Physics, 26.07.2019 04:00 IsabelAyshi

Aflywheel with a radius of 0.700 m starts from rest and accelerates with a constant angular acceleration of 0.800 rad/s2 . part a part complete compute the magnitude of the tangential acceleration of a point on its rim at the start. atan = 0.560 m/s2 submitprevious answers correct part b part complete compute the magnitude of the radial acceleration of a point on its rim at the start. arad = 0 m/s2 submitprevious answers correct part c part complete compute the magnitude of the resultant acceleration of a point on its rim at the start. a = 0.560 m/s2 submitprevious answers correct part d compute the magnitude of the tangential acceleration of a point on its rim after it has turned through 60.0 â.

Answers: 1

Show answers

Answers

1) For a point on the rim of the flywheel, the distance from the center of the motion is equal to the radius of the wheel:

.

The tangential acceleration for a point on the rim is given by
$a_t = \alpha r$
where $\alpha=0.800 rad/s^2$ is the angular acceleration while r is the radius. Substituting the numbers, we get
a_t = (0.800 rad/s^2)(0.700 m)=0.560 m/s^2

a_t = (0.800 rad/s^2)(0.700 m)=0.560 m/s^2

2) The radial acceleration for a point on the rim (r=0.700 m) at time t=0 is given by:
$a_r = \frac{v_t^2}{r}$
where v_t

is the tangential velocity at time t=0.

The tangential velocity is given by
$v_t = \omega r$
where $\omega$ is the angular speed; however, at the start of the motion (t=0) the flywheel is at rest, so $\omega=0$ and v_t=0

, so the radial acceleration is a_r = 0

.

3) The magnitude of the acceleration at time t=0 is given by:
$|a| = \sqrt{a_t^2 + a_r^2}= \sqrt{(0.560 m/s^2)^2+(0)^2} =0.560 m/s^2$

4) As we said at point 1), the tangential acceleration is given by
$a_t = \alpha r$
but $\alpha$ , the angular acceleration, is constant, so the tangential acceleration after an angle of $\theta=60.0^{\circ}$ is just equal to the tangential acceleration at the beginning of the motion:
a_t = (0.800 rad/s^2)(0.700 m)=0.560 m/s^2

Answer from: Quest

the results is a decrease in air temperature with an increase of altitude.

explanation:

Answer from: Quest

force is based upon mass and acceleration.the statement is not always true.

it can be explained from newton's second law of motion.

as per newton's second law of motion the rate of change of momentum is directly proportional to the applied force and takes place along the direction of force.

mathematically it can be written as -

$f\alpha \frac{dp}{dt}$

⇒ $f= k \frac{dp}{dt}$

here k is the proportionality constant whose value is 1.

$f=\frac{dp}{dt}$

$f= \frac{d[mv]}{dt}$

$f= m\frac{dv}{dt} +v\frac{dm}{dt}$

$f=ma+v\frac{dm}{dt}$ [a is the acceleration=dv/dt]

hence we see that force not only depends on mass and acceleration but also on velocity and rate of change of mass. force f will depend on mass and acceleration only when mass is constant.

hence the statement that force is based upon mass and acceleration is false.

Another question on Physics

Physics, 22.06.2019 05:30

Anybody wanna ? (picture included? )

Answers: 1

Answer

Physics, 22.06.2019 11:50

The scalar triple product computes the magnitude m of the moment of a force vector f about a specified line. it is m = (r × f) · n, where r is the position vector from the line to the point of application of the force and n is a unit vector in the direction of the line. use matlab to compute the magnitude m for the case where f = [12, −5, 4] n, r = [−3, 5, 2] m, and n = [6, 5, −7].

Answers: 3

Answer

Physics, 22.06.2019 12:50

The heliocentric and the geocentric models of the solar system included these central principles

Answers: 1

Answer

Physics, 22.06.2019 15:10

Auniform crate c with mass mc is being transported to the left by a forklift with a constant speed v1. what is the magnitude of the angular momentum of the crate about point a, that is, the point of contact between the front tire of the forklift and the ground

Answers: 3

Answer

You know the right answer?

Aflywheel with a radius of 0.700 m starts from rest and accelerates with a constant angular accelera...

Questions