Ask question

Ask question

All categories

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.

You might be interested in

A 200g air-track glider is attached to a spring. The glider is pushed in 10cm and released. A student with a stopwatch finds tha

cricket20 [7]

The spring constant will be k= 5.5N/m for a 200g air track glider attached to a spring.

<h3>What is spring constant?</h3>

The spring constant, k, is a measure of the stiffness of the spring. It is different for different springs and materials.

Calculation for What is the spring constant

First step is to calculate the time period

T = 12 second/10

T = 1.2 second

Now let calculate the spring constant using this formula

$k=\dfrac{4\pi ^2m}{T^2}$

Where,

m=0.2kg

T=1.2second

k represent spring constant=?

Let plug in the formula

$k=\dfrac{4 \pi \times 0.2}{(1.2)^2}$

$k=\dfrac{39.48\times 0.2}{1.44}$

$k=\dfrac{7.90}{1.44}$

k=5.48 N/m

k=5.5 N/m ( Approximately)

Therefore the spring constant will be 5.5 N/m

To know more about spring constant follow

brainly.com/question/1968517

#SPJ4

4 0

2 years ago

Newton's law of gravity was inconsistent with Einstein's special relativity because

Lubov Fominskaja [6]

Answer:

Mass and thus force depends on the reference frame chosen

Explanation:

This can be explained as Newton's law of gravity provides action which are instantaneous at a distance and involves the evaluation of all the quantities at present time or at the instant they occur.

If the body undergoes a change in its mass distribution there will be an immediate change in its gravitational force without any lag.

Now, if we talk about special relativity, it would be absurd to say that an information can travel faster than light. The effect is in synchronization with the cause in one reference frame where the effect occurs after the cause for some observer in some other reference frame.

In order to observe Newton's law of gravity all the observer's in different reference frames must observe the same phenomena which could only be possible if time were absolute and in special relativity, time is not absolute.

Therefore, Newton's law of gravity was inconsistent with the Einstein's Special Relativity.

3 0

3 years ago

Physics - Electricity and Magnetism

Orlov [11]

Answer:

<h2>480</h2>

Explanation:

<h2>R=120÷0.25</h2><h2>R=480 ohms </h2>

because the unit for resistance is in ohms

4 0

3 years ago

State the principle of floatation

WARRIOR [948]

Explanation:

<em><u>Principle of Floatation</u></em>

Principle of Floatation states that weight of floating body is equal to weight of water displaced by it

7 0

3 years ago

An automobile engine can produce 153 N · m of torque. Calculate the angular acceleration (in rad/s^2) produced if 85.2% of this

galina1969 [7]

Answer:

46.2 rad/s2

Explanation:

Angular acceleration works very similar to linear acceleration, it follows this equation:

$\gamma = \frac{Mt}{J}$

Where:

γ: angular acceleration

Mt: torque

J: moment of inertia of the load from its turning axis

Since we have the torque we just need the moment of inertia. We have to add together the moments of the drive shaft, tires, wheel walls and wheels.

The wheels act like disks. For disks the moment of inertia is:

$J = \frac{1}{2} * m * r^2$

$Jwheel = \frac{1}{2} = 15 * 0.18^2 = 0.243 kg*m^2$

The wheel walls act like annular rings, for these the moment of inertia is:

$J = \frac{1}{2} * m * (re^2 - ri^2)$

$Jwall = \frac{1}{2} * 2 * (0.32^2 - 0.18^2) = 0.07 kg * m^2$

The tread acts like a hoop, as in mass concentrated into a circunference, for these:

$J = m * r^2$

$Jtread = 10 * 0.33^2 = 1.09 kg*m^2$

The axle acts like a rod, which is the same as the disk:

$Jaxle = \frac{1}{2} * 14.1 * 0.02^2 = 0.0028 kg*m^2$

The drive shaft acts like a rod too:

$Jshaft = \frac{1}{2} * 31.7 * 0.032^2 = 0.016 kg*m^2$

SO, the total moment of inertia is:

J = 2*Jwheel + 2*Jwall + 2*Jtread + Jaxle + Jshaft

J = 2*0.243 + 2*0.07 + 2*1.09 + 0.0028 + 0.016 = 2.82 kg*m2

Finally the angular acceleration is:

$\gamma = \frac{0.852 * 153}{2.82} = 46.2 \frac{rad}{s^2}$

4 0

3 years ago