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3 years ago

13

Smooth surfaces produce a __________________ reflection.

1 answer:

Galina-37 [17]3 years ago

6 0

Answer:

I really hope this is right I think this is Diffuse I'm sorry if its worng

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Is it true that acceleration occurs when there is a change in speed

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Yes it does. But not always

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2 years ago

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An electric dipole of dipole moment 'p' is placed in the position of stable equilibrium in a uniform field of intensity 'E'. The

larisa86 [58]

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Torque by electric dipole = pEcos thita

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3 years ago

An 85-kg man plans to tow a 109 000-kg airplane along a runway by pulling horizontally on a cable attached to it. Suppose that h

Lelu [443]

Answer:

The greatest acceleration the man can give the airplane is 0.0059 m/s².

Explanation:

Given that,

Mass of man = 85 kg

Mass of airplane = 109000 kg

Distance = 9.08

Coefficient of static friction = 0.77

We need to calculate the greatest friction force

Using formula of friction

$F=\mu mg$

Where, m = mass of man

g = acceleration due to gravity

Put the value into the formula

$F = 0.77\times85\times9.8$

$F= 641.41\ N$

We need to calculate the acceleration

Using formula of newton's second law

$F = ma$

$a=\dfrac{F}{m}$

Put the value into the formula

$a=\dfrac{ 641.41}{109000}$

$a=0.0059\ m/s^2$

Hence, The greatest acceleration the man can give the airplane is 0.0059 m/s².

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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

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

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

$h=127.55\ m$

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3 years ago

When a star is moving away the light waves appear to be....

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When a star is moving away from earth it appears blue

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2 years ago