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Anarel [89]
3 years ago
14

An object moves in a circular path at a constant speed. What is the relationship between the directions of the object's velocity

and acceleration vectors? (a)-The velocity and acceleration vectors point in opposite directions. (b)-The velocity and acceleration vectors are perpendicular. (c)-The velocity vector points in a direction tangent to the circular path. The acceleration is zero.
(d)-The velocity vector points toward the center of the circular path. The acceleration is zero. (e)-The velocity and acceleration vectors point in the same direction.
Physics
1 answer:
Reil [10]3 years ago
3 0

Answer:

b

Explanation:

When an object is in circular motion, the direction of velocity is in the direction of tangent to the circle and acceleration is always directed towards the radial direction. This means that velocity is always perpendicular to acceleration of the object.

Hence option  (b)-The velocity and acceleration vectors are perpendicular is correct.

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kirill115 [55]

Answer:

Efriction = 768.23 [kJ]

Explanation:

In order to solve this problem we must use the principle of energy conservation. Where it tells us that the energy of a system plus the work applied or performed by that system, will be equal to the energy in the final state. We have two states the initial at the time of the balloon jump and the final state when the parachutist lands.

We must identify the types of energy in each state, in the initial state there is only potential energy, since the reference level is in the ground, at the reference point the potential energy is zero. At the time of landing the parachutist will only have potential energy, since it reaches the reference level.

The friction force acts in the opposite direction to the movement, therefore it will have a negative sign.

E_{pot}-E_{friction}=E_{kin}

where:

E_{pot}=m*g*h\\E_{kin}=\frac{1}{2}*m*v^{2}

m = mass = 56 [kg]

h = elevation = 1400 [m]

v = velocity = 5.6 [m/s]

(56*9.81*1400)-E_{friction}=\frac{1}{2}*56*(5.6)^{2}\\769104 -E_{friction}= 878.08 \\E_{friction}=769104-878.08\\E_{friction}=768226[J] = 768.23 [kJ]

4 0
3 years ago
In this electric motor, an electric current causes the coil to rotate, changing
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There is a dropping of a coconut of 2kg mass. the speed of the coconut at the moment before it hits the ground is 40m s-1.
igor_vitrenko [27]
  • Mass=2kg
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We know

\boxed{\sf K.E=\dfrac{1}{2}mv^2}

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\\ \sf\longmapsto K.E=40^2

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