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3 years ago

How does Scientific theories differ from scientific laws

1 answer:

7 0

Hey there!

$\ggg \ scientific \ laws$ : Scientific law's would be a law that was proven and test and examined by scientist.These laws would basically be fact's, proven that what ever they say would technically be true.

$\ggg \ Scientific \ theories$ : These are "theories" that are made by scientist usually hypothesis to see what law would actually be true. These "theories" are of course not true, they are not quite laws, they are experiment's that could be laws, but they're theories, thing's that are technically false.

Hope this helps you!

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A brick is dropped with zero initial speed from the roof of a building and strikes the ground in 1.90 s. How tall is the buildin

irinina [24]

Answer:

17.69 m

Explanation:

The time it takes the brick to strike the ground is 1.90 seconds.

We can apply one of Newton's equation of linear motion to find the height of the building:

$s = ut + 0.5gt^2$

where s = distance (in this case height)

u = initial velocity = 0 m/s

t = time = 1.90 s

g = acceleration due to gravity = 9.8 m/s^2

Therefore:

s = (0 * 1.9) + (0.5 * 9.8 * 1.9 * 1.9)

s = 0 + 17.68

s = 17.69 m

The height of the building is 17.69 m.

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3 years ago

A ball is launched vertically with an initial speed of y˙0= 50 m/s, and its acceleration is governed by y¨=-g-cDy˙2, where the a

stira [4]

Answer:

Explanation:

Given

acceleration is given by

$a=-g-c_Dv^2$

where $\ddot{y}=a$

$\dot{y}=v$

Also acceleration is given by

$a=v\frac{\mathrm{d} v}{\mathrm{d} s}$

$ds=\frac{v}{a}dv$

$\int ds=\int \frac{v}{-g-0.001v^2}dv$

$\Rightarrow Let -g-0.001v^2=t$

$-0.001\times 2vdv=dt$

$vdv=-\frac{dt}{0.002}$

$at\ v_0=50\ m/s,\ t=-g-0.001(50)^2$

$t=-g-2.5$

at $v=0,\ t=-g$

$\int_{0}^{s}ds=\int_{-g}^{-g-2.5}\frac{-dt}{0.002t}$

$\int_{0}^{s}ds=\int^{-g}_{-g-2.5}\frac{dt}{0.002t}$

$s=\frac{1}{0.002}lnt|_{-g}^{-g-2.5}$

$s=\frac{1}{0.002}\ln (\frac{g+2.5}{g})$

$s=113.608\ m$

when air drag is neglected maximum height reached is

$h=\frac{v_0^2}{2g}$

$h=\frac{50^2}{2\times 9.8}$

$h=127.55\ m$

3 0

3 years ago

I need help by today guys plz help

NISA [10]

radio waves bc they have the longest wave lenthgs in a magnetic spectrum

5 0

3 years ago

An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr

Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

$F= mg$

Where, f = restoring force

$kx=mg$

$k=\dfrac{mg}{x}$

Put the value into the formula

$k=\dfrac{2.15\times9.8}{0.0895}$

$k=235.41\ N/m$

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Here, $v = A\omega$

$K.E=\dfrac{1}{2}m\times(A\omega)^2$

Here, $\omega=\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}kA^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2$

$K.E=0.06500\ J$

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

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3 years ago

Pablo's engineering team has developed a new material that is strong enough to withstand major impacts, including bullets, witho

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Answer:

the answer is c

Explanation:

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3 years ago