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iris [78.8K]
3 years ago
15

Identify the prefixes used in the International System of

Engineering
1 answer:
lesya [120]3 years ago
3 0

Answer:

i need points 425677

Explanation:

yurrrrrr  awnser C

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iogann1982 [59]
It’s engineering gnzvxhfhbkchufkh
6 0
3 years ago
Read 2 more answers
An uncharged capacitor and a resistor are connected in series to a source of voltage. If the voltage = 7.41 Volts, C = 11.5 µFar
Musya8 [376]

Answer:

a) RC = 1.03 mseg.

b) Qmax = CV = 85.2 μC

c) Q = 53.9 μC

Explanation:

a) In a RC circuit, during the transient period, the capacitor charges exponentially (starting from 0 due to the voltage in the capacitor can´t change instantaneously) with time, being the exponent -t/RC.

This product RC, which defines the rate at which the capacitor charges, is called the time constant of the circuit.

In this case , it can be calculated as follows:

ζ = R C = 89.4 Ω . 11.5 μF = 1.03 mseg.

b) As the charge begins to build up the capacitor plates, a voltage establishes between plates, that opposes to the battery voltage. When this voltage is equal to the battery one, the capacitor reaches to the maximum charge, which is, by definition, as follows:

Q = C V = 11.5 μF . 7.41 V = 85.2 μC

c) During the charging process, the charge increases following this equation:

Q = CV (1 - e⁻t/RC)

When t = RC, the expression for Q is as follows:

Q = CV ( 1- e⁻¹) = 0.63 x CV = 53.9 μC

6 0
4 years ago
(30 pts) A simply supported beam with a span L=20 ft and cross sectional dimensions: b=14 in; h=20 in; d=17.5 in. is reinforced
Nat2105 [25]

Answer:

Zx = 176In³

Explanation:

See attached image file

5 0
3 years ago
A rocket is launched vertically from rest with a constant thrust until the rocket reaches an altitude of 25 m and the thrust end
hjlf

Answer:

a) v=19.6 m/s

b) H=19.58 m

c) v_{f}=29.57 m/s  

Explanation:

a) Let's calculate the work done by the rocket until the thrust ends.

W=F_{tot}h=(F_{thrust}-mg)h=(35-(2*9.81))*25=384.5 J

But we know the work is equal to change of kinetic energy, so:

W=\Delta K=\frac{1}{2}mv^{2}

v=\sqrt{\frac{2W}{m}}=19.6 m/s

b) Here we have a free fall motion, because there is not external forces acting, that is way we can use the free-fall equations.

v_{f}^{2}=v_{i}^{2}-2gh

At the maximum height the velocity is 0, so v(f) = 0.

0=v_{i}^{2}-2gH

H=\frac{19.6^{2}}{2*9.81}=19.58 m  

c) Here we can evaluate the motion equation between the rocket at 25 m from the ground and the instant before the rocket touch the ground.

Using the same equation of part b)

v_{f}^{2}=v_{i}^{2}-2gh

v_{f}=\sqrt{19.6^{2}-(2*9.81*(-25))}=29.57 m/s

The minus sign of 25 means the zero of the reference system is at the pint when the thrust ends.

I hope it helps you!

6 0
3 years ago
Malia is working with aluminum wire. while working with this type of wire, she must remember to a. always use pressure-type term
Y_Kistochka [10]

Answer:

B

Explanation:

Aluminium doesn't rust unless exposed to copper for a duration of time.

4 0
3 years ago
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