Answer:
![125\sqrt[4]{8}](https://tex.z-dn.net/?f=125%5Csqrt%5B4%5D%7B8%7D)
Explanation:
A number of the form

can be re-written in the radical form as follows:
![\sqrt[n]{a^m}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%5Em%7D)
In this problem, we have:
a = 1,250
m = 3
n = 4
So, if we apply the formula, we get
![1,250^{\frac{3}{4}}=\sqrt[4]{(1,250)^3}](https://tex.z-dn.net/?f=1%2C250%5E%7B%5Cfrac%7B3%7D%7B4%7D%7D%3D%5Csqrt%5B4%5D%7B%281%2C250%29%5E3%7D)
Then, we can rewrite 1250 as

So we can rewrite the expression as
![=\sqrt[4]{(2\cdot 5^4)^3}=5^3 \sqrt[4]{2^3}=125\sqrt[4]{8}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B4%5D%7B%282%5Ccdot%205%5E4%29%5E3%7D%3D5%5E3%20%5Csqrt%5B4%5D%7B2%5E3%7D%3D125%5Csqrt%5B4%5D%7B8%7D)
The total amount of energy stays the same, but throughout the ride, the kinetic energy and the potential energy change, still adding up to the same number. At the top of the ride it has potential energy, and as it goes down the potential energy decreases and the kinetic energy increases. When it’s at the bottom of the first drop it has maxed out its kinetic energy, and minimized its potential energy. Friction slows down the car, and pushes on the cart with a force that is equal and opposite to the force being exerted in the track. The reason the track keeps going is because though it exerts and equal and opposite force the momentum of the objects is different, allowing the car to continue moving, however friction will slow it down until eventually it comes to a stop.
Answer:
Sorry to say but where is the photo???
The positive effect of technology being used might be using sethescope or checking BP rate if it is good.
Explanation:
Answer:
The magnitude of the force that each wire exerts on the other will increase by a factor of two.
Explanation:
force on parallel current carrying wire, F = BILsinθ
where;
B is the strength of the magnetic field
L is the length of the wire
I is the magnitude of current on the wire
θ is the angle of inclination of the wire
Assuming B, L and θ is constant, then F ∝ I
F = kI

When the amount of current is doubled in one of the wires, lets say the second wire;

Also, if will double the amount of current on the first wire, then
F₁ = 2F₂
Therefore, the magnitude of the force that each wire exerts on the other will increase by a factor of two.