Physics, 02.12.2019 15:31 zahriamarie10

It is common to see birds of prey rising upward on thermals. the paths they take may be spiral-like. you can model the spiral motion as uniform circular motion combined with a constant upward velocity. assume a bird completes a circle of radius 6.00m every 5.00s and rises vertically at a rate of 3.00m/s.

1. find the speed of the bird relative to the ground.
2. find the magnitude of the bird's acceleration.
3. find the direction of the bird's acceleration.
4. find the angle between the bird's velocity vector and the horizontal.

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Answer from: kmontanab00

1. 8.11 m/s

In order to find the speed of the bird, we need to combine both the horizontal and the vertical component of its velocity.

The horizontal component is equal to the tangential velocity of the circular motion of the bird. We have:

r = 6.00 m radius of the circle

T = 5.00 s period of the circular motion

So the tangential velocity (=horizontal component of the velocity) is

$v_x=\frac{2\pi r}{T}=\frac{2\pi (6.00 m)}{5.00 s}=7.54 m/s$

The vertical component of the velocity is given by the problem

v_y = 3.00 m/s

So the speed of the bird is the magnitude of its velocity:

$v=\sqrt{v_x^2 + v_y^2}=\sqrt{(7.54 m/s)^2+(3.00 m/s)^2}=8.11 m/s$

2. 9.48 m/s^2

The only component of the acceleration of the bird is the one that determine the circular motion in the horizontal plane - so it is the centripetal acceleration.

The centripetal acceleration of the bird is given by:

$a=\frac{v_x^2}{r}$

where

v_x = 7.54 m/s is the tangential velocity of the bird

r = 6.00 m is the radius of the circular trajectory

Substituting,

$a=\frac{(7.54 m/s)^2}{6.00 m}=9.48 m/s^2$

3. Towards the centre of the circle

The bird is moving upward at constant velocity and in a straight line, so there is no component of the acceleration in the vertical direction.

In the horizontal plane, however, the bird is moving in a uniform circular motion: this means that the tangential acceleration is zero, while there is a centripetal acceleration (calculated in the previous part of the exercise). The direction of the centripetal acceleration is always towards the centre of the circular trajectory: so, the acceleration of the bird has exactly the same direction as the centripetal acceleration, i.e. towards the centre of the circle.

4. $21.7^{\circ}$

We said that the components of the velocity of the bird are:

v_x = 7.54 m/s in the horizontal direction

v_y = 3.00 m/s in the vertical direction

So we can find the angle of the velocity with the horizontal by using the equation:

$\theta = tan^{-1} (\frac{v_y}{v_x})$

Substituting the values in, we find

$\theta = tan^{-1} (\frac{3.00 m/s}{7.54 m/s})=21.7^{\circ}$

Answer from: Quest

stands for electromotive force and it is measured in volts. despite its name, it is not actually the force in the sense of being a "pushing" or "pulling" (not measured in newtons).an equation that relates emf, ℰ, to a battery's terminal voltage, v, current, i, and internal resistance, r, is ℰ = v + irin this case, the emf can be interpreted as a voltage—a difference in electric potential.i would not say that it would be considered an energy. if emf can be defined as the energy/work per a unit charge from a battery or other electric source, then it is not strictly energy.

Answer from: Quest

Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge, $v = \dfrac{e_p}{q}$ then the following gives us the potential difference, $\delta v = v_b - v_a$ $\begin{aligned} \delta v & = \dfrac{e_p_b}{q} - \dfrac{e_p_a}{q} \\ & = \dfrac{\delta e_p}{q} \end{aligned}$ the formula $\delta v = \delta e_p / q$ may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get $\delta e_p = q\delta v$ using q = 8.00 × 10⁻¹⁹ c and δv = 2.00 v, we get $\begin{aligned} \delta e_p & = (8.00\times10^{-19} \text{ c})(2.00\text{ v}) \\ & = 1.60\times 10^{-18}\text{ j} \end{aligned}$ the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.

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