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Lilit [14]
2 years ago
11

At a certain instant, the speedometer of the car indicates 80 km/h.

Physics
1 answer:
Scorpion4ik [409]2 years ago
4 0

Answer:

  • <em>It</em><em> </em><em>tells</em><em> </em><em>the</em><em> </em><em>speed</em><em> </em><em>of</em><em> </em><em>car</em><em>.</em><em> </em><em>At</em><em> </em><em>this</em><em> </em><em>time</em><em> </em><em>the</em><em> </em><em>speed</em><em> </em><em>of</em><em> </em><em>car</em><em> </em><em>is</em><em> </em><em>8</em><em>0</em><em>k</em><em>m</em><em>/</em><em>h</em><em> </em><em>,</em><em> </em><em>means</em><em> </em><em>if</em><em> </em><em>the</em><em> </em><em>car</em><em> </em><em>runs</em><em> </em><em>constantly</em><em> </em><em>at</em><em> </em><em>this</em><em> </em><em>speed</em><em> </em><em>then</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>cover</em><em> </em><em>8</em><em>0</em><em> </em><em>kilometres</em><em> </em><em>in</em><em> </em><em>1</em><em> </em><em>hour</em><em>.</em><em> </em>

<em>.</em><em> </em>

<em>.</em><em> </em>

<em>.</em>

{\bold{\red{HOPE\:IT\:HELPS!}}}

\color{yellow}\boxed{\colorbox{black}{MARK\: BRAINLIEST!❤}}

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A 9.0 kg test rocket is fired vertically from Cape Canaveral. Its fuel gives it a kinetic energy of 1905 J by the time the rocke
mrs_skeptik [129]

Answer:

21.6m

Explanation:

Since the rocket engine burns all the fuel hence the kinetic energy will be converted to potential energy

Potential Energy = mass × acceleration due to gravity × height

Given

PE = 1905J

Mass = 9.0kg

Acceleration due to gravity =9.8m/s²

Required

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Substitute into the formula

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1905 = 88.2h

h =1905/88.3

h = 21.6m

Hence the required height is 21.6m

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To measure the coefficient of kinetic friction by sliding a block down an inclined plane the block must be in equilibrium.
lozanna [386]

Answer:

a)

Explanation:

  • A block sliding down an inclined plane, is subject to two external forces along the slide.
  • One is the component of gravity (the weight) parallel to the incline.
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        F_{gp} = m*g* sin \theta (1)

       (taking as positive the direction of the movement of the block)

  • The other force, is the friction force, that adopts any value needed to meet the Newton's 2nd Law.
  • When θ is so large, than the block moves downward along the incline, the friction force can be expressed as follows:

       F_{f} = \mu_{k} * N  (2)

  • The normal force, adopts the value needed to prevent any vertical movement through the surface of the incline:

       N = m*g* cos \theta (3)

  • In equilibrium, both forces, as defined in (1), (2) and (3) must be equal in magnitude, as follows:

        m*g* sin \theta =  \mu_{k} * m*g* cos \theta

  • As the block is moving, if the net force is 0, according to Newton's 2nd Law, the block must be moving at constant speed.
  • In this condition, the friction coefficient is the kinetic one (μk), which can be calculated as follows:

        \mu_{k}  = tg \theta

8 0
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