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3 years ago

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nata0808 [166]3 years ago

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<h3>Answer</h3>

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On the seashore, pebbles that are rough and uneven become smooth and rounded. Explain how the become smooth and rounded.

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Pebbles usually become smooth and rounded by being polished by water, sand polished by winds, or by tumbling and exposing other sides to erosive effects.

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3 years ago

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Please help, very confused!

PtichkaEL [24]

D. the last choice because the info above tells u so

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In a pool game, the cue ball, which has an initial speed of 3.0 m/s, make an elastic collision with the eight ball, which is ini

zhenek [66]

Explanation:

Given

initial speed(u)=3 m/s

mass of each ball is m

Let the cue ball is moving in x direction initially

In elastic collision Energy and momentum is conserved

Let u be the initial velocity and $v_1 , v_2$ be the final velocity of 8 ball and cue ball respectively

$\frac{mu^2}{2}+0=\frac{mv^2_1}{2}+\frac{mv^2_2}{2}$

The angle after which cue ball is deflected is given by

$\theta _1=90-40=50^{\circ}$

Conserving momentum in x direction

$mu=mv_1cos40+mv_2cos50$

$3=v_1cos40+v_2cos50$

Along Y axis

$0+0=v_1sin40-v_2sin50$

$v_1sin40=v_2sin50$

substitute the value of $v_1$

we get $v_2=1.912 m/s$

$v_1=2.27 m/s$

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3 years ago

Suppose a 96.10 kg hiker has ascended to a height of 1841 m above sea level in the process of climbing Mt. Washington. By what p

lbvjy [14]

Answer:

0.029%

Explanation:

Weight on the surface is given by,

$F_W = mg = (96.1Kg)(9.8m/s^2)=941.78N$

The acceleration due to gravity at given height is,

$g_h=\frac{gR^2}{R+h^2}$

The weight in another height is equal to

$F_h=mg_h$

$F_h= m*\frac{gR^2}{R+h^2}$

$F_h = \frac{F_WR^2}{(R+h)^2}$

$F_h=\frac{(941.78)(6.371*10^6)}{(6.371*10^6+1841)}$

$F_h = 941.50N$

The change of the Weight is,

$\frac{F-F_h}{F}*100=\frac{941.78-941.50}{941.78}*100=0.029\%$

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3 years ago

Multiple-Concept Example 7 and Interactive LearningWare 26.1 provide some helpful background for this problem. The drawing shows

Vlada [557]

Answer:

n = 1.4266

Explanation:

Given that:

refractive index of crystalline slab n = 1.665

let refractive index of fluid is n.

angle of incidence θ₁ = 37.0°

Critical angle $\theta _c = sin^{-1} (\frac{n}{n_{slab}} )$

$sin \theta _ c =\frac{n}{n_{slab}}$

According to Snell's law of refraction:

$n sin \theta _1 = n_{slab} \ sin \ (90- \theta_c)$

At point P ; $90 - \theta _2 \leq \theta _c$

$\theta _2 = 90 - \theta _c$

Therefore:

$n \ sin \theta_1 = n_{slab} \sqrt{(1-sin^2 \theta _c)} \\ \\ n \ sin \theta_1 = n_{slab} \sqrt{(1- \frac{n}{n_{slab}} )}$

Then maximum value of refractive index n of the fluid is:

$n = \frac{n_{slab}}{\sqrt{1+ sin^2 \theta _1 } }$

$n = \frac{1.665}{\sqrt{1+ sin^2 \ 37} }$

n = 1.4266

3 0

3 years ago