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kotegsom [21]
3 years ago
9

A toy car is given a quick push so that it rolls up an inclined ramp. After it is released, it rolls up, reaches its highest poi

nt and rolls back down again. Friction is so small it can be ignored. What net force acts on the car?(A) net force of zero(B) Net constant force up the ramp(C) net increasing force up the ramp(D) net increasing force down the ramp(E) net constant force down the ramp(F) net decreasing force down the ramp(G) net decreasing force up the ramp
Physics
1 answer:
kicyunya [14]3 years ago
5 0

Answer: Option (E)

Explanation:

There are only two forces acting on the car: The force of gravity, which point down, and the normal force generated by the interaction of the car and the ramp, which points perpendicular to the ramp.

N=mgcos(\alpha ),

m:mass of the car ; \alpha:angle of the ramp.

g: gravity's acceleration.

N is constante since all parameters are constants.

So, since the force of gravity is also constant, the net force acting on the car is constant. Since velocity decreases over time and the car stars moving down, the force is pointing down the ramp.

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A rock band playing an outdoor concert produces sound at 120 dB 5.0 m away from their single working loudspeaker. What is the so
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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>

<em />

3 0
4 years ago
5
Ronch [10]

Answer:

Current in a parallel circuit = 0.61 amps (Approx)

Explanation:

Given:

Voltage V = 6 volt

Two resistors = 17.2 , 22.4 in parallel circuit

Find:

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Computation:

1/R = 1/r1 + 1 / r2

1/R = 1/17.2 + 1 / 22.4

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Current in a parallel circuit = V / R

Current in a parallel circuit = 6 / 9.73

Current in a parallel circuit = 0.61 amps (Approx)

4 0
3 years ago
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