The complete queston is The amount of a radioactive element A at time t is given by the formula
A(t) = A₀e^kt
Answer: A(t) =N e^( -1.2 X 10^-4t)
Explanation:
Given
Half life = 5730 years.
A(t) =A₀e ^kt
such that
A₀/ 2 =A₀e ^kt
Dividing both sides by A₀
1/2 = e ^kt
1/2 = e ^k(5730)
1/2 = e^5730K
In 1/2 = 5730K
k = 1n1/2 / 5730
k = 1n0.5 / 5730
K= -0.00012 = 1.2 X 10^-4
So that expressing N in terms of t, we have
A(t) =A₀e ^kt
A₀ = N
A(t) =N e^ -1.2 X 10^-4t
Answer:
In a tennis match, the racket exerts the action force on the ball and, as the ball hits it, it exerts an equal and opposite reaction force on the racket. The rocket launches because it pushes on the gas coming out the back end for the action force, while the gas pushes the rocket upward with a reaction force.
It’s very big and very small numbers
Answer:
1. Light is electromagnetic waves while sound is mechanical.
Light wave is transverse and sound is longitudinal.
Explanation:
2.
<i = <r
3.
when distance between source and obstacle and distance between obstacle and screen is finite then the diffraction is fresnal diffraction
when distance between source and obstacle and distance between obstacle and screen is infinite then the diffraction is Fraunhofer's diffraction
We are given the following:
Bobo's swimming speed = 2.0 m/s
Width of the river = 100 m
Flowrate of the river = 6.0 m/s due east
First, we need to illustrate the problem. Draw the river with a width of 100 meters. Then, the flow of the river, east at 6 meters per second. Lastly, draw Bobo at one side of the river facing north and an arrow representing swimming speed at 2 meters per second.
Now, we can use the Pythagorean theorem to solve this rate problem.
c^2 = a^2 + b^2
c = speed of Bobo needed
a = speed of Bobo facing north
b = flow rate of the river going east
c^2 = 2^2 + 6^2
c = 6.32 m / s should be his speed to overcome the current and make a landing at the desired location.
