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nikitadnepr [17]
3 years ago
12

An electric saw uses a circular spinning blade to slice through wood. When you start the saw, the motor needs 2.00 seconds of co

nstant angular acceleration to bring the blade to its full angular velocity. If you change the blade so that the rotating portion of the saw now has 3.00 times its original rotational mass, how long will the motor need to bring the blade to its full angular velocity
Physics
1 answer:
liberstina [14]3 years ago
7 0

Answer:

6 seconds

Explanation:

So suppose the new saw has the same mass distribution, only difference in mass density, then the moment of the new saw would be 3 times the original ones.

Suppose both saws are under the same torque, by Newton's 2nd law, we have the following equation of angular acceleration related to torque T and moment of inertia I:

\alpha = \frac{T}{I}

So if I is tripled while T stays the same, α would have been reduced 3 times

So to achieve the full angular velocity:

\omega = \alpha t

Now that angular acceleration is reduced 3 times. The time it takes would have to be 3 times as much, or 2 * 3 = 6 seconds.

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Lightwaves travel from the air into a lens made of glass. Their velocity decreases as they enter the glass. How does this affect
Papessa [141]
The waves become longer but slower
4 0
3 years ago
Calculate the rate of heat conduction through a layer of still air that is 1 mm thick, with an area of 1 m, for a temperature of
max2010maxim [7]

Answer:

The rate of heat conduction through the layer of still air is 517.4 W

Explanation:

Given:

Thickness of the still air layer (L) = 1 mm

Area of the still air = 1 m

Temperature of the still air ( T) = 20°C

Thermal conductivity of still air (K) at 20°C = 25.87mW/mK

Rate of heat conduction (Q) = ?

To determine the rate of heat conduction through the still air, we apply the formula below.

Q =\frac{KA(\delta T)}{L}

Q =\frac{25.87*1*20}{1}

Q = 517.4 W

Therefore, the rate of heat conduction through the layer of still air is 517.4 W

6 0
3 years ago
Read 2 more answers
Can someone check my answers and tell me if their correct?
Otrada [13]

Seven

The magnitude is pointing towards the origin and is at - 20 degrees. The combination makes 160 with the x axis: C answer

Eight

They keep doing this. They use distance where they should use displacement but they use distance to try and fool you. It's a mighty poor practice.

The distance between the start and end points is the displacement. That "distance" is 180*sqrt(25) = 900 . The actual distance should be 180*4 + 180*3 = 720 + 540 = 1260. That's what a car's odometer or a bicycle odometer would read.  the difference is 360.

I really do object to the wording, but what can I do?

Nine

Nine is the same thing as 8.

Displacement = sqrt(400^2 + 80^2)= sqrt(166400) = 408

The actual distance is 400 + 80 = 480

The difference is the answer = 480 - 408 = 72 <<<< Answer

Ten

This is just the displacement magnitude.

dis = sqrt(30^2 + 80^2)

dis = sqrt(900 + 6400)

dis = sqrt(7300)

dis = 85.44 <<<< Answer D

Twelve

Vi =  2.15*Sin(30) = 1.075 m/s

vf = 0

a = - 9.81

t = ?

<u>Formula</u>

a = (vf - vi)/t

<u>Solve</u>

-9.81 =  (0 - 1.075)/t

- 9.81 * t = -1.075

t = 0.11 seconds

Thirteen

I'm leaving this last one to you. You need the initial height xo to answer it properly. Judging by the other questions, this one is right.

Edit

That is a surprise! Really quickly

d = 3.2 m

a = - 9.82

vf = 0

vi = ?

vf^2 = vi^2 - 2*a*d

0 = vi^2 - 2*9.81*3.2

vi = sqrt(19.62*3.2)

vi = 8.0  m/s   But that is the vertical component of the speed

v = vi/sin(25)

v = 8.0/sin(25) = 11


6 0
3 years ago
A particle travels clockwise on a circular path of diameter​ R, monitored by a sensor on the circle at point​ P; the other endpo
kotykmax [81]

We make a graphic of this problem to define the angle.

The angle we can calculate through triangle relation, that is,

sin\theta = \frac{c}{QP}\\sin\theta = \frac{c}{R}\\\theta=sin^{-1}\frac{c}{R}

With this function we should only calculate the derivate in function of c

\frac{d\theta}{dc} = \frac{1}{\sqrt{1-\frac{c^2}{R^2}}}(\frac{c}{R})'\\\frac{d\theta}{dc} = \frac{1}{\sqrt{R^2-c^2}}

That is the rate of change of \theta.

b) At this point we need only make a substitution of 0 for c in the equation previously found.

\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{\sqrt{R^2-0}}\\\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{R}

Hence we have finally the rate of change when c=0.

6 0
3 years ago
John’s mass is 95.6 kg, and Barbara’s is 55.3 kg. He is standing on the x axis at xJ = +10.9 m, while she is standing on the x a
Anna11 [10]

Answer:Shifted towards Left by distance of 2.243 m

Explanation:

Given

Mass of john m_1=95.6 kg

Mass of barbara m_2=55.3 kg

John is standing at x=10.9 m

Barbara is standing at x=2.50 m

x_{com}=\frac{m_1x_1+m_2x_2}{m_1+m_2}

x_{com}=\frac{95.6\times 10.9+55.3\times 2.5}{95.6+55.3}

x_{com}=\frac{1180.29}{150.9}

x_{com}=7.821 m

Now if they change their Position then

x'_{com}=\frac{95.6\times 2.5+55.3\times 10.9}{95.6+55.3}

x'_{com}=\frac{841.77}{150.9}

x'_{com}=5.578

Thus we can see that center of mass shifted towards left by a distance of 2.243 m because heavier is shifted towards left

8 0
3 years ago
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