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Contact [7]
3 years ago
11

A car (mass 1200 kg, speed 100 km/h) and a truck (mass 2800 kg, speed 50 km/h) are moving in the same direction along a highway.

If the car slams into the truck from behind and the two vehicles remain locked, what is the speed of the wreck immediately after the collision?
Physics
1 answer:
Sloan [31]3 years ago
5 0

Answer:

Speed of the wreck after the collision is 65 km/h

Explanation:

When a car hits truck and sticks together, the  collision would be totally inelastic.  Since the both the vehicles  locked  together, they have the same final velocity.

Mass of car  m_{1}=1200 kg

Mass of truck m_{2}=2800 kg

Initial speed of the car u_{1}=100 km /h

Initial speed  of the truck u_{2}=50 km /h

The final velocity of the wreck will be

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

since final speed are same, v_{1}=v_{2}=v

m_{1}u_{1}+m_{2}u_{2}=(m_{1}+m_{2})v

1200\times 100 +2800 \times 50 =(1200+2800)v\\v=\frac{260000}{4000} \\v=65 km/h

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Aleks04 [339]

Answer:

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Explanation:

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3 years ago
What are the<br>2 factors that<br>increase the<br>electric force<br>between<br>objects?​
harkovskaia [24]

The two factors that affect the electric force are the charges and the separation between the objects

Explanation:

The magnitude of the electrostatic force between two objects is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where:

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r is the separation between the two objects

Looking at the equation, we see that the electric force between objects depends on two factors:

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Learn more about electric force:

brainly.com/question/8960054

brainly.com/question/4273177

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3 years ago
Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103
son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

PE = -\frac{GMm}{R}

As M=m, then

PE = -\frac{Gm^2}{R}

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

U = -\frac{2Gm^2}{R}

dU = -\frac{2Gm^2}{R}

dKE = -dU

Therefore replacing we have that,

\frac{1}{2}mv^2 =\frac{Gm^2}{2R}

Re-arrange to find v,

v= \sqrt{\frac{Gm}{R}}

v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}

v = 816.7m/s

Therefore the  velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

2r = 2*10^3m

Therefore

dU = Gm^2(\frac{1}{R}-\frac{1}{2r})

v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}

v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}

v = 1.83*10^7m/s

Therefore the velocity when they are about to collide is 1.83*10^7m/s

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