If the wavelength (λ) of this wave is 0.012m, what is the speed (v) of this wave on graph A?

Freight car A with a gross weight of 200,000 lbs is moving along the horizontal track in a switching yard at 4 mi/hr. Freight ca

zhenek [66]

Answer: a) 4.7 mi/hr. b) 86,500 lbs. mi²/Hr²

Explanation:

As in any collision, under the assumption that no external forces exist during the very small collision time, momentum must be conserved.

If the collision is fully inelastic, both masses continue coupled each other as a single mass, with a single speed.

So, we can write the following:

p₁ = p₂ ⇒m₁.v₁ + m₂.v₂ = (m₁ + m₂). vf

Replacing by the values, and solving for vf, we get:

vf = (200,000 lbs. 4 mi/hr + 100,000 lbs. 6 mi/hr) / 300,000 lbs = 4.7 mi/hr

If the track is horizontal, this means that thre is no change in gravitational potential energy, so any loss of energy must be kinetic energy.

Before the collision, the total kinetic energy of the system was the following:

K₁ = 1/2 (m₁.v₁² + m₂.v₂²) = 3,400,000 lbs. mi² / hr²

After the collision, total kinetic energy is as follows:

K₂ = 1/2 ((m₁ + m₂) vf²) = 3,313,500 lbs. mi²/hr²

So we have an Energy loss, equal to the difference between initial kinetic energy and final kinetic energy, as follows:

DE = K₁ - K₂ = 86,500 lbs. mi² / hr²

This loss is due to the impact, and is represented by the work done by friction forces (internal) during the impact.

8 0

3 years ago

The water in a river flows uniformly at a constant speed of 2.50 m/s between parallel banks 80.0 m apart. You are to deliver a p

NISA [10]

Answer:

a) The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) 133.33 m

c) 53.13°

d) 106.67 m

Explanation:

a) The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) velocity = distance * time

Let the velocity of the swimmer be $v_{s}$ = 1.5 m/s

The separation of the two sides of the river, d = 80 m

The time taken by the swimmer to get to the other end of the river bank,

$t = \frac{d}{v_{s} }$

t = 80/1.5

t = 53.33 s

The swimmer will be carried downstream by the river through a distance, s

Let the velocity of the river be $v_{r}$ = 2.5 m/s

$S = v_{r} t$

S = 53.33 * 2.5

S = 133.33 m

c) To minimize the distance traveled by the swimmer, his resultant velocity must be perpendicular to the velocity of the swimmer relative to water

That is ,

$cos \theta = \frac{v_{s} }{v_{r} } \\cos \theta = 1.5/2.5\\cos \theta = 0.6\\\theta = cos^{-1} 0.6\\\theta = 53.13^{0}$

d) Downstream velocity of the swimmer, $v_{y} = v_{s} sin \theta\\$

$v_{y} = 1.5 sin 53.13\\v_{y} = 1.2 m/s$

The vertical displacement is given by, $y = v_{y} t$

80 = 1.2 t

t = 80/1.2

t = 66.67 s

the horizontal speed,

$v_{x} = 2.5 - 1.5cos53.13\\v_{x} = 1.6 m/s$

The downstream horizontal distance of the swimmer, $x = v_{x} t$

x = 1.6 * 66.67

x = 106.67 m

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