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Oksi-84 [34.3K]
3 years ago
5

A car of mass 2200 kg collides with a truck of mass 4500 kg, and just after the collision the car and truck slide along, stuck t

ogether, with no rotation. The car's velocity just before the collision was <33, 0, 0> m/s, and the truck's velocity just before the collision was <-18, 0, 22> m/s. (a) Your first task is to determine the velocity of the stuck-together car and truck just after the collision. What system and principle should you use
Physics
1 answer:
katovenus [111]3 years ago
7 0

Answer:

37.7m/s: principle of conservation of momentum

Explanation:

The principle to make use of is the principle of conservation of momentum which States that the sum of momentum of bodies before collision is equal to the sum of momentum of bodies after collision. This bodies will move with the same velocity after collision.

Momentum = Mass × velocity

For car of mass 2200kg moving with velocity 33m/s:

Momentum of car before collision = 2200×33

= 72,600kgm/s

For the truck of mass 4500kg;

Momentum = 4500 ×(22-(-18)

= 4500×40

= 180000kgm/s

After collision, their momentum is:

Momentum after collision = (2200+4500)v

= 6700v

Using the principle above to get the common velocity v we have

72600+180000 = 6700v

252600 = 6700v

v = 252600/6700

v = 37.7m/s

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calculate the distance a plane flies on a 7.95 hour flight from Chicago to London. Assume a constant speed of 800.0 km/h
soldier1979 [14.2K]

Answer:

6360 km

Explanation:

Use the kinematics equation x=v_ot+\frac{1}{2}at^2.  We are given t = 7.95 hours and a = 0 m/s^2 (constant speed means there is no acceleration).  Solve for x.

x=(800)(7.95)+\frac{1}{2}(0)(7.95^2)\\x=6360 \ km

4 0
3 years ago
Later, while taking your groceries back to the car, you accidentally let go of your cart. It rolls straight down a grassy slope
VladimirAG [237]

Car is moving on the glassy slope with constant speed

Now we know that

a = \frac{dv}{dt}

so acceleration is rate of change in velocity

as we know that velocity is constant here so acceleration is zero

so here

a = 0

now as we know by Newton's II law

F = ma

since a = 0

F = 0

so net force will be ZERO on it during this motion

6 0
3 years ago
Is cold fusion science or pseudoscience
kondor19780726 [428]

The experiments that claimed to demonstrate cold fusion were found
to have been faulty by others who reviewed them.  Also, nobody else
was able to reproduce the finding in other laboratories.  In the world
of Science, this pretty much says that the initial claims were unfounded.

5 0
3 years ago
A vertical spring has a spring constant of 2900 N/m. The spring is compressed 80 cm and a 8 kg spider is placed on the spring. T
Serga [27]

Answer:

a)  k_{e} = 928 J , b)U = -62.7 J , c) K = 0 , d) Y = 11.0367 m,  e)  v = 15.23 m / s  

Explanation:

To solve this exercise we will use the concepts of mechanical energy.

a) The elastic potential energy is

      k_{e} = ½ k x²

      k_{e} = ½ 2900 0.80²

      k_{e} = 928 J

b) place the origin at the point of the uncompressed spring, the spider's potential energy

     U = m h and

     U = 8 9.8 (-0.80)

     U = -62.7 J

c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also

      K = ½ m v²

      K = 0

d) write the energy at two points, maximum compression and maximum height

     Em₀ = ke = ½ m x²

     E_{mf} = mg y

     Emo = E_{mf}

     ½ k x² = m g y

     y = ½ k x² / m g

     y = ½ 2900 0.8² / (8 9.8)

     y = 11.8367 m

As zero was placed for the spring without stretching the height from that reference is

     Y = y- 0.80

     Y = 11.8367 -0.80

     Y = 11.0367 m

Bonus

Energy for maximum compression and uncompressed spring

     Emo = ½ k x² = 928 J

     E_{mf}= ½ m v²

     Emo = E_{mf}

     Emo = ½ m v²

      v =√ 2Emo / m

     v = √ (2 928/8)

     v = 15.23 m / s

8 0
3 years ago
un astronauta cuya masa es 80kg permanece inmovil en el espacio exterior, al activar la unidad propulsora que lleva en la espald
beks73 [17]

Answer:

Explanation:

80(0) = 2(15) + (80-2)v

v = - 0.38 m/s

5 0
3 years ago
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