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SIZIF [17.4K]
3 years ago
12

Part AFind the x- and y-components of the vector d⃗ = (4.0 km , 29 ∘ left of +y-axis).Express your answer using two significant

figures. Enter the x and y components of the vector separated by a comma.d⃗ = km Part BFind the x- and y-components of the vector v⃗ = (2.0 cm/s , −x-direction).Express your answer using two significant figures. Enter the x and y components of the vector separated by a comma.v⃗ = cm/s Part CFind the x- and y-components of the vector a⃗ = (13 m/s2 , 36 ∘ left of −y-axis).Express your answer using two significant figures. Enter the x and y components of the vector separated by a comma.a⃗ x = m/s2
Physics
1 answer:
Hoochie [10]3 years ago
3 0

Solution :

Part A .

Given : The x and y components of the vector, d = \text{4 km 29} degree left of y-axis.

So the x component is = -4 x sin (29°) = -1.939 km

           y component is = 4 x cos (29°) = 3.498 km

Part B

Given : The x and y components of the vector, \text{v = 2 cm/s} , \text{-x direction}

So the x component is = -2 cm/s

           y component is = 0

Part C

Given : The x and y components of the vector, \text{a = 13 m/s, 36 degree} left of y-axis.

So the x component is = -13 x sin (36°) = -7.6412 m/S^2

           y component is = -13 x cos (36°) = -10.517 m/S^2

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joja [24]

ANSWER:100dB

f<em>rom the  sound intensity level,the sound intensity is calculated as:</em>

<em />\beta<em>₁=</em>\beta<em>=(10dB)㏒₁₀(l₁/l₀)</em>

<em>inserting numbers:</em>

<em>120dB=(10dB)㏒₁₀[l₁/10⁻¹²W/m⁸] or 12=㏒₁₀[l₁/(10⁻¹²)W/m²</em>

<em>Getting the antilog of both sides and obtain 10¹²=l₁(10⁻¹²W/m²)which </em>

<em>can be used to solve for l₁ and get</em>

<em>          l₁=(10⁻¹²W/m²)(10¹²)=1 W/m²</em>

<em>since the sound  intensity is related to the power and that the power does not change,the sound intensity at any other point can be solved.Plugging-in</em>

<em>ᵃ = 4πr²,into P=l₁ₐ₁=l₂ₐ₂ and get:</em>

<em>l₂=l₁(r₁/r₂)² =(1W/m²)(5/35) =2.04×10⁻²W/m²</em>

<em>since we know the sound intensity at the sound point 2r,the sound intensity level at the point can be solved.We have:</em>

<em>     </em>\beta<em>₂=</em>\beta<em>=(10dB)㏒₁₀(l₂/l₀)=(10dB)㏒₁₀(2.04×10⁻²/1×10⁻²)</em>

<em>    </em>\beta<em>₂=(10dB)㏒₁₀(2.04×10¹⁰)</em>

<em />\beta<em> =(10dB)[㏒₁₀(2.04)+㏒₁₀(10¹⁰)]=10dB[0.32+10]=103dB=100dB</em>

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4 years ago
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relation between potential difference and electric field is given as

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so here we know that

d = 3 cm

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the potential difference between them will change but the electric field between them will remain constant

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3 years ago
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4 years ago
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kirza4 [7]
Hi =D
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C, is correct.

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