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Damm [24]
3 years ago
8

A 2000 kg truck traveling north at 34 km/h turns east and accelerates to 58 km/h. (a) What is the change in the truck's kinetic

energy? (b) What is the magnitude of the change in its momentum?
Physics
1 answer:
barxatty [35]3 years ago
6 0

Explanation:

It is given that,

Mass of the truck, m = 2000 kg

Initial velocity of the truck, u = 34 km/h = 9.44 m/s

Final velocity of the truck, v = 58 km/h = 16.11 m/s

(a) Change in truck's kinetic energy, \Delta E=\dfrac{1}{2}m(v^2-u^2)

\Delta E=\dfrac{1}{2}\times 2000\ kg\times (16.11^2-9.44^2)

\Delta E=170418.5\ J

\Delta E=1.7\times 10^5\ J

(b) Change in momentum of the truck, \Delta p=m(v-u)

\Delta p=2000\ kg\times (16.11-9.44)

\Delta p=13340\ kg-m/s

Hence, this is the required solution.

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When she rides her bike, she gets to her first classroom building 36 minutes faster than when she walks. Of her average walking
postnew [5]

Answer:

d=2.4\ miles

Explanation:

Given:

  • average walking speed, v_w=3\ mph
  • average biking speed, v_b=12\ mph

<u>According to given condition:</u>

t_w=t_b+\frac{36}{60}

where:

t_w= time taken to reach the building by walking

t_b= time taken to reach the building by biking

We know that,

\rm time=\frac{distance}{speed}

so,

\frac{d}{v_w} =\frac{d}{v_b} +\frac{36}{60}

\frac{d}{3}=\frac{d}{12} +\frac{3}{5}

d=2.4\ miles

7 0
3 years ago
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If mechanical energy is conserved in a system the energy at any point in time can be in the form of
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potential, kinetic, elastc energies

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Which statement about the effects of medium on the speed of a mechanical wave is true?
Anastasy [175]

C. A mechanical wave generally travels faster in gases than liquids.

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3 years ago
A 4-lb ball b is traveling around in a circle of radius r1 = 3 ft with a speed (vb)1 = 6 ft&gt;s. if the attached cord is pulled
Leya [2.2K]
Position #1:
radius, r₁ = 3 ft
Tangential speed, v₁ = 6 ft/s

By definition, the angular speed is
ω₁ = v₁/r₁ = (3 ft/s) / (3 ft) = 1 rad/s

Position #2:
Radius, r₂ = 2 ft

By definition, the moment of inertia in positions 1 and 2 are respectively
I₁ = (4 lb)*(3 ft)² = 36 lb-ft²
I₂ = (4 lb)*(2 ft)² = 16 lb-ft²

Because momentum is conserved,
I₁ω₁ = I₂ω₂
Therefore the angular velocity in position 2 is
ω₂ = (I₁/I₂)ω₁
      = (36/16)*1 = 2.25 rad/s
The tangential velocity in position 2 is
v₂ = r₂ω₂ = (2 ft)*(225 rad/s) = 4.5 ft/s

At each position, there is an outward centripetal force.
In position 1, the centripetal force is
F₁ = m*(v²/r₂) = (4)*(6²/3) = 48 lbf
In position 2, the centripetal force is
F₂ = (4)*(4.5²/2) = 40.5 lbf

The radius diminishes at a rate of 2 ft/s.
Therefore the force versus distance curve is as shown below.

The work done is the area under the curve, and it is
W = (1/2)*(48.0+40.5 ft)*(3-2 ft) = 44.25 ft-lb

Answer:  44.25 ft-lb


6 0
3 years ago
I dont know what pfp to put on today so I'm asking u guys....
Mandarinka [93]

Answer:

i think number 2 should be your pfp

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