Physics, 09.04.2020 20:21 goodluck99

A high school physics instructor catches one of his students chewing gum in class. He decides to discipline the student by asking that he stick the gum to a fan and calculate how fast the fan is moving when the gum gets thrown off. The label says that the diameter of the fan is d = 29 cm, and at full speed it turns at a rate of f = 35 rev/s, and that the fan is guaranteed to accelerate uniformly. The fan takes t = 11 s to go from rest to full speed. a. Calculate the maximum the angular velocity of the fan ωmax, in radians per second. b. Surprisingly, the gum seems to remain stuck to the fan at this speed. Calculate the angular acceleration of the gum α, in radians per square second, as the fan is speeding up. sig. gif?tid=7M79-EB-88-49-B44F-20531c. Calculate the tangential component of the acceleration of the gum atan, in meters per square second, as the fan is speeding up. d. What is the magnitude of the centripetal acceleration of the gum arad, in meters per square second, when the fan reaches full speed?e. What is the direction of the centripetal acceleration of the gum, as the fan is turning at top speed?f. Calculate the tangential component of the acceleration of the gum atan, f, in meters per square second, when the fan is at full speed. g. Soon after reaching this speed, the gum becomes un-stuck from the fan blade. Determine the linear speed of the gum v, in meters per second, immediately after it leaves the fan.

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Answer from: Siebee23

a) 219.8 rad/s

b) 20.0 rad/s^2

c) 2.9 m/s^2

d) 7005 m/s^2

e) Towards the axis of rotation

f) 0 m/s^2

g) 31.9 m/s

Explanation:

The angular velocity of an object in rotation is the rate of change of its angular position, so

$\omega=\frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time elapsed

In this problem, we are told that the maximum angular velocity is

$\omega_{max}=35 rev/s$

The angle covered during 1 revolution is

$\theta=2\pi$ rad

Therefore, the maximum angular velocity is:

$\omega_{max}=35 \cdot 2\pi = 219.8 rad/s$

The angular acceleration of an object in rotation is the rate of change of the angular velocity:

$\alpha = \frac{\Delta \omega}{t}$

where

$\Delta \omega$ is the change in angular velocity

t is the time elapsed

Here we have:

$\omega_0 = 0$ is the initial angular velocity

$\omega_{max}=219.8 rad/s$ is the final angular velocity

t = 11 s is the time elapsed

Therefore, the angular acceleration is:

$\alpha = \frac{219.8-0}{11}=20.0 rad/s^2$

For an object in rotation, the acceleration has two components:

- A radial acceleration, called centripetal acceleration, towards the centre of the circle

- A tangential acceleration, tangential to the circle

The tangential acceleration is given by

$a_t = \alpha r$

where

$\alpha$ is the angular acceleration

r is the radius of the circle

Here we have

$\alpha =20.0 rad/s^2$

d = 29 cm is the diameter, so the radius is

r = d/2 = 14.5 cm = 0.145 m

So the tangential acceleration is

a_t=(20.0)(0.145)=2.9 m/s^2

The magnitude of the radial (centripetal) acceleration is given by

$a_c = \omega^2 r$

where

$\omega$ is the angular velocity

r is the radius of the circle

Here we have:

$\omega_{max}=219.8 rad/s$ is the angular velocity when the fan is at full speed

r = 0.145 m is the distance of the gum from the centre of the circle

Therefore, the radial acceleration is

a_c=(219.8)^2(0.145)=7005 m/s^2

The direction of the centripetal acceleration in a rotational motion is always towards the centre of the axis of rotation.

Therefore also in this case, the direction of the centripetal acceleration is towards the axis of rotation of the fan.

The magnitude of the tangential acceleration of the fan at any moment is given by

The tangential acceleration is given by

$a_t = \alpha r$

where

$\alpha$ is the angular acceleration

r is the radius of the circle

When the fan is rotating at full speed, we have:

$\alpha=0$ , since the fan is no longer accelerating, because the angular velocity is no longer changing

r = 0.145 m

Therefore, the tangential acceleration when the fan is at full speed is

a_t=(0)(0.145)=0 m/s^2

The linear speed of an object in rotational motion is related to the angular velocity by the formula:

$v=\omega r$

where

v is the linear speed

$\omega$ is the angular velocity

r is the radius

When the fan is rotating at maximum angular velocity, we have:

$\omega=219.8 rad/s$

r = 0.145 m

Therefore, the linear speed of the gum as it is un-stucked from the fan will be:

v=(219.8)(0.145)=31.9 m/s

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

the answer would be a

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