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andrezito [222]

2 years ago

4

A 72.0-kg person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface of the door. The doorknob is loc

ated 0.800 m from axis of the frictionless hinges of the door. The door begins to rotate with an angular acceleration of 0.52 rad/s2. What is the moment of inertia of the door about the hinges

1 answer:

marissa [1.9K]2 years ago

4 0

Answer:

$7.69kgm^2$

Explanation:

We are given that

Mass,m=72 kg

Force,F=5 N

Distance,r=0.8 m

Angular acceleration, $\alpha=0.52rad/s^2$

We have to find the moment of inertia of the door about hinges.

We know that

Torque, $\tau=Fr=5\times 0.8=4Nm$

Moment of inertia, $I=\frac{\tau}{\alpha}$

Using the formula

$I=\frac{4}{0.52}=7.69Kgm^2$

Hence, the moment of inertia of the door about hinges= $7.69kgm^2$

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Water evaporating from a pond does so as if it were diffusing across an air film 0.15 cm thick. The diffusion coefficient of wat

QveST [7]

Answer:

The water level will drop by about 1.24 cm in 1 day.

Explanation:

Here Mass flux of water vapour is given as

$j_{H_2O}=\frac{D}{l} \bigtriangleup c$

where

$j_{H_2O}$ is the mass flux of the water which is to be calculated.

D is diffusion coefficient which is given as $0.25 cm^2/s$

l is the thickness of the film which is 0.15 cm thick.
$\bigtriangleup c$ is given as

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT}$

In this

$P_{sat}$ is the saturated water pressure, which is look up from the saturated water property at 20°C and 0.5 saturation given as 2.34 Pa

$P_a$ is the air pressure which is given as 0.5 times of $P_{sat}$

R is the universal gas constant as $8.314 kJ/kmol-K$

T is the temperature in Kelvin scale which is $20+273= 293K$

By substituting values in the equation

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT} \\ \bigtriangleup c= \frac{P_{sat}-0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5 \times 2.34}{8.314 \times 293} \\\bigtriangleup c= 0.48 mol/m^3$

Converting $\bigtriangleup c$ into $cm^3/cm^3$

As 1 mole of water 18 $cm^3$ so

$\bigtriangleup c= 0.48 mol/m^3 \\ \bigtriangleup c= 0.48 \times 18 \times 10^{-6} cm^3/cm^3 \\ \bigtriangleup c= 8.64 \times 10^{-6} cm^3/cm^3$

Putting this in the equation of mass flux equation gives

$j_{H_2O}=\frac{D}{l} \bigtriangleup c \\ j_{H_2O}=\frac{0.25}{0.15} \times 8.64 \times 10^{-6} \\ j_{H_2O}=14.4 \times 10^{-6} cm/s$

For calculation of water level drop in a day, converting mass flux as

$j_{H_2O}=14.4 \times 10^{-6} \times 24 \times 3600 cm/day\\ j_{H_2O}=1.24 cm/day$

So the water level will drop by about 1.24 cm in 1 day.

7 0

2 years ago

Rod AB is held in place by the cord AC. Knowing that the tension in the cord is 1350 N and that c 5 360 mm, determine the moment

riadik2000 [5.3K]

Answer:

291.598 N-m

291.6 N-m

Explanation:

Let's first take a look at the free bodily diagrammatic representation.

The first diagram will aid us in answering question (a), so as the second diagram will facilitate effective understanding when solving for question (b).

Let's first determine our angle θ from the diagram

To find angle θ ; we have :

tan θ = $\frac{360+240}{450}$

tan θ = $\frac{600}{450}$

tan θ = 1.333

θ = tan⁻¹ (1.333)

θ = 53.13°

Now, to determine the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point A.

We have:

$M__B}=(Fcos \theta *240)-(Fsin \theta *450)$

where Force(F) = Force in the cord AC = 1350 N and θ = 53.13° ; we have:

$M__B}=(1350&cos 53.13^0 *240)-(1350sin 53.13^0 *450)$

$M__B}= 194400.463-485999.348$

$M__B}=-291598.885 N-mm\\$

$M__B}=-291.598 N-m$

Since the negative sign illustrates just the clockwise movement ; then the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point A = 291.598 N-m

b) From the second diagram, taking the moment at point B $(M__B})$ ,

we have:

$M__B}=-Fcos \theta *360 - Fsin \theta * 0$

$M__B}=-Fcos \theta *360 - 0$

$M__B}=-Fcos \theta *360$

where Force(F) = 1350 N and θ = 53.13° ; we have:

$M__B}= -1350*cos53.13^0*360$

$M__B}= -291600 N-mm$

$M__B}= -291.6 N-m$

Since the negative sign illustrates just the clockwise movement ; then the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point C = 291.6 N-m

6 0

2 years ago

Fred wants to summarize mitosis in the cell cycle. Which statement describes mitosis?

salantis [7]

The statement that most accurately describes mitosis simply is <span>that mitosis is a type of cell division in which one cell (the mother) divides to produce two new cells (the daughters) that are genetically identical to itself</span>. This is the most basic textbook definition, that is summary, of what the mitosis is.

6 0

2 years ago

Read 2 more answers

An example of potential energy is a ball sitting _____ of the stairs.

expeople1 [14]

Answer:

at the top

Explanation:

Potential energy is the stored energy, mechanical energy,

or energy possessed by by virtue of the position of an object.an example of potential energy is the energy that a ball possesses by virtue of its sitting at the top of the stairs it being about to roll down the stairs.

3 0

2 years ago

A circular loop of wire with a radius of 12.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne

Ulleksa [173]

(a) 34 V

The average emf induced in the loop is given by Faraday-Newmann-Lenz law:

$\epsilon = -\frac{\Delta \Phi_B}{\Delta t}$ (1)

where

$\Delta \Phi_B$ is the variation of magnetic flux through the coil

$\Delta t = 2.0 ms = 0.002 s$ is the time interval

We need to find the magnetic flux before and after. The magnetic flux is given by:

$\Phi_B = BA$

where

B is the magnetic field intensity

A is the area of the coil

The radius of the coil is r = 12.0 cm = 0.12 m, so its area is

$A=\pi r^2 = \pi (0.12 m)^2 = 0.045 m^2$

At the beginning, the magnetic field is

$B_i = 1.5 T$

so the flux is

$\Phi_i = B_i A = (1.5 T)(0.045 m^2)=0.068 Wb$

while after the removal of the coil, the magnetic field is zero, so the flux is also zero:

$\Phi_f = 0$

so the variation of magnetic flux is

$\Delta \Phi = 0-0.068 Wb=-0.068 Wb$

And substituting into (1) we find the average emf in the coil

$\epsilon=-\frac{-0.068 Wb}{0.002 s}=34 V$

(b) Counterclockwise

In order to understand the direction of the induced current, we have to keep in mind the negative sign in Lenz's law (1), which tells that the direction of the induced current must be such that the magnetic field produced by this current opposes the variation of magnetic flux in the coil.

In this situation, the magnetic flux through the coil is decreasing, since the coil is removed from the field. So, the induced current must be such that it produces a magnetic field whose direction is the same as the direction of the external magnetic field, which is upward along the positive z-direction.

Looking down from above and using the right-hand rule on the loop (thumb: direction of the current, other fingers wrapped: direction of magnetic field), we see that in order to produce at the center of the coil a magnetic field which is along positive z-direction, the induced current must be counterclockwise.

4 0

2 years ago