg

M₂ = 5 kg

$\mu_k$ = 0.5

height, h = 5 m

distance traveled, s = 2 m

spring constant, k = 250 N/m

now,

the initial velocity of the first block as it approaches the second block

u₁ = √(2 × g × h)

u₁ = √(2 × 9.8 × 5)

u₁ = 9.89 m/s

let the velocity of second ball be v₂

now from the conservation of momentum, we have

M₁ × u₁ = M₂ × v₂

on substituting the values, we get

10 × 9.89 = 5 × v₂

v₂ = 19.79 m/s

now,

let the velocity of mass 2 when it reaches the spring be v₃

from the work energy theorem, we have

Work done by the friction force = change in kinetic energy of the mass 2

$0.5\times5\times9.8\times2 = \frac{1}{2}\times5\times( v_3^2-19.79^2)$

v₃ = 20.27 m/s

now, let the spring is compressed by the distance 'x'

therefore, from the conservation of energy

we have

Energy of the spring = Kinetic energy of the mass 2

$\frac{1}{2}kx^2=\frac{1}{2}mv_3^2$

on substituting the values, we get

$\frac{1}{2}\times250\times x^2=\frac{1}{2}\times5\times20.27^2$

x = 2.86 m

8 0

3 years ago