Physics, 09.01.2020 06:31 bhale2030

Acar accelerates uniformly from rest and reaches a speed of 21.9 m/s in 9.01 s. assume the diameter of a tire is 59.0 cm.(a) find the number of revolutions the tire makes during this motion, assuming that no slipping occurs. rev(b) what is the final angular speed of a tire in revolutions per second? /s

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To solve this problem we will apply the linear motion kinematic equations. There we will find the acceleration, which will allow us to find the total displacement. From the relationship given between the circumference of a circle and the total length, we can determine the number of revolutions.

For the second part we will use the relationship between angular and linear velocities given through the radius of the circumference to calculate the final value.

PART A) Calculate the acceleration of the car

v= v_0+at

Where,

a = Acceleration

v_0 = Initial velocity

v = Final velocity

t = time

Replacing we have that

21.9=0+a*9.01s

a = 2.43m/s^2

Calculate the displacement

$x = v_0t+\frac{1}{2} at^2$

Replacing

$x = 0*9.01 + \frac{1}{2} (2.43)(9.01)$

x = 98.65m

Calculate the circumference of wheel:

$\phi = \pi d$

Where,

d = Diameter

$\phi$ = Circumference

$\phi = \pi (59*10^{-2})$

$\phi = 1.85m^2$

Calculate the number of revolutions

$N = \frac{x}{\phi}$

$N = \frac{98.65m}{1.85m^2}$

N = 53.32rev

Therefore the number of revolutions is 53.32rev

PART B )

Calculate the angular velocity

$v = r\omega$

$\omega = \frac{v}{r} \rightarrow r = \frac{d}{2}$

$\omega = \frac{21.9}{29.5*10^{-2}}$

$\omega = 74.23rad/s$

Therefore the final angular speed is 74.23rad/s

Answer from: Quest

i think it because equal matrices have equal dimensions, only square matrices can be symmetric. for all indices and . every square diagonal matrix is symmetric, since all off-diagonal elements are zero. in linear algebra, a real symmetric matrix represents a self-adjoint operator over a real inner product space.

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.

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Acar accelerates uniformly from rest and reaches a speed of 21.9 m/s in 9.01 s. assume the diameter...

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