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2 years ago

What is the half life-life of your 100 atoms of Carbon-14?

Physics

1 answer:

Naddika [18.5K]2 years ago

3 0

Answer:

Answer: It takes 5,730 years for half the carbon-14 to change to nitrogen; this is the half-life of carbon-14. After another 5,730 years only one-quarter of the original carbon-14 will remain

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A ball is projected horizontally from the top of a bertical building 25.0m above the ground level with an initial velocity of 8.

kirill115 [55]

Answer:

Solution given:

height [H]=25m

initial velocity [u]=8.25m/s

g=9.8m/s

now;

a. How long is the ball in flight before striking the ground?

Time of flight =?

Now

Time of flight= $\sqrt{\frac{2H}{g}}$

substituting value

= $\sqrt{\frac{2*25}{9.8}}$
=2.26seconds

<h3>the ball is in flight before striking the ground for 2.26seconds.</h3>

b. How far from the building does the ball strike the ground?

Horizontal range=?

we have

Horizontal range=u* $\sqrt{\frac{2H}{g}}$

=8.25*2.26
=18.63m

<h3>The ball strikes 18.63m far from building. </h3>

7 0

3 years ago

A train at rest emits a sound at a frequency of 1057 Hz. An observer in a car travels away from the sound source at a speed of 2

il63 [147K]

Answer:

993.52 Hz

Explanation:

The frequency of sound emitted by the stationery train is 1057 Hz.

The car travels away from the train at 20.6 m/s.

The frequency the observer hears is given by the formula:

$f_o = \frac{v - v_o}{v}f$

where v = velocity of sound = 343 m/s

vo = velocity of observer

f = frequency from source

This phenomenon is known as Doppler's effect.

Therefore:

$f_o = \frac{343 - 20.6}{343} * 1057\\ \\f_o = 322.4 / 343 * 1057\\\\f_o = 993.52 Hz$

The frequency heard by the observer is 993.52 Hz.

4 0

4 years ago

A uniform, 4.5 kg, square, solid wooden gate 2.0 mm on each side hangs vertically from a frictionless pivot at the center of its

True [87]

Answer:

The angular velocity is $w = 1.43\ rad/sec$

Explanation:

From the question we are told that

The mass of wooden gate is $m_g = 4.5 kg$

The length of side is L = 2 m

The mass of the raven is $m_r = 1.2 kg$

The initial speed of the raven is $u_r = 5.0m/s$

The final speed of the raven is $v_r = 1.5 m/s$

From the law of conservation of angular momentum we express this question mathematically as

Total initial angular momentum of both the Raven and the Gate = The Final angular momentum of both the Raven and the Gate

The initial angular momentum of the Raven is $m_r * u_r * \frac{L}{2}$

Note: the length is half because the Raven hit the gate at the mid point

The initial angular momentum of the Gate is zero

Note: This above is the generally formula for angular momentum of square objects

The final angular velocity of the Raven is $m_r * v_r * \frac{L}{2}$

The final angular velocity of the Gate is $\frac{1}{3} m_g L^2 w$

Substituting this formula

$m_r * u_r * \frac{L}{2} = \frac{1}{3} m_g L^2 w + m_r * v_r * \frac{L}{2}$

$\frac{1}{3} m_g L^2 w = m_r * v_r * \frac{L}{2} - m_r * u_r * \frac{L}{2}$

$\frac{1}{3} m_g L^2 w = m_r * \frac{L}{2} * [u_r - v_r]$

Where $w$ is the angular velocity

Substituting value

$\frac{1}{3} (4.5)(2)^2 w = 1.2 * \frac{2}{2} * [5 - 1.5]$

$6w = 4.2$

$w = \frac{6}{4.2}$

$w = 1.43\ rad/sec$

5 0

3 years ago

How fast does water flow from a hole at the bottom of a very wide, 3.2 m deep storage tank filled with water

Nezavi [6.7K]

Answer:

the velocity of the water flow is 7.92 m/s

Explanation:

The computation of the velocity of the water flow is as follows

Here we use the Bernouli equation

As we know that

$V_1 = \sqrt{2g(h_2 - h_1)}\\\\=\sqrt{2g(\Delta h)} \\\\= \sqrt{2\times9.8\times3.2} \\\\= \sqrt{62.72}$

= 7.92 m/s

Hence, the velocity of the water flow is 7.92 m/s

We simply applied the above formula so that the correct value could come

And, the same is to be considered

5 0

3 years ago

A 55-kg woman is wearing high heels.

Grace [21]

Answer:

Pressure, $P=1.90\times 10^7\ Pa$

Explanation:

It is given that,

Mass of the woman, m = 55 kg

Diameter of the circular cross section, d = 6 mm

Radius, r = 3 mm = 0.003 m

Let P is the pressure exerted on the floor. It is equal to the force acting on woman per unit area. It is given by :

$P=\dfrac{F}{A}$

$P=\dfrac{mg}{\pi r^2}$

$P=\dfrac{55\times 9.8}{\pi (0.003)^2}$

$P=1.90\times 10^7\ Pa$

So, the pressure exerted on the floor is $1.90\times 10^7\ Pa$ . Hence, this is the required solution.

7 0

3 years ago