When you first pull back on the pendulum, and when you pull it back really high the Potential Energy is high and the Kinetic Energy is low, But when up let go, and it gets right around the middle, that's when the Potential energy transfers to Kinetic, at that point the kinetic Energy is high and the potential Energy is low. But when it comes back up at the end. The same thing will happen, the Potential Energy is high, and the Kinetic Energy is low. Through all of that the Mechanical Energy stays the same.
I hope this helps. :)
Brainliest?
100000 Pascal
Explanation:
pressure= force/area
Max pressure= force/min area
so f=5
min area= 5×10^-5
5÷5*10^-5 = 100000pascal
Part a:
= 56
= 60
= 63
The quartiles are found by finding the medium of the data, and then the mediums of the two different data sets on either side of the medium. The
is the overall medium,
is the medium of the first half, and
is the medium of the second half.
-> How is the medium found? When finding the medium we put the values in order least to greatest and pick the middle value.
[] See attached
Part b:
The range is 7.
The interquartile range is the range of numbers between
and
. In other words, it is 50% of the data, directly in the middle.
This becomes 63 - 56 = 7
Part c:
79 is an outlier.
It is an outlier because it is 1.5 above or below (in this case, above) the interquartile range.
-> 63 + (7 +
) ≤ 79
-> 63 + 10.5 ≤ 79
-> 73.5 ≤ 79
Have a nice day!
I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly.
- Heather
Answer:
The time it takes the proton to return to the horizontal plane is 7.83 X10⁻⁷ s
Explanation:
From Newton's second law, F = mg and also from coulomb's law F= Eq
Dividing both equations by mass;
F/m = Eq/m = mg/m, then
g = Eq/m --------equation 1
Again, in a projectile motion, the time of flight (T) is given as
T = (2usinθ/g) ---------equation 2
Substitute in the value of g into equation 2

Charge of proton = 1.6 X 10⁻¹⁹ C
Mass of proton = 1.67 X 10⁻²⁷ kg
E is given as 400 N/C, u = 3.0 × 10⁴ m/s and θ = 30°
Solving for T;

T = 7.83 X10⁻⁷ s
Answer:
The transverse wave will travel with a speed of 25.5 m/s along the cable.
Explanation:
let T = 2.96×10^4 N be the tension in in the steel cable, ρ = 7860 kg/m^3 is the density of the steel and A = 4.49×10^-3 m^2 be the cross-sectional area of the cable.
then, if V is the volume of the cable:
ρ = m/V
m = ρ×V
but V = A×L , where L is the length of the cable.
m = ρ×(A×L)
m/L = ρ×A
then the speed of the wave in the cable is given by:
v = √(T×L/m)
= √(T/A×ρ)
= √[2.96×10^4/(4.49×10^-3×7860)]
= 25.5 m/s
Therefore, the transverse wave will travel with a speed of 25.5 m/s along the cable.