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Neko [114]
3 years ago
11

Prove : cos(3x)=sin(2x)​

Physics
2 answers:
ahrayia [7]3 years ago
8 0

Explanation:

cos3x=sin2x

cos3x=cos(2π−2x)

⇒3x=2nπ+2π−2x

⇒5x=2nπ+2π

⇒x=52nπ+10π

brainliest plzz

ss7ja [257]3 years ago
6 0

Answer:

this is the answer do it help

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Which example shows an endothermic reaction?
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reactants --> products + thermal energy

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Heat is absorbed in endothermic reaction.

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The nearest star to our planet other than the sun , is 4.4 light year away.One light-year is the distance light travels in a yea
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3 years ago
A puck of mass 0.70 kg approaches a second, identical puck that is stationary on frictionless ice. The initial speed of the movi
natali 33 [55]

Answer:

  • v_1  =  \ 5.196 \frac{m}{s}
  • v_2 =  3 \frac{m}{s}

Explanation:

For this problem, we just need to remember conservation of momentum, as there are no external forces in the horizontal direction:

\vec{p}_i = \vec{p}_f

where the suffix i  means initial, and the suffix f means final.

The initial momentum will be:

\vec{p}_i = m_1 \ \vec{v}_{1_i} + m_2 \ \vec{v}_{2_i}

as the second puck is initially at rest:

\vec{v}_{2_i} = 0

Using the unit vector \vec{i} pointing in the original line of motion:

\vec{v}_{1_i} = 6.0 \frac{m}{s} \hat{i}

\vec{p}_i = 0.70 \ kg  \ 6.0 \frac{m}{s} \ \hat{i} + 0.70 \ kg \ 0

\vec{p}_i = 4.2 \ \frac{kg \ m}{s} \ \hat{i}

So:

\vec{p}_i =  4.2 \ \frac{kg \ m}{s} \ \hat{i} = \vec{p}_f

\vec{p}_f =  4.2 \ \frac{kg \ m}{s} \ \hat{i}

Knowing the magnitude and directions relative to the x axis, we can find Cartesian representation of the vectors using the formula

\ \vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )

So, our velocity vectors will be:

\vec{v}_{1_f} = v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ )

\vec{v}_{2_f} = v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )

We got

\vec{p}_f = 0.7 \ kg \ \vec{v}_{1_f} + 0.7 \ kg \ \vec{v}_{2_f}

4.2 \ \frac{kg \ m}{s} \ \hat{i} = 0.7 \ kg \   v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ )  + 0.7 \ kg \ v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )

So, we got the equations:

4.2 \ \frac{kg \ m}{s}  = 0.7 \ kg \   v_1 \  cos(30 \°) + 0.7 \ kg \ v_2 \  cos(-60 \°)

and

0  = 0.7 \ kg \   v_1 \  sin(30 \°) + 0.7 \ kg \ v_2 \  sin(-60 \°).

From the last one, we get:

0  = 0.7 \ kg \  ( v_1 \  sin(30 \°) +  \ v_2 \  sin(-60 \°) )

0  =  v_1 \  sin(30 \°) +  \ v_2 \  sin(-60 \°)

v_1 \  sin(30 \°) = -  \ v_2 \  sin(-60 \°)

v_1  =  \ v_2 \  \frac{sin(60 \°)}{ sin(30 \°) }

and, for the first one:

4.2 \ \frac{kg \ m}{s}  = 0.7 \ kg  \ (  v_1 \  cos(30 \°) + v_2 \  cos(60 \°) )

\frac{4.2 \ \frac{kg \ m}{s}}{ 0.7 \ kg} =    v_1 \  cos(30 \°) + v_2 \  cos(60 \°)

\frac{4.2 \ \frac{kg \ m}{s}}{ 0.7 \ kg} =    v_1 \  cos(30 \°) + v_2 \  cos(60 \°)

6 \ \frac{m}{s} =    (\ v_2 \  \frac{sin(60 \°)}{ sin(30 \°) } ) \  cos(30 \°) + v_2 \  cos(60 \°)

6 \ \frac{m}{s} = v_2     (\   \frac{sin(60 \°)}{ sin(30 \°) } ) \  cos(30 \°) +   cos(60 \°)

6 \ \frac{m}{s} = v_2  * 2

so:

v_2 = 6 \ \frac{m}{s} / 2 = 3 \frac{m}{s}

and

v_1  =  \ 3 \frac{m}{s}  \  \frac{sin(60 \°)}{ sin(30 \°) }

v_1  =  \ 5.196 \frac{m}{s}

3 0
3 years ago
What is the logical oreder for these words : verse, interlude,verse,refrain, introduction,refrain,coda,refrain,verse
Nikolay [14]

Sorry to say but I know that t(e introduction is first and the coda is last


3 0
3 years ago
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