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Dimas [21]
3 years ago
8

Lasers can be used in in place of a knife.

Physics
1 answer:
Anna11 [10]3 years ago
4 0

Answer: No

Explanation:

Laser can cut many things,

its cutting is so popular largely because of its versatility.  Laser cutterscan cut a wide variety of materials – anything from paper, wood, cork, acrylic and foam to different types of metals. Still, there are some materials that would  seemingly fit well into this list but should never be cut using a laser cutter.

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1. You drop a rock off the top of a building. It takes 6,0 s. hit the ground. How tall is the building?
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The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz. Find the possible range of wavelengths in ai
taurus [48]

Answer:

The possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

Explanation:

Given that,

The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz.

The speed of sound in air is 343 m/s.

To find,

The wavelength range for the corresponding frequency.

Solution,

The speed of sound is given by the following relation as :

v=f_1\lambda_1

Wavelength for f = 45 Hz is,

\lambda_1=\dfrac{v}{f_1}

\lambda_1=\dfrac{343}{45}=7.62\ m

Wavelength for f = 375 Hz is,

\lambda_2=\dfrac{v}{f_2}

\lambda_2=\dfrac{343}{375}=0.914\ m/s

So, the possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

6 0
3 years ago
An ore sample weighs 17.50 N in air. When the sample
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Answer with Explanation:

We are given that

Weight of an ore sample=17.5 N

Tension in the cord=11.2 N

We have to find the total volume and the density of the sample.

We know that

Tension, T=W-F_b

F_b=buoyancy force

T=Tension force

W=Weight

By using the formula

11.2=17.5-F_b

F_b=17.5-11.2=6.3 N

F_b=V_{object}\times \rho_{water}\cdot g

Where

V_{object}=Volume of object

\rho_{water}=1000 kgm^{-3}=Density of water

g=9.8 ms^{-2}=Acceleration due to gravity

Substitute the values then we get

6.3=9.8\times 1000\times V_{object}

V_{object}=\frac{6.3}{9.8\times 1000}=6.43\times 10^{-4} m^3

Volume of sample=6.43\times 10^{-4} m^3

Density of sample,\rho_{object}=\frac{Mass}{volume_{object}}

Where mass of ore sample=1.79 kg

Substitute the values then, we get

\rho_{object}=\frac{1.79}{6.43\times 10^{-4}}=2.78\times 10^3 kg/m^3

Density of the sample=2.78\times 10^{3} kgm^{-3}

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