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klasskru [66]
3 years ago
10

Which of the following statements accurately describe a scientific law?

Physics
1 answer:
Rufina [12.5K]3 years ago
3 0

Answer:

The first statement appears most accurate.

The last statement also appears accurate, but it does not preclude any contradictory evidence.

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Which of the following is an example of conformity?
Phoenix [80]

I think choice B.Malcolm acts, dresses, and speaks like the teens he hangs out with at school is a good example of conformity

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4 years ago
Once carbon begins burning in the core of a high-mass star, the outer layers begin to fall inward, driving up the fusion rates a
sammy [17]

Answer:

Fusion rates and star evolution increase rapidly because of the conversion of hydrogen molecules to helium.

Explanation:

At the early stage of a star mass, hydrogen atoms are produced. After a few billion years, these hydrogen atoms formed at the core of the star mass begin to fuse on the external side of the core. Hydrogen atoms get rapidly fused, resulting to a contraction of the star core, and as more hydrogen atoms get fused, helium atoms are formed.

This action results into rapid into a rapid and violent burning action of hydrogen atoms around the core.

6 0
3 years ago
An Alaskan rescue plane traveling 39 m/s drops a package of emergency rations from a height of 198 m to a stranded party of 3 ex
MAVERICK [17]

Answer:

y = 0m

y0 = 166m

v0y = 0 m/s

g = 9.8 m/s^2

t = ?

Solve for t:

y = y0 + v0y*t - (0.5)gt^2

0 = 166 - (0.5)(9.8)t^2

t = 5.82 s

Now, using time, we can solve for the range using the equation:

x = vx(t)

x = (40)(5.82)

x = 232.8 m

The impact horizontal component of velocity will be 40 m/s as velocity in terms of x is always constant. To find the impact vertical component of velocity, we use the equation:

v = v0y - gt

v = 0 - (9.8)(5.82)

v = -57.04 m/s

4 0
3 years ago
Read 2 more answers
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

5 0
4 years ago
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6 0
3 years ago
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