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grigory [225]
3 years ago
11

8. A car starts from rest and accelerates at 5.5ms in an easterly direction.

Physics
2 answers:
snow_lady [41]3 years ago
8 0
  • Initial velocity=0m/s=u
  • Acceleration=a=5.5m/s^2

#8.1

  • t=12s

\\ \sf\longmapsto v=u+at

\\ \sf\longmapsto v=0+5.5(12)

\\ \sf\longmapsto v=66m/s

#8.2

  • u=0m/s
  • v=18m/s
  • a=5.5m/s^2

Use third equation of kinematics

\\ \sf\longmapsto v^2-u^2=2as

\\ \sf\longmapsto s=\dfrac{v^2-u^2}{2a}

\\ \sf\longmapsto s=\dfrac{18^2-0^2}{2(5.5)}

\\ \sf\longmapsto s=\dfrac{324}{11}

\\ \sf\longmapsto s=29.4m

wel3 years ago
4 0

Answer:

<u>Question</u><u> </u><u>8</u><u>.</u><u>1</u>

• from first newtons equation of motion:

\dashrightarrow \: { \tt{v = u + at}}

• u → 0

• a → 5.5 m/s²

• t → 12 s

\dashrightarrow \:  { \tt{v = 0 + (5.5 \times 12)}} \\  \\ { \boxed{ \tt{v = 66 \:  {ms}^{ - 1} }}}

<u>Question</u><u> </u><u>8</u><u>.</u><u>2</u>

• from third newton equation of motion:

{ \tt{ {v}^{2} =  {u}^{2}   + 2as }}

• s → displacement

• u → 0

• a → 5.5 m/s²

• v → 18 m/s

\dashrightarrow \: { \tt{ {18}^{2} =  {0}^{2}  + (2 \times 5.5 \times s) }} \\  \\ { \tt{324 = 11s}} \\  \\ { \boxed{ \tt{displacement = 29.5 \: m}}}

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Weight, on the other hand is the downward force you exert on the ground. Weight is calculated by multiplying the mass by the gravitational field strength and changes in different places with different gravitational strength. E.g. The moon's gravitational strength is 1/5 of Earth's so the mass of the object would stay the same but the weight would be only 20% of the weight is had on earth.

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Draw a schematic diagram of a circuit consisting of a battery of five 2 V cells, a 5 ohm resistor, a 10 ohm resistor, a 15 ohm r
MakcuM [25]

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Explanation:

  • For diagram refer the attachment.

It is given that five cells of 2V are connected in series, so total voltage of the battery:

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Three resistor of 5\Omega, 10\Omega, 15\Omega are connected in Series, so the net resistance:

\dashrightarrow \: \: \sf R_{n} = R_{1} + R_{2} + R_{3}

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\dashrightarrow  \sf\: \: V = IR

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