A single point on a distance time graph tells the

You stand on a frictional platform that is rotating at 1.8 rev/s. Your arms are outstretched, and you hold a heavy weight in eac

dusya [7]

Answer:

20.62361 rad/s

489.81804 J

Explanation:

$I_i$ = Initial moment of inertia = 9.3 kgm²

$I_f$ = Final moment of inertia = 5.1 kgm²

$\omega_i$ = Initial angular speed = 1.8 rev/s

$\omega_f$ = Final angular speed

As the angular momentum of the system is conserved

$I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_f=\dfrac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\dfrac{9.3\times 1.8}{5.1}\\\Rightarrow \omega_f=3.28235\ rev/s=3.28235\times 2\pi=20.62361\ rad/s$

The resulting angular speed of the platform is 20.62361 rad/s

Change in kinetic energy is given by

$\Delta K=\dfrac{1}{2}(I_f\omega_f^2-I_i\omega_i^2)\\\Rightarrow \Delta K=\dfrac{1}{2}(5.1\times (20.62361)^2-9.3\times (1.8\times 2\pi)^2)\\\Rightarrow \Delta K=489.81804\ J$

The change in kinetic energy of the system is 489.81804 J

As the work was done to move the weight in there was an increase in kinetic energy

6 0

3 years ago

is this true or false? an object with a mass of 9.0kg has an acceleration of 3m/s2. the force acting on it is 3N

ludmilkaskok [199]

The answer is "False". The force acting on the object is 27 N.

According to Newton's second law, when a force F acts on am object of mass m, it produces an acceleration a. The force is given by the expression,

$F=ma$

Thus, if the body has a mass of 9.0 kg and if it has an acceleration of 3 m/s², then, on substituting the values in the equation for force,

$F=ma\\ =(9.0kg)(3m/s^2)\\ =27N$

Thus, it can be seen that the force acting on the body is 27 N and not 3 N as is mentioned in the statement. Hence the statement is false.

5 0

3 years ago

What is the speed of sound in air when the temperature is 20°C?

Anarel [89]

The answer to your question is 343 m/s

5 0

3 years ago

Read 2 more answers

Two point charges are fixed on the y axis: a negative point charge q1 = -25 μC at y1 = +0.18 m and a positive point charge q2 at

dedylja [7]

Answer:

$50.91 \mu C$

Explanation:

The magnitude of the net force exerted on q is known, we have the values and positions for $q_{1}$ and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted by $q_{1}$ on q. Then we can know the magnitude of the force exerted by $q_{2}$ about q, finally this will allow us to know the magnitude of $q_{2}$

$q_{1}$ exerts a force on q in +y direction, and $q_{2}$ exerts a force on q in -y direction.

$F_{1}=\frac{kq_{1} q }{d^2}\\F_{1}=\frac{(8.99*10^9)(25*10^{-6}C)(8.4*10^{-6}C)}{(0.18m)^2}=58.26 N\\$

The net force on q is:

$F_{T}=F_{1} - F_{2}\\25N=58.26N-F_{2}\\F_{2}=58.26N-25N=33.26N\\\mid F_{2} \mid=\frac{kq_{2}q}{d^2}$

Rewriting for $q_{2}$ :

$q_{2}=\frac{F_{2}d^2}{kq}\\q_{2}=\frac{33.26N(0.34m)^2}{8.99*10^9\frac{Nm^2}{C^2}(8.4*10^{-6}C)}=50.91*10^{-6}C=50.91 \mu C$

8 0

3 years ago

A rotating space station is said to create "artificial gravity" - a loosely-defined term used for an acceleration that would be

FrozenT [24]

Answer:

$\omega=0.31\frac{rad}{s}$

Explanation:

The artificial gravity generated by the rotating space station is the same centripetal acceleration due to the rotational motion of the station, which is given by:

$a_c=\frac{v^2}{r}(1)$