A half-life is how long it takes for half of an element's particles to decay.
In 0 half-lives there will be 100% or 4/4 of the element left
After 1 half-life there will be 50% or 2/4 of the element's original particles left
After 2 half lives there will be 25% or 1/4 of the element's original particles left.
Since there is only 1/4 the original amount of the radium-226, we know that 2 half-lives have passed. Simply divide 3200 by 2 to get the time of 1 half-life.
3200/2 = 1,600 years is the half life of radium-226.
I really hope this helps you =)
The momentum of a neutron p = 586.25 kg m / s.
<u>Explanation:</u>
The product of mass and the velocity gives the momentum of an object and it is a vector quantity. It is denoted by the letter p. The unit of momentum is kilogram meter per second (or) kg m / s.
Given mass m = 1.675 10, velocity v = 3.500 10
Momentum, p = mv
where m represents the mass,
v represents the velocity.
momentum p = (1.675 10) (3.500 10)
momentum p = 586.25 kg m / s.
Answer:
Harish went 1.92 Km in 2 minutes.
Explanation:
Harish is running at 16 m/s (Meters per second) and we need to find how far he goes in 2 minutes.
We can use this formula to calulate the distance:
m/s x time
We know that Harish is going 16m/s so let's plug that in:
16 (m/s) x time
We also should know that 2 minutes is 120 seconds - Let's plug that in too:
16 (m/s) x 120 (Seconds)
Now we calculate:
16 x 120 = 1920
Harish went 1,920 meters in 2 minutes.
Even though we have a correct answer, we can still make it easier to read by converting meters into kilometers.
1 Kilometer (Km) is equal to 1,000 meters.
All we need to do now is divide 1,920 by 1,000 and that will gove us our simplified answer of:
1.92 Km
To find the tangent plane to the surface f(x,y,z)=0 at a point (X,Y,Z) we use the following method:
<span>Calculate grad f = (f_x, f_y, f_z). The normal vector to the surface at the point (X,Y,Z) is grad f(X,Y,Z). The equation of a plane with normal vector n which passes through the point p is (r-p).n=0, where r=(x,y,z) is the position vector. So the equation of the tangent plane to the surface through the point (X,Y,Z) is ((x,y,z)-(X,Y,Z)).grad f(X,Y,Z)=0. </span>
<span>Now in your case we have f(x,y,z)=y-x^2-z^2, so grad f=(-2x,1,-2z), and the equation of the tangent plane at the point (X,Y,Z) is </span>
<span>((x,y,z)-(X,Y,Z)).(-2X,1,-2Z)=0, </span>
<span>that is </span>
<span>-2X(x-X)+1(y-Y)-2Z(z-Z)=0, </span>
<span>i.e. </span>
<span>-2Xx+y-2Zz = -2X^2+Y-2Z^2. (1) </span>
<span>Now compare this equation with the plane </span>
<span>x + 2y + 3z = 1. (2) </span>
<span>The two planes a_1x+b_1y+c_1z=d_1, a_2x+b_2y+c_2z=d_2 are parallel when (a_1,b_1,c_1) is a multiple of (a_2,b_2,c_2). So the two planes (1),(2) are parallel when (-2X,1,-2Z) is a multiple of (1,2,3), and we have </span>
<span>(-2X,1,-2Z)=1/2(1,2,3) </span>
<span>for X=-1/4 and Z=-3/4. On the paraboloid the corresponding y coordinate is Y=X^2+Z^2=1^4+9^4=5/2. </span>
<span>So the tangent plane to the given paraboloid at the point (-1/4,5/2,-3/4) is parallel to the given plane.</span>