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vladimir2022 [97]
3 years ago
12

If a person traveled from Sacramento to Los Angeles by car, their distance would be 384.9 miles. What would their displacement b

e? _____________________ If that same person returned to Sacramento from Los Angeles by the same route in the same car, what would the total round trip distance be? 400 miles What would their total displacement be? _____________________
Physics
1 answer:
kumpel [21]3 years ago
3 0

Answer:

Displacement from Sacramento to Los Angeles is 384.9 miles

Total path length is given as 769.8 miles

Displacement of complete round trip is ZERO

Explanation:

As we know that the displacement is the change in the position from initial to final in a straight line

Here we know that the car travel from Sacramento to Los Angeles

So the straight line distance between two is the displacement

so we have

displacement = 384.9 miles

Now we know that he returns from Los Angeles to Sacramento by same path

So total distance moved by it is same as total path length

d = 384.9 + 384.9

d = 769.8 miles

Also we know that initial and final positions are same so displacement is ZERO for complete round trip.

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Answer:

habituation; adaptation

Explanation:

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How would decreasing the amplitude of the sound waves from a television affect its intensity
Rainbow [258]
<span>The  sound will decrease in volume.</span>
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A 4000-kg car bumps into a stationary 6000kg truck. The Velocity of the car before the collision was +4m/s and -1m/s after the c
Goryan [66]

Answer:

<em>The velocity of the truck is 3.33 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and velocity v is  

P=mv.  

If we have a system of bodies, then the total momentum is the sum of the individual momentums:

P=m_1v_1+m_2v_2+...+m_nv_n

If some collision occurs, the velocities change to v' and the final momentum is:

P'=m_1v'_1+m_2v'_2+...+m_nv'_n

In a system of two masses:

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

There are two objects: The m1=4000 Kg car and the m2=6000 Kg truck. The car was moving initially at v1=4 m/s and the truck was at rest v2=0. After the collision, the car moves at v1'=-1 m/s. We need to find the velocity of the truck v2'. Solving for v2':

\displaystyle v'_2=\frac{m_1v_1+m_2v_2-m_1v'_1}{m_2}

Substituting:

\displaystyle v'_2=\frac{4000*4+6000*0-4000(-1)}{6000}

\displaystyle v'_2=\frac{16000+4000}{6000}

\displaystyle v'_2=3.33

The velocity of the truck is 3.33 m/s

7 0
3 years ago
These has two questions, any help would be much appreciated. Ignore the current answers.
Sati [7]
Bro sorry I don’t know these answer but keeep trying
4 0
3 years ago
A 0.717kg block is attached to a spring with spring constant 19.32N/m. While the block is sitting at rest, a student hits it wit
hjlf

Answer:

15.38 m.

Explanation:

The kinetic energy of the block is equal to potential energy of spring at maximum compression

1/ 2 m V² = 1 /2 K X²

m is mass of block , V is its velocity , K is spring constant and  X is maximum compression or its amplitude.

X = V\times\sqrt{\frac{m}{K} }

Putting the values

x = 79.856\times\sqrt{\frac{.717}{19.32} }

= 15.38 m.

7 0
3 years ago
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