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sukhopar [10]
3 years ago
5

Is "Noble gas" A compound or an element? I'm confused :/

Physics
2 answers:
madam [21]3 years ago
6 0
Compound is a noble gas
vlabodo [156]3 years ago
4 0
Noble gas is a compound.

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A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm
bogdanovich [222]

Answer:

A) 2.75 m/s  B) 0.1911 m    C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = \sqrt\frac{k}{M}

                                                       = \sqrt\frac{21}{0.1}

                                                       = 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic  Motion is given as:

                                x = A sin (\omega t + \phi)\\

Differentiating this to find speed of the block immediately before the collision:

                    v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\

As bullet strikes at equilibrium position so,

                                  φ = 0

                                   t= 2nπ

                             ⇒ cos (ω₀t + φ) = 1

                             ⇒ v= A\omega_{o}

                                       v=(.19)(14.49)\\v= 2.75 ms^{-1}

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

                              x= Bsin(\omega t + \phi)

To find B, consider law of conservation of energy

K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2}  \\P.E = \frac{1}{2} kB^{2}

\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec

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4 years ago
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hodyreva [135]

Answer:

mass number = protons + neutrons

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2 years ago
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Help me plz answering this ​
IceJOKER [234]

Answer:

a Chemical change

b. reversed.

c permanent

d. React

e. Physical change

Explanation:

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4 years ago
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Which tool can be used to separate white light into different colors? *
swat32

Answer:

prisms

Explanation:

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3 years ago
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A car’s bumper is designed to withstand a 4.0-km/h (1.1-m/s) collision with an immovable object without damage to the body of th
poizon [28]

Answer:

The force bumper at 0.200m

F=2722.5 N

Explanation:

Using the energy theorem of work

W=K_{f}- K_{i}

W=ΔK

W=F*d

ΔK=\frac{1}{2}*m*v_{f} ^{2} -\frac{1}{2}*m*v_{i} ^{2}

v_{i} =0

ΔK=F*d=\frac{1}{2}*m*v_{f} ^{2} -\frac{1}{2}*m*v_{i} ^{2}

F*d=\frac{1}{2}*m*0 ^{2}- \frac{1}{2}*m*v_{i} ^{2}\\F*d=-\frac{1}{2}*m*v_{i} ^{2}=\frac{1}{2}*900kg*(1.1\frac{m}{s})^{2}\\F*d=544.5\frac{kg*m}{s^{2}}\\ F=\frac{544.5\frac{kg*m}{s^{2}}}{0.2m} \\F=2722.5 N

3 0
3 years ago
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