Two charges q1 = 5 µC, q2 = -26 µC, are L = 19 cm apart. A third charge is to be placed on the line between the two charges. How

V = Volume of gas sample = 1.00 L = 0.001 m³T = temperature of gas = 25.0 °C = 25 + 273 = 298 K P = pressure = 1.00 atm = 101325 Pa n = number of moles of gas using ideal gas law:PV = n RT101325 (0.001) = n (8.314) (298)n = 0.041 n₁ = moles of heliumn₂ = moles of neonm₁ = mass of helium = n₁ (4) = 4 n₁m₂ = mass of neon = n₂ (20.2) = 20.2 n₂given that:m₁ = m₂4 n₁ = 20.2 n₂n₁ = 5.05 n₂also n₁ + n₂ = n5.05 n₂ + n₂ = 0.041n₂ = 0.0068mole fraction of neon is mole fraction = n₂ /n = 0.0068/0.041 = 0.166P₂ = partial pressure of neon =(mole fraction) P P₂ = (0.166) (1)P₂ = 0.166 atm

Answer:

The distance covered by the minutes hand is 39.60 cm.

Explanation:

Note: A clock has a circular shape, where the minutes hand acts as the radius, and its motion creates an arc.

Length of an arc is calculated as ∅/360(2πr)

L = ∅/360(2πr).................... Equation 1π

Here, L represents the arc’s length, ∅ is the angle made by the arc, and r is the arc’s radius.

Given: ∅ = 252°, r = 9 cm, π = 3.143.

Upon substituting these values into equation 1,

L = 252/360(2×3.143×9)

L = 0.7×2×3.143×9

L = 39.60 cm.

Thus, the distance traversed by the minutes hand is 39.60 cm.

Calculating the average speed is straightforward by using the formula involving distance and time:

average speed = distance / time

 

Thus, we have:

average speed = 4875 ft / 6.85 minutes

<span>average speed = 711.68 ft / min</span>

I'm unaware of the term accident?

Response:

The change in momentum experienced by the goose during this encounter is 33.334 m/s

Explanation:

Provided;

goose's mass, m₁ = 2.0 kg

airliner mass, m₂ = 160,000 kg

goose's initial speed, u₁ = 60 km/hr  = 16.667 m/s

airliner initial speed, u₂ = 870 km/hr = 241.667 m/s

The alteration in momentum is defined as;

ΔP = mv - mu

where;

u being the foremost velocity of the bird

v representing the concluding velocity of the bird

Using linear momentum conservation principles;

Total momentum pre-collision = Total momentum post-collision

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the bird and airliner speed after collision;

(2 x 16.667) + (160,000 x 241.667) = v (2 + 160,000)

38,666,753.334 = 160,002v

v = 38,666,753.334 / 160,002

v = 241.664 m/s

Thus, the final velocity of the bird is minimal in comparison to the final velocity of the airliner.

ΔP = mv - mu

ΔP = m(v - u)

ΔP = 2(0 - 16.667)

ΔP = -33.334 m/s

The negative value indicates a slowdown in the bird following the collision.

Thus, the change in momentum of the goose during this interaction is 33.334 m/s