A computer is connected across a 110 V power supply. The computer dissipates 123 W of power in the form of electromagnetic radia

Bernoulli's equation at a point on the streamline is
p/ρ + v²/(2g) = constant
where
p = pressure
v = velocity
ρ = air density, 0.075 lb/ft³ (under standard conditions)
g = 32 ft/s²

Point 1:
p₁ = 2.0 lb/in² = 2*144 = 288 lb/ft²
v₁ = 150 ft/s

Point 2 (stagnation):
The velocity at the stagnation point is zero.

The density stays constant.
Let p₂ denote the pressure at the stagnation location.
Then,
p₂ = ρ(p₁/ρ + v₁²/(2g))
p₂ = (288 lb/ft²) + [(0.075 lb/ft³)*(150 ft/s)²]/[2*(32 ft/s²)
     = 314.37 lb/ft²
     = 314.37/144 = 2.18 lb/in²

Thus, the answer is 2.2 psi

Electromagnetism, which deals with interactions between charged particles, has a considerable range and can create forces that repel like charges while attracting opposites. In contrast, the weak nuclear force is very short-ranged and isn't classified as a force.

Response:

Explanation:

Let T denote the tension.

By employing Newton's second law to analyze the bucket's downward motion, we have:

mg - T = ma

A torque, TR, acts on the drum, inducing an angular acceleration α in it. If I refers to the moment of inertia of the drum, then:

TR = Iα

Rearranging gives: TR = Ia/R

This leads to T =  Ia/R²

Substituting this expression for T back into the previous equation yields:

mg - T = ma

mg - Ia/R² = ma

Consequently, we find that mg =  Ia/R² + ma

Therefore, a (I/R² + m) = mg

This results in: a = mg / (I/R² + m)

Next, we aim to express T as:

mg - T = ma

which simplifies to mg - ma  = T

Rearranging gives mg - m²g / (I/R² + m) = T

Thus, we arrive at: mg - mg / (1 + I / m R²) = T

For part (b), T =  Ia/R²

and for part (c), the moment of inertia of a hollow cylinder calculates to:

I = 1/2  M (R² - (R² / 4))

This simplifies to 3/4 x 1/2 MR², yielding 3/8 MR²

Thus, I / R² = 3/8 M

When we substitute, we find a = mg / (3/8 M + m)

and subsequently T =  Ia/R²

= 3/8 MR² × mg / (3/8 M + m) × 1/R²

Results in: