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finlep [7]
2 years ago
14

Name two everyday examples in which stored elastic potential energy is made use of. In each case state the energy transfer which

occurs
Physics
1 answer:
trasher [3.6K]2 years ago
7 0

Answer:

A raised weight.

Water that is behind a dam.

Energy transfer takes place when energy moves from one place to another. Energy can move from one object to another, like when the energy from your moving foot is transferred to a soccer ball, or energy can change from one form to another.

Explanation:

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The \: parts \: of \: the \: model \\ airplane \: are \: in \: the \\ same \: proportions \: as \: the \\ actual \: airplane.

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2 years ago
Two teams are playing tug-of-war. The team on the right is pulling with 4320 N. The team on the left is pulling with 4380 N. Whi
zmey [24]
60 N to the left because 4380-4320is 60
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3 years ago
(b) Cable of a microwave oven has three wires inside it which have insulation of different colours black, green and red. Mention
MakcuM [25]
The green is ground.  Ideally, no current travels in this one.  The red and black are the power and neutral wires but which colors they are depends on a convention.  In the US, you will actually have a black (power) and a white (neutral)  Here it's red and black and usually in a red/black system the red is the power.  Either way there is a potential of 120V rms between them.
8 0
3 years ago
Graham and hunter are circus performers. a cable lifts graham into the air at a constant speed of 1.5 ft/s. when graham’s arms a
siniylev [52]

Answer:

The equations are :

h = 18 + 1.5t

h = 5 + 24t - 16t^2

Explanation:

Speed = distance/time

1.5ft/s = 18ft/t

1.5t=18

The equation to determine height h= 18 + 1.5t

Graham moved another 5ft further with a velocity of 24ft/s

The equation to determine his height will be:

h= 5 + 24t -16t^2

3 0
3 years ago
Read 2 more answers
A 2028 kg Oldsmobile traveling south on Abbott Road at 14.5 m/s is unable to stop on the ice covered intersection for a red ligh
Svetradugi [14.3K]

Answer:

v = 8.1 m/s

θ = -36.4º (36.4º South of East).

Explanation:

  • Assuming no external forces acting during the collision (due to the infinitesimal collision time) total momentum must be conserved.
  • Since momentum is a vector, if we project it along two axes perpendicular each other, like the N-S axis (y-axis, positive aiming to the north) and W-E axis (x-axis, positive aiming to the east), momentum must be conserved for these components also.
  • Since the collision is inelastic, we can write these two equations for the momentum conservation, for the x- and the y-axes:
  • We can go with the x-axis first:

        p_{ox} = p_{fx}  (1)

         ⇒ m_{tr} * v_{tr}= (m_{olds} + m_{tr}) * v_{fx}  (2)

  • Replacing by the givens, we can find vfx as follows:

       v_{fx}  = \frac{m_{tr}*v_{tr} }{(m_{tr} + m_{olds)} }  = \frac{4146kg*9.7m/s}{2028kg+4146 kg} = 6.5 m/s (3)

  • We can repeat the process for the y-axis:

        p_{oy} = p_{fy}  (4)

        ⇒m_{olds} * v_{olds}= (m_{olds} + m_{tr}) * v_{fy}  (5)

  • Replacing by the givens, we can find vfy as follows:

       v_{fy}  = \frac{m_{olds}*v_{olds} }{(m_{tr} + m_{olds)} }  = \frac{2028kg*(-14.5)m/s}{2028kg+4146 kg} = -4.8 m/s (6)

  • The magnitude of the velocity vector of the wreckage immediately after the impact, can be found applying the Pythagorean Theorem to vfx and vfy, as follows:

       v_{f} = \sqrt{v_{fx} ^{2} +v_{fy} ^{2} }} = \sqrt{(6.5m/s)^{2} +(-4.8m/s)^{2}} = 8.1 m/s  (7)

  • In order to get the compass heading, we can apply the definition of tangent, as follows:

       \frac{v_{fy} }{v_{fx} } = tg \theta (8)

      ⇒ tg θ = vfy/vfx = (-4.8m/s) / (6.5m/s) = -0.738 (9)

      ⇒ θ = tg⁻¹ (-0.738) = -36.4º

  • Since it's negative, it's counted clockwise from the positive x-axis, so this means that it's 36.4º South of East.
5 0
2 years ago
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