Answer:
Explanation:
Given
N0 = 20kg (original substance)
decay constant λ = 0.179/sec
time t = 300s
We are to find N(t)
Using the formula;
n(t) = N0e^-λt
Substitute the given values
N(t) = 20e^-(0.179)(300)
N(t) = 20e^(-53.7)
N(t) = 20(4.7885)
N(t) =143.055
To know how much of the original material that is active, we will find N(t)/N0 = 143.055/20 = 7.152
About 7 times the original material is still radioactive
Your answer is going to be 0 degrees Kelvin!
Attenuation is the correct answer.
Answer:
1. The bird close to the center
2. 4/25 of the original force.
Explanation:
1. Tangential velocity is v=w*d (in m/s), where w is the rotational speed, commonly denoted as the letter omega (in radians per second). d is the distance from the center of the rotating object to the position of where you would like to calculate the velocity (in meters).
As we can note, the furthest from the center we are calculating the velovity the higher it is, because the rotational velocity is not changing but the distance of the object with respect to the center is. If v=w*d, then the lower the d (distance) the lower the tangential velocity.
2. Take a look at the picture:
We have the basic equation for the gravitational force.
We have to forces: Fg1, which is the original force, and Fg2, the force when the mass and the distance changes.
If we consider that mass 2 didn't change (m2'=m2), mass 1 is four times its original (m1'=4*m1) and distance is 5 times the original (r'=5*r), then next step is just plugging it into the equation for Fg2.
Dividing the original force Fg1 by the new force Fg2 (notice you can just as well do the inverse, Fg2 divided by Fg1) gives us the relation between the forces, cancelling all the variables and being left only with a simple fraction!
Answer:
a)0.674 kg b) 2.2 s c) 0.9 m/s²
Explanation:
The amplitude of the ball (xo) = 11.0cm, half way between its equilibrium point its maximum displacement x = 11 cm / 2 = 5.5 cm = 5.5 / 100 in meters = 0.055 meters, speed at this point = 27.2 cm /s = (27.2 / 100) in m/s = 0.272 m/s,
spring constant K = 5.5 N/m
a) The mass of the ball (m) can be calculated using the formula below
v =√ (x²o - x²)K/m
make m subject of the formula
v² = (xo² - x²) K/m
m = K ( xo² - x²) / v²
m = 0.674kg
b) The period of the oscillation can be calculated by the following formula
T = 2π√ (m /K)
substitute the values into the formula
T = 2 × 3.142 × √ (0.674/ 5.5) = 2.2s
c) The maximum acceleration of the ball which occurs at the maximum displacement of the ball can be calculated by the following formula
a = K / m × x ( maximum displacement of the body) = 5.5 / 0.674 × 0.11 = 0.9 m/s²