Via the half-life equation:
Where the time elapse is 11,460 year and the half-life is 5,730 years.
Therefore after 11,460 years the amount of carbon-14 is one fourth (1/4) of the original amount.
Answer:
B. He should change the lengths of the vectors that point tangent to the circle so that each is the same length.
Explanation:
A uniform circular motion is a motion in a circle where the tangential speed of the object is constant.
In the motion map:
- The arrows pointing towards the centre of the circle represent the centripetal acceleration, and their length represent the magnitude of the acceleration
- The arrows pointing tangential to the circle represent the tangential speed, and their length represent the magnitude of the speed
In this motion map, we see that the length of the vectors pointing tangent to the circle is not constant: this means that the speed is not constant. In order to have a uniform circular motion, the speed must be constant, therefore the lengths of the vectors that point tangent to the circle must be the same.
Answer:
a) 4.9*10^-6
b) 5.71*10^-15
Explanation:
Given
current, I = 3.8*10^-10A
Diameter, D = 2.5mm
n = 8.49*10^28
The equation for current density and speed drift is
J = I/A = (ne) Vd
A = πD²/4
A = π*0.0025²/4
A = π*6.25*10^-6/4
A = 4.9*10^-6
Now,
J = I/A
J = 3.8*10^-10/4.9*10^-6
J = 7.76*10^-5
Electron drift speed is
J = (ne) Vd
Vd = J/(ne)
Vd = 7.76*10^-5/(8.49*10^28)*(1.60*10^-19)
Vd = 7.76*10^-5/1.3584*10^10
Vd = 5.71*10^-15
Therefore, the current density and speed drift are 4.9*10^-6
And 5.71*10^-15 respectively
True! The mechanical advantage of the wheel and axle is equal to the ratio of the radius of the wheel over the radius of the axle.
Answer:
1 greater distances fallen in successive seconds
Explanation:
When a body falls freely it is subjected to the action of the force of gravity, which gives an acceleration of 9.8 m / s2, consequently, we are in an accelerated movement
If we use the kinematic formula we can find the position of the body
Y = Vo t + ½ to t2
Where the initial velocity is zero or constant and the acceleration is the acceleration of gravity
Y = - ½ g t2 = - ½ 9.8 t2 = -4.9 t2
Let's look for the position for successive times
t (s) Y (m)
1 -4.9
2 -19.6
3 -43.2
The sign indicates that the positive sense is up
It can be clearly seen that the distance is greatly increased every second that passes
