Raphael refers to a wave by noting its wavelength. lucinda refers to a wave by noting its frequency. which student is correct an

Assuming that the mass of the empty wagon is "M," according to Newton's second law, we can derive the following relationships. Given that the empty wagon accelerates at 1.4 m/s², we proceed with this information. If a child weighing three times the mass of the wagon is on it, we can establish the relevant equations.

assuming north-south is along the Y-axis and east-west along the X-axis

X = total X-displacement

from the graph, total displacement in the X-direction is computed as

X = 0 - 20 + 60 Cos45 + 0

X = 42.42 - 20

X = 22.42 m

Y = total Y-displacement

from the graph, total displacement in the Y-direction is computed as

Y = 40 + 0 + 60 Sin45 + 50

Y = 90 + 42.42

Y = 132.42 m

To calculate the magnitude of the net displacement vector, we apply the Pythagorean theorem, yielding

magnitude: Sqrt(X² + Y²) = Sqrt(22.42² + 132.42²) = 134.31 m

Direction: tan⁻¹(Y/X) = tan⁻¹(132.42/22.42) = 80.4 deg north of east

a)

b) 803.4 N

Explanation:

a) At point C (top of the loop), the pilot experiences weightlessness, leading to the normal force from the seat being zero:

N = 0

Consequently, the force balance equation at position C becomes:

where the left term signifies the pilot's weight and the right term represents the centripetal force, with:

= acceleration due to gravity

= jet's velocity at the top

= loop radius

By solving for v,

Thus, this is the jet's speed at C.

The speed at position A (bottom) can be derived from

The distance traveled by the jet corresponds to half the circumference of the circle with radius r, therefore

Given the plane's deceleration is consistent, we can obtain it using the following equation:

b) The pilot experiences a force equal to the normal force from the seat. At point B (half-way through the loop), we find:

- The normal force from the seat, N, directed towards the center of the loop

- As there are no further forces acting toward the central axis, N must equal the centripetal force:

(1)

where

represents the speed at position B.

To deduce the velocity at B, we note that the distance covered by the jet between positions A and B is a quarter of a circle:

With knowledge of the deceleration, we can implement the equation of motion to find the velocity at the midway point B:

Thus, we then apply eq.(1) to determine the normal force acting on the pilot at B:

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.

Answer:

The partial pressure of H2 is 0.375 atm.

The partial pressure of Ne also stands at 0.375 atm.

Explanation:

Mass of H2 = 1 g

Mass of Ne = 1 g

Mass of Ar = 1 g

Mass of Kr = 1 g

Overall mass of the gas mixture totals 4 g.

Pressure in the sealed container is 1.5 atm.

Calculating the partial pressure for H2 yields: (mass of H2/total mass of gas mixture) × pressure of sealed container = 1/4 × 1.5 = 0.375 atm.

Calculating the partial pressure for Ne similarly gives: (mass of Ne/total mass of gas mixture) × pressure of sealed container = 1/4 × 1.5 = 0.375 atm.