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Rasek [7]
3 years ago
6

A motor is operating at 3450 RPM and is producing a torque of 32 lb-in. What is the output horsepower of the motor?

Physics
1 answer:
ra1l [238]3 years ago
6 0

Answer:

11561W

Explanation:

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Convert 800 cm to meters.
Rasek [7]

Answer:8 meters

Explanation:100cm is a meter

8 0
2 years ago
Read 2 more answers
The electric potential at the origin of an xy-coordinate system is 40 V. A -8.0-μC charge is brought from x = +∞ to that point.
vredina [299]

Answer:

-320 μJ.

Explanation:

Consider a point with an electrical charge of q. Assume that V is the electrical potential at the position of that charge. The electrical potential of that point charge will be equal to:

\text{Potential Energy} = q \cdot V.

Keep in mind that since both q and V might not be positive, the size of the electrical potential energy might not be positive, either.

For this point charge,

  • q = \rm -8.0\; \mu C; (that's -8.0 microjoules, which equals to \rm -8.0\times 10^{-6}\; J)
  • V = \rm 40\; V.

Hence its electrical potential energy:

\text{Potential Energy} = q\cdot V = \rm (-8.0\; \mu C) \times 40\; V = -320\; \mu J.

Why is this value negative? The electrical potential energy of a charge is equal to the work needed to bring that charge from infinitely far away all the way to its current position. Also, negative charges are attracted towards regions of high electrical potential. Bringing this \rm -8.0\; \mu C negative charge to the origin will not require any external work. Instead, this process will release 320 μJ of energy. As a result, the electrical potential energy is a negative value.

7 0
3 years ago
A 2,000 g quantity of C-14 is left to undergo radioactive decay. The half-life of C-14 is approximately 5,700 years. After going
Katyanochek1 [597]
Start with 2,000 grams.
After 1 half-life, 1,000 grams are left.
After another half-life, 500 are left.
After another half-life, 250 are left.
After another half-life, 125 are left.

That was FOUR half-lifes.

X = 4 .
8 0
3 years ago
Read 2 more answers
Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer radius R2. The electric field at a po
gregori [183]

Answer:

E = 0    r <R₁

Explanation:

If we use Gauss's law

      Ф = ∫ E. dA = q_{int} / ε₀

in this case the charge is distributed throughout the spherical shell and as we are asked for the field for a radius smaller than the radius of the spherical shell, therefore, THERE ARE NO CHARGES INSIDE this surface.

Consequently by Gauss's law the electric field is ZERO

           E = 0    r <R₁

6 0
2 years ago
PLEASE HELP MEE PLEASE I BEG
TEA [102]

Explanation: (I think)

Plug your values into the momentum equation.

So m1= 63kg

m2 = 10 kg

V1 = 12 m/s

And then plug in your values and solve for your unknown (v2)

8 0
2 years ago
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