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gavmur [86]
3 years ago
9

What is the difference between distance and displacement? Give an example of a situation where distance and displacement both ha

ve the same magnitude. Give another example, where distance and displacement have different values. Make sure to address the terms "vector" and "scalar" in your response.
Physics
1 answer:
Oksana_A [137]3 years ago
4 0

Explanation :

Distance is total path travelled by an object during its entire journey. It is a scalar quantity i.e only magnitude.

Displacement is the shortest distance covered by an object. It is basically the change in position of object. It is a vector quantity i.e direction as well as magnitude.

When an object is travelling in a straight line and stops at the end point, then both distance and displacement are same.

When an object is travelling in a straight line and then changes its direction or we can say come backwards then the magnitude of distance and displacement are different.

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A 0.155 kg arrow is shot upward
solniwko [45]

Answer:

2.43J

Explanation:

Given parameters:

Mass of the arrow = 0.155kg

Velocity = 31.4m /s

Unknown:

Kinetic energy when it leaves the bow = ?

Solution:

The kinetic energy of a body is the energy in motion of the body;

 it can be derived using the expression below:

 

   K.E  = \frac{1}{2}  m v²

m is the mass

v is the velocity

 Solve for K.E;

    K.E  =  \frac{1}{2}  x 0.155 x 31.4 = 2.43J

3 0
2 years ago
Read 2 more answers
A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before
Darina [25.2K]

{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{First \: penetrating \: length\:(s_{1}) = 3 \: cm}

\\

{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Left \: Penetration \: length \: before  \: it \: comes \: to \: rest \:( s_{2} )}

\\

{\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Let \: Initial \: velocity   = v\:m/s} \\\\

\:\:\:\:\bullet\:\:\:\sf{Left \: velocity \: after \:  s_{1} \: penetration =  \dfrac{v}{2}  \:m/s} \\\\

\:\:\:\:\bullet\:\:\:\sf{s_{1} =  \dfrac{3}{100}  = 0.03 \: m}

\\

☯ As we know that,

\\

\dashrightarrow\:\: \sf{ {v}^{2}  =  {u}^{2} + 2as }

\\

\dashrightarrow\:\: \sf{  \bigg(\dfrac{v}{2} \bigg)^{2}  =  {v}^{2}   + 2a s_{1}}

\\

\dashrightarrow\:\: \sf{  \dfrac{ {v}^{2} }{4}  =  {v}^{2}  + 2 \times a \times 0.03  }

\\

\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4}  -  {v}^{2}  = 0.06 \times a  }

\\

\dashrightarrow\:\: \sf{\dfrac{ -  3{v}^{2} }{4}  = 0.06 \times a  }

\\

\dashrightarrow\:\: \sf{a =  \dfrac{ - 3 {v}^{2} }{4 \times 0.06}  }

\\

\dashrightarrow\:\: \sf{ a =  \dfrac{ - 25 {v}^{2} }{2}\:m/s^{2} ......(1) }

\\

\:\:\:\:\bullet\:\:\:\sf{  Initial\:velocity=v\:m/s} \\\\

\:\:\:\:\bullet\:\:\:\sf{ Final \: velocity = 0 \: m/s }

\\

\dashrightarrow\:\: \sf{  {v}^{2}  =  {u}^{2}  + 2as}

\\

\dashrightarrow\:\: \sf{{0}^{2}  =  {v}^{2}  + 2 \times  \dfrac{ - 25 {v}^{2} }{2}  \times s  }

\\

\dashrightarrow\:\: \sf{ -  {v}^{2}  =  - 25 {v}^{2}  \times s  }

\\

\dashrightarrow\:\: \sf{  s =  \dfrac{ -  {v}^{2} }{ - 25 {v}^{2} }}

\\

\dashrightarrow\:\: \sf{  s =  \dfrac{1}{25} }

\\

\dashrightarrow\:\: \sf{ s = 0.04 \: m }

\\

☯ For left penetration (s₂)

\\

\dashrightarrow\:\: \sf{s =  s_{1} +  s_{2}  }

\\

\dashrightarrow\:\: \sf{  0.04 = 0.03 +  s_{2}}

\\

\dashrightarrow\:\: \sf{ s_{2} = 0.04 - 0.03 }

\\

\dashrightarrow\:\: \sf{s_{2} = 0.01 \: m = {\boxed{\sf{\purple{1 \: cm }}} }}

\\

\star\:\sf{Left \: penetration \: before  \: it \: come \: to \: rest \: is \:{\bf{ 1 \: cm}}} \\

4 0
2 years ago
How does a Freebody diagram tell you about the net force an object?
Sloan [31]
So you subtract the numbers that are on the same axis. So if your gravitational force is 10 and your normal force is 5 you do 5-10 to get -5 since gravity acts downward
6 0
3 years ago
Sound is an example of a ______ wave.
tester [92]
Sound is a longitudinal wave.
7 0
3 years ago
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Express the distance between the moon and earth in meters with a meter prefix (km).
tankabanditka [31]

Answer;

= 3.86 × 10^8 Meters

Explanation;

-The distance between the Earth and the moon is 386000 km

But; 1 km = 1000 m

Therefore; 386000 km will be equivalent to;

= 386000 × 1000

= 386000000 m

= 3.86 × 10^8 meters

3 0
2 years ago
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