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3 years ago

This is an example of A) alpha decay B) beta decay C) gamma decay D) half-life decay

Physics

1 answer:

bulgar [2K]3 years ago

5 0

Answer:

The answer is option A.

alpha decay

Hope this helps

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f the CD rotates clockwise at 500 rpmrpm (revolutions per minute) while the last song is playing, and then spins down to zero an

vazorg [7]

Answer:

The magnitude of the angular acceleration is $a = 20.14 rad/s^2$

Explanation:

From the question we are told that

The angular speed of CD is $w_{CD} = 500 rpm = \frac{500 rpm}{\frac{60 \ s }{1 \ min} } * \frac{2 \pi }{ 1 \ rev} = 52.37 rad/s$

time taken to decelerate is $t_{CD} = 2.60\ s$

The final angular speed is $w_f= 0 \ rad/s$

The angular acceleration is mathematically represented as

$a = \frac{w_f - w_{CD}}{t}$

substituting values

$a = \frac{0 - 52.37}{2.60}$

$a = - 20.14 rad/s^2$

The negative sign show that the CD is decelerating but the magnitude is

$a = 20.14 rad/s^2$

3 0

3 years ago

What unit is used for the value of G in Newton's Law of Universal Gravitation? *

Helga [31]

Answer:

In SI units, its value is approximately 6.674×10−11 m3⋅kg−1⋅s−2. The modern notation of Newton's law involving G was introduced in the 1890s by C. V. Boys. The first implicit measurement with an accuracy within about 1% is attributed to Henry Cavendish in a 1798 experiment.

Explanation:

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3 0

2 years ago

To test the performance of its tires, a car

velikii [3]

Answer:

The coefficient of static friction between the tires and the road is 1.987

Explanation:

Given:

Radius of the track, r = 516 m

Tangential Acceleration $a_r$ = 3.89 m/s^2

Speed,v = 32.8 m/s

To Find:

The coefficient of static friction between the tires and the road = ?

Solution:

The radial Acceleration is given by,

$a_{R = \frac{v^2}{r}$

$a_{R = \frac{(32.8)^2}{516}$

$a_{R = \frac{(1075.84)}{516}$

$a_{R = 2.085 m/s^2$

Now the total acceleration is

$\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2$

=> $= \sqrt{ (a_r)^2+(a_R)^2}$

=> $\sqrt{ (3.89 )^2+( 2.085)^2}$

=> $\sqrt{ (15.1321)+(4.347)^2}$

=> $19.4791 m/s^2$

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of static friction is

$\mu =\frac{f}{W}$

From (1) and (2)

$\mu =\frac{ma}{mg}$

$\mu =\frac{a}{g}$

Substituting the values, we get

$\mu =\frac{19.4791}{9.8}$

$\mu =1.987$

8 0

3 years ago

An experiment is performed to test the effect of three different types of water on the growth of a plant. The test is done by us

natima [27]

I think the correct answer would be D. The tap water in the experiment is one the three test conditions of the independent variable, the type of water. The independent variable in a experiment is the one being manipulated or the one being changed. In this case, it is the type of water.

6 0

3 years ago

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PLS HELP :) GIVING BRAINLIEST SIMPLE SCIENCE QUESTION HELPS PLSSSS

raketka [301]

OMG ITS B ITS B I HOPE I HELPED U

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3 years ago