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Blizzard [7]
3 years ago
5

Which of the following STEM discoverers is known for inventing the circular saws used in sawmills? Babbitt Edison Gates Hawking

Physics
2 answers:
Jet001 [13]3 years ago
5 0
Among the STEM discoverers, the one who is known for the invention of the circular saws that are used in sawmills is Babbitt or better known as Sarah "Tabitha<span>" Babbitt, an American inventor and Shaker tool maker. The answer to this is the first option. Hope this helps.</span>
lara31 [8.8K]3 years ago
5 0

Answer:

Babbitt

Explanation:

Large circular saws are found in sawmills and are used for cutting wood in most applications. In 1813, Shaker's sister, Tabitha Babbitt invented the first circular saw, noted in a sawmill that was working on the spinning house in the Shaker Harvard community in Massachusetts, when she decided to invent an improvement for two-handed saws. men, for the production of lumber. Tabitha Babbitt is also known as the creator of an improved version of clipped nails, a new method of making false teeth, and a better spinning head.

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two charges having the same charge magnitude experiencing an attracting force of 3.60N when the charges are 30cm apart.what is t
Tomtit [17]

The charges have opposite sign and magnitude 6 \mu C

Explanation:

The magnitude of the electrostatic force between two electric charges is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the two charges

r is the separation between the two charges

In this problem, we have:

F = 3.60 N is the force between the two charges

r = 30 cm = 0.30 m is their separation

The two charges have same magnitude, so

q_1 = q_2 = q

So we can rewrite the equation as

F=\frac{kq^2}{r^2}

And solving for q:

q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(3.60)(0.30)^2}{8.99\cdot 10^9}}=6\cdot 10^{-6} C = 6\mu C

Moreover, the force between the charges is attractive: we know that charges of same sign repel each other while charges of opposite sign attract each other, therefore the charges in this problem have opposite sign, so

q_1 = 6 \mu C\\q_2 = -6 \mu C

Learn more about electric force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

3 0
2 years ago
What will change the velocity of a periodic wave?
Gwar [14]
The one that will change the velocity of a periodic wave is : 
B. Changing the medium of the wave
Waves is always determined by the properties of the medium, which means that changing the medium will change the velocity of the wave

hope this helps
7 0
2 years ago
Read 2 more answers
The distance from Earth to the North Star is about 430 light-years. Which of the following statements describes why scientists u
vampirchik [111]

Answer:

D

Explanation:

D

7 0
3 years ago
A pipe that is open at both ends has a fundamental frequency of 320 Hz when the speed of sound in air is 331 m/s.
fenix001 [56]

Question

What is the length of the pipe?

Answer:

(a) 0.52m

(b) f2=640 Hz and f3=960 Hz

(c) 352.9 Hz

Explanation:

For an open pipe,  the velocity is given by

v=\frac {2Lf}{n}

Making L the subject then

L=\frac {nV}{2f}

Where f is the frequency,  L is the length,  n is harmonic number,  v is velocity

Substituting 1 for n,  320 Hz for f and 331 m/s for v then

L=\frac {1*331}{2*320}=0.5171875\approx 0.52m

(b)

The next two harmonics is given by

f2=2fi

f3=3fi

f2=3*320=640 Hz

f3=3*320=960 Hz

Alternatively, f2=2\times \frac {v}{2L} and f3=3\times \frac {v}{2L}

f2=2\times \frac {331}{2*0.52}=636.5 Hz\\f3=3\times \frac {331}{2*0.52}=954.8 Hz

(c)

When v=367 m/s then

f1= \frac {v}{2L}\\f1= \frac {367}{2*0.52}=352.9 Hz

5 0
3 years ago
An X-ray tube emits X-rays with a wavelength of 1.0 x 10-11 m. Calculate the potential that must be applied across the X-ray tub
sergey [27]

Answer:

1.24 x 10 to the 5 ev = 124,000 ev          its B

Explanation:

E = hc/lambda = 1.24 ev-micrometer/1.0x10 to the -5 micrometers = 1.24 x 10 to the 5 ev = 124,000 ev

h = Planck's constant = 6.626 × 10 to the -34 joule·s

c = speed of light = 2.998 × 10 to the 8 m/s

lambda is the given wavelength

E is the desired photon energy

3 0
3 years ago
Read 2 more answers
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