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2 years ago

What are close-toed shoes least likely to provide protection against?

Physics

2 answers:

Dennis_Churaev [7]2 years ago

8 0

Hydrogen gas is harmless to your feet so since you don’t need protection against it that seems the best answer.

VLD [36.1K]2 years ago

6 0

Answer: Hydrogen gas produced during a reaction

Explanation:

A chemical lab is a place where potential hazards oriented to the use of the chemicals can happen. These chemicals can cause injuries to the skin when come in contact directly or indirectly. Therefore, protective safety measures should be taken so as to protect oneself while working in a chemical lab.

Among the options given, hydrogen gas produced during a reaction is the correct option. This is because of the fact that close-toed shoes can provide protection against all types of chemical fluids. But the gases like hydrogen gas are reactive and they are permeable inside the shoe material hence, may cause damage.

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Study be comftoable try to be organized a lott and have fun

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2 years ago

1) Air modeled as an ideal gas enters a well-insulated diffuser operating at steady state at 270 K with a velocity of 180 m/s an

GarryVolchara [31]

Answer:

Explanation:

Kinetic energy of a gas molecule = 3/2 k T

1/2 m v² = 3/2 k T

v² ∝ T

(v₁ / v₂)² = T₁ / T₂

( 180 / 48.4 )² = 270 / T₂

13.83 = 270 / T₂

T₂ = 19.52 K .

Exit temp = 19.52 K .

5 0

2 years ago

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postnew [5]

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3 0

2 years ago

Read 2 more answers

Ultrasound with a frequency of 4.257 MHz can be used to produce images of the human body. If the speed of sound in the body is t

shutvik [7]

Answer:

2.49 * 10^(-4) m

Explanation:

Parameters given:

Frequency, f = 4.257 MHz = 4.257 * 10^6 Hz

Speed of sound in the body, v = 1.06 km/ = 1060 m/s

The speed of a wave is given as the product of its wavelength and frequency:

v = λf

Where λ = wavelength

This implies that:

λ = v/f

λ = (1060) / (4.257 * 10^6)

λ = 2.49 * 10^(-4) m

The wavelength of the sound in the body is 2.49 * 10^(-4) m.

4 0

2 years ago

A cat dozes on a stationary merry-go-round, at a radius of 5.5 m from the center of the ride. Then the operator turns on the rid

bixtya [17]

Answer:

$u_{s}=0.56$

Explanation:

For the cat to stay in place on the merry go round without sliding the magnitude of maximum static friction must be equal to magnitude of centripetal force

$F_{s.max}=\frac{mv^2}{r} \\$

Where the r is the radius of merry-go-round and v is the tangential speed

but

$F_{s.max}=u_{s}F_{N}=u_{s}mg$

So we have

$u_{s}mg=\frac{mv^2}{r}\\ u_{s}=\frac{v^2}{gr}\\ Where\\v=\frac{2\pi R}{T} \\So\\u_{s}=\frac{(\frac{2\pi R}{T} )^2}{gr} \\u_{s}=\frac{4\pi^2 r}{gT^2}$

Substitute the given values

$u_{s}=\frac{4\pi^2 5.5m}{(9.8m/s^2)(6.3s)^2} \\u_{s}=0.56$

6 0

2 years ago