Which statement about the carbon cycle is most true?

When you push a 1.85-kg book resting on a

Margaret [11]

Answer:

Given:

mb=2.05 kgmb=2.05 kg Mass of the book

fs=2.45 Nfs=2.45 N Static friction

fk=1.50 Nfk=1.50 N Kinetic friction

6 0

3 years ago

A piece of charcoal used for cooking is found at the remains of an ancient campsite. A 1.09 kg sample of carbon from the wood ha

fgiga [73]

Answer:

t = 17199 years

Explanation:

given,

mass of sample = 1.09 Kg

Activity of living material = 15 decays / min /g

Activity of living material = 15 x 1000 decays /min /kg

Activity of living material per 1.09 kg A = 1.09 x 15 x 1000 decays / min

Activity of after time t is A ' = 2020

half life = 57300 years

desegregation constant

λ = 0.693 / 5700

$A'= A e^{-\lambda\ t}$

$A'= 1.09 \times 15 \times 1000 e^{-\lambda\ t}$

$2020= 1.09 \times 15 \times 1000 e^{-\dfrac{0.693}{5700}\times t}$

$0.124=e^{-\dfrac{0.693}{5700}\times t}$

taking ln both side

$\dfrac{0.693}{5700}\times t = 2.09$

t = 17199 years

5 0

3 years ago

A heavy object and a light object are dropped from the same height. If we neglect air resistance, which will hit the ground firs

Maksim231197 [3]

Answer:

None, both objects will hit ground at the same time.

Explanation:

Assuming no air resistance present, and that both objects start from rest, we can apply the following kinematic equation for the vertical displacement:

$\Delta h = \frac{1}{2}*g*t^{2} (1)$

As the left side in (1) is the same for both objects, the right side will be the same also.
Since g is constant close to the surface of the Earth, it's also the same for both objects.
So, the time t must be the same for both objects also.

6 0

3 years ago

The beautiful Multnomah Falls in Oregon are approximately 206m high. If the Columbia river is flowing horizontally at 2.90m/s ju

zhenek [66]

Answer:

The overall velocity of the water when it hits the bottom is:

$v_f=63.61\ \frac{m}{s}$

Explanation:

Use the law of conservation of energy.

Call it instant [1] to the moment when the water is just before reaching the falls.

At this moment its height h is 206 meters and its velocity horizontally $v_i$ is $v_i = 2.90$ m/s.

At the instant [1] the water has gravitational power energy $E_g$

$E_g = mgh$

The water also has kinetic energy Ek.

$E_k = 0.5mv_i ^ 2$

Then the Total E1 energy is:

$E_1 = mgh + 0.5mv_i ^ 2$

In the instant [2] the water is within an instant of touching the ground. At this point it only has kinetic energy, since the height h = 0. However at time [2] the water has maximum final velocity $v_f$

So:

$E_2 = 0.5mv_f ^ 2$

As the energy is conserved then $E_1 = E_2$

$mgh + 0.5mv_i ^ 2 = 0.5mv_f ^ 2$

Now we solve for $v_f$ .

$gh + 0.5v_i ^ 2 = 0.5v_f ^ 2\\\\9.8(206) + 0.5(2.90) ^ 2 = 0.5v_f 2\\\\v_f^2 = \frac{9.8(206) + 0.5(2.90) ^ 2}{0.5}\\\\v_f = \sqrt{\frac{9.8(206) + 0.5(2.90) ^ 2}{0.5}}\\\\v_f=63.61\ \frac{m}{s}$

7 0

3 years ago

How much does a oil tanker weigh?

Novosadov [1.4K]

About <span>250,000 DWT(Dead Weight Tons)</span>

4 0

3 years ago