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3 years ago

The half-life of a certain isotope is 12 years. How much of a 600 g sample will remain after 36 years?

Physics

1 answer:

miskamm [114]3 years ago

8 0

36/12= 3 3 IS THE AMOUT OF HALFLIVES

ONE HALF LIFE = 300

TWO HALF LIFE = 150

THREE HALF LIFE = 75

*Divide 600 three times

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A stone is dropped into a well. The sound of the splash is heard 6.25 s later. What is the depth of the well? (Take the speed of

Naya [18.7K]

Answer:

depth of well is 163.30 m

Explanation:

Given data

speed of sound = 343 m/s

timer = 6.25 s

to find out

depth of well

solution

let us consider depth d

so equation will be

depth = 1/2 ×g ×t² ..............1

and

depth = velocity of sound × time .................2

here we have given time 6.25 that is sum of 2 time

when stone reach at bottom that time

another is sound reach us after stone strike on bottom

so time 1 + time 2 = 6.25 s

so from equation 1 and 2 we get

1/2 ×g ×t² = velocity of sound × time

1/2 ×9.8 × t1² = 343 × (6.25 - t1 )

t1 = 5.77376 sec

so height = 1/2 ×g ×t²

height = 1/2 ×9.8 × (5.773)²

height = 163.30 m

3 0

3 years ago

A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω. The material constant, β, for this therm

Karo-lina-s [1.5K]

Answer:

the thermistor temperature = $325.68 \ ^0 \ C$

Explanation:

Given that:

A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω.

i.e Temperature

$T_1 = 100^0C\\T_1 = (100+273)K\\\\T_1 = 373\ K$

Resistance of the thermistor $R_1 =$ 20,000 ohms

Material constant $\beta$ = 3650

Resistance of the thermistor $R_2$ = 500 ohms

Using the equation :

$R_1 = R_2 \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

$\frac{R_1}{ R_2} = \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

Taking log of both sides

$In \ \frac{R_1}{ R_2} = In \ \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

$In \ \frac{R_1}{ R_2} = {\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

$\frac{ In \ \frac{R_1}{ R_2}}{ {\beta}} = (\frac{1}{T_1}- \frac{1}{T_2})$

$\frac{1}{T_2} = \frac{1}{T_1} - \frac{ In \ \frac{R_1}{ R_2}}{ {\beta}}$

${T_2} = \frac{\beta T_1}{\beta - In (\frac{R_1}{R_2})T}$

Replacing our values into the above equation :

${T_2} = \frac{3650*373}{3650 - In (\frac{20000}{500})373}$

${T_2} = \frac{1361450}{3650 - 3.6888*373}$

${T_2} = \frac{1361450}{3650 - 1375.92}$

${T_2} = \frac{1361450}{2274.08}$

${T_2} = 598.68 \ K$

${T_2} = 325.68 \ ^0 \ C$

Thus, the thermistor temperature = $325.68 \ ^0 \ C$

4 0

3 years ago

The sun is 150,000,000 km from the earth.

aniked [119]

Answer:

<h3>What is the angular speed of the earth around the sun? </h3>

It takes the Earth approximately 23 hours, 56 minutes and 4.09 seconds to make one complete revolution (360 degrees). This length of time is known as a sidereal day. The Earth rotates at a moderate angular velocity of

$7.2921159 × 10 {}^{ - 5} \: \: \frac{radians}{second}$

<h3>What is the tangential speed of the earth? </h3>

The earth rotates once every 23 hours, 56 minutes and 4.09053 seconds, called the sidereal period, and its circumference is roughly 40,075 kilometers. Thus, the surface of the earth at the equator moves at a speed of 460 meters per second--or roughly 1,000 miles per hour.

8 0

3 years ago

A wheelbarrow is pushed with a force of 40 N. If 6,000 J of work is

stepladder [879]

Answer:

Distance = 150 meters

Explanation:

Given the following data;

Work done = 6,000 Joules

Force = 40 Newton

To find the total distance covered by the wheelbarrow;

Workdone = force * distance

Substituting into the formula, we have;

6000 = 40 * distance

Distance = 6000/40

Distance = 150 meters

Therefore, the total distance the wheelbarrow was pushed is 150 meters.

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2 years ago

The electric current leaves the battery passes through the it then travels through the next through the and finally padses throu

miss Akunina [59]

<span>The answers are:

bulb, motor, buzzer and swtich.

As seen in the picture attached,

The electric current leaves the battery passes through the bulb. It then travels through the motor, next through the buzzer and finally passes through the switch before returning to the battery.</span>

8 0

3 years ago