Physics, 11.07.2019 07:30 darlenelandaverde

Suppose an object moves in an elliptical orbit parametrized by x = 4 cos t and y = 6 sin t, where t ≥ 0. use (4) and (5) in section 6.8 in order to find a formula for the velocity v(t) of the object at a given time t. then determine the value of the velocity at the indicated times:

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Answer from: Meiyuh1

The velocity of an object depends on its displacement within a given time interval.

The velocity of the object is $v(t)= \sqrt{16 sin^2t+36cos^2t}$ .

What is the velocity?

Velocity can be defined as the change in the displacement x of an object with respect to a given time interval t.

$v = \dfrac {\Delta x}{\Delta t}$

The displacement of object in x-direction is x = 4 cos(t) and the displacement of object in y-direction is y = 6 sin(t).

The velocity in the x-direction is calculated as given below.

$v_x = \dfrac {dx}{dt}$

$v_x = \dfrac {d}{dt} (4cos (t))$

v_x = - 4sin (t)

The velocity in the y-direction is calculated as given below

$v_y = \dfrac {dy}{dt}$

$v_y = \dfrac {d}{dt} (6 sin (t))$

v_y = 6cos (t)

The magnitude of the net velocity can be calculated as given below.

$v(t) = \sqrt{v_x^2 + v_y^2}$

$v(t) = \sqrt{(-4sin(t))^2 + (6cos(t))^2}$

$v(t)= \sqrt{16 sin^2t+36cos^2t}$

Hence we can conclude that the velocity of the object is $v(t)= \sqrt{16 sin^2t+36cos^2t}$ .

To know more about the velocity, follow the link given below.

link.

Answer from: jadeafrias

As we know that position is given as

x = 4 cost , y = 6 sint

now in order to find the velocity

$\frac{dx}{dt} = v_x$

so here we will have

$v_x = \frac{d}{dt} 4cost = -4sint$

similarly in y direction

$v_y = \frac{dy}{dt}$

$v_y = \frac{d}{dt}(6sint) = 6 cost$

now the net velocity is given as

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{(-4sint)^2+(6cost)^2}$

$v = \sqrt{16sin^2t + 36 cos^2t}$

so above is the velocity as a function of time

Answer from: Quest

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

$_a^z\textrm{x}\rightarrow _{a-2}^{z-4}+_2^4\alpha$

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

$_a^z\textrm{x}+_{-1}^0\textrm{e}\rightarrow _{z-1}^a\textrm{y}$

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases $\gamma -radiations$ . this process does not change the mass number.

$_a^z\textrm{x}^*\rightarrow _a^z\textrm{x}+_0^0\gamma$

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as $\beta ^+$ -decay. in this the mass number remains same.

$_a^z\textrm{x}\rightarrow _{z-1}^a\textrm{y}+_{+1}^0e$

from the above information, we see that only alpha-decay is producing a change in the mass number of the nucleus.

Answer from: Quest

Answerwavelength;

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Suppose an object moves in an elliptical orbit parametrized by x = 4 cos t and y = 6 sin t, where t...

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