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Galina-37 [17]
1 year ago
13

Grace, Erin, and Tony are on a seesaw. Grace has a mass of 45kg and is seated 0.7m to the left of the fulcrum. Nicole has a mass

of 50kg and is
seated 0.6m to the left of the fulcrum. Tony has a mass of 50kg. How far to the right of the fulcrum must Tony be seated for the seesaw to be in equilibrium? Show work
Physics
1 answer:
salantis [7]1 year ago
5 0
Use Scoratic it works with any time of subject
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The loudness of a sound is inversely proportional to the square of your distance from the source of the sound. if your friend is
kirill [66]
Let loudness be L, distance be d, and k be the constant of variation such that the equation that would best represent the given above is,
                    L = k/(d^2)
For Case 1,
                       L1 = k/(d1^2)
For Case 2,
                       L2 = k/((d1/4)^2)
For k to be equal, L1 = 16L2. 
Therefore, the loudness at your friend's position is 16 times that of yours. 
5 0
2 years ago
Read 2 more answers
A machine is currently set to a feed rate of 5.921 inches per minute (IPM). Te machinist changes this setting to 6.088 IPM. By h
lukranit [14]

Answer:

By 16.7% or 0.167 IPM

Explanation:

Substracting the final IPM (6.088) to the initial IPM (5.921) gives us the net difference, which is how much did it increase in IPM. Multiplying this number by 100 gives us the percentual increase in the feed rate.

4 0
1 year ago
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A circular saw blade with radius 0.175 m starts from rest and turns in a vertical plane with a constant angular acceleration of
ANEK [815]

Answer:

The distance the piece travel in horizontally axis is

L=3.55m

Explanation:

a=2 \frac{rev}{s^{2}} \\h=0.820m\\r = 0.125 m
\\d=150rev

d= 155 rev = 155(2\pi ) = 310\pi rad

a= 2.0 \frac{rev}{s^{2} } = 2.0(2\pi )  = 4.0\pi \frac{rev}{s^{2} }

d=d_{i}+vo*t+\frac{1}{2}*a*t^{2} \\ di=0\\vo=0\\d=\frac{1}{2}*a*t^{2}\\t=\sqrt{\frac{2*d}{a}}\\t=\sqrt{\frac{2*310 rad}{4\frac{rad}{s^{2}}}} \\t=12.449

w=a*t\\w=4\frac{rad}{s^{2}}*12.449s\\ w=49.79 \frac{rad}{s}

Now the angular velocity is the blade speed so:

V=w*r\\V=49.79 \frac{rad}{s}*0.175m\\V=8.7 \frac{m}{s}

assuming no air friction effects affect blade piece:

time for blade piece to fall to floor

t=\sqrt{\frac{2*h}{g}}\\t=\sqrt{\frac{2*0.820m}{9.8\frac{m}{s^{2} } }}\\t=0.409s

Now is the same time the piece travel horizontally

L=t*V\\L=0.409s*8.7\frac{m}{s}\\L=3.55m

blade piece travels  HORIZONTALLY = (24.5)(0.397) = 9.73 m  ANS

6 0
2 years ago
Assume the motions and currents mentioned are along the x axis and fields are in the y direction. (a) does an electric field exe
matrenka [14]
<span> (a) does an electric field exert a force on a stationary charged object? 
Yes. The force exerted by an electric field of intensity E on an object with charge q is
</span>F=qE
<span>As we can see, it doesn't depend on the speed of the object, so this force acts also when the object is stationary.

</span><span>(b) does a magnetic field do so?
No. In fact, the magnetic force exerted by a magnetic field of intensity B on an object with  charge q and speed v is
</span>F=qvB \sin \theta
where \theta is the angle between the direction of v and B.
As we can see, the value of the force F depends on the value of the speed v: if the object is stationary, then v=0, and so the force is zero as well.

<span>(c) does an electric field exert a force on a moving charged object? 
Yes, The intensity of the electric force is still
</span>F=qE
<span>as stated in point (a), and since it does not depend on the speed of the charge, the electric force is still present.

</span><span>(d) does a magnetic field do so?
</span>Yes. As we said in point b, the magnetic force is
F=qvB \sin \theta
And now the object is moving with a certain speed v, so the magnetic force F this time is different from zero.

<span>(e) does an electric field exert a force on a straight current-carrying wire?
Yes. A current in a wire consists of many charges traveling through the wire, and since the electric field always exerts a force on a charge, then the electric field exerts a force on the charges traveling through the wire.

</span><span>(f) does a magnetic field do so? 
Yes. The current in the wire consists of charges that are moving with a certain speed v, and we said that a magnetic field always exerts a force on a moving charge, so the magnetic field is exerting a magnetic force on the charges that are traveling through the wire.

</span><span>(g) does an electric field exert a force on a beam of moving electrons?
Yes. Electrons have an electric charge, and we said that the force exerted by an electric field is
</span>F=qE
<span>So, an electric field always exerts a force on an electric charge, therefore on an electron beam as well.

</span><span>(h) does a magnetic field do so?
Yes, because the electrons in the beam are moving with a certain speed v, so the magnetic force
</span>F=qvB \sin \theta
<span>is different from zero because v is different from zero.</span>
6 0
2 years ago
What is the tangential velocity at the edge of a disk of radius 10cm when it spins with a frequency of 10Hz? Give your answer wi
Nina [5.8K]

Answer:

630cm/s

Explanation:

In simple harmonic motion, the tangential velocity is expressed mathematically as v = ὦr

ὦ is the angular velocity = 2πf

r is the radius of the disk

f is the frequency

Given the radius of disk = 10cm

frequency = 10Hz

v = 2πfr

v = 2π×10×10

v = 200π

v = 628.32 cm/s

The tangential velocity = 630cm/s ( to 2 significant figures)

8 0
2 years ago
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