It’s voltage can easily be modified and it allows power to be transmitted at a high voltage rate before being lowered to a smaller voltage rate to be safe.
That depends on a few things that you haven't told us about the setup.
So I'm going to assume one of them, and then give you the answer
in terms of another one:
-- Assume a Class-I lever . . . the fulcrum is between the load and the effort.
-- Then the effort needed to lift the load is
(the weight of the load) x (13 / the distance between the fulcrum and the effort)
Answer:
Full range of motion of a human arm.
Explanation:
Excavator arms have three articulations. Just like a shoulder, elbow, and wrist.
12-15 billion years i think
There are 4 questions related to this problem:
1 If the half-life of the drug is 7.3 hours, what fraction of the drug remains in the patient after 24 hours?The amount of the drug is halved every 7.3-hour period, and 24 hours equals 24/7.3 of these halving periods.
So the portion of the drug left over after 24 hours is (1/2) ^ (24/7.3) = 0.10224 2 Write a general expression for the amount of the drug in the patient immediately after taking the nth dose of the drug
One method is to combine the residual amounts from each amount, when the nth dose arises; this will contain adding a finite geometric series
So the total amount of the drug immediately after the nth dose, in mg, is An = 40+ 40(0.10224) + 40(0.10224)^2 + ... + 40(0.10244)^(n-1)
An = 40[1 - (0.010224)^n]/(1 - 0.10224)
3 Write a broad expression for the quantity of the drug in the patient directly before taking the nth dose of the drug
Pn = An – 40
= 40(0.10224) + 40(0.10224)^2 + ... + 40(0.10244)^(n-1)
= 40(0.10224) [1 - (0.10224)^(n-1)]/(1 – 0.10224)
= 4.0895 [1 - (0.10224)^(n-1)]
4 What is the long-term minimum amount of drug in the patient?
= lim n-->infinity of Pn
= lim n-->infinity of 4.0895[1 - (0.10224)^(n-1)]
= 4.0895(1 - 0)
= 4.0895 mg.
