Answer:
The sound intensity level in the car is 57.2 dB.
Explanation:
Sound intensity level in decibels, β = 10 log (I/I₀); where I = 0.525 × 10⁻⁶ W/m², I₀ = 1.0 × 10⁻¹² W/m²
β (dB) = 10 log ((0.525 × 10⁻⁶)/(1.0 × 10⁻¹²)) = 10 × 5.72 = 57.2 dB
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Answer:
vb = 22.13 m/s
So, the only thing that was measured here was the height of point A relative to point B. And the Law of Conservation of Energy was used.
Explanation:
In order to find the speed of roller coaster at Point B, we will use the law of conservation of Energy. In this situation, the law of conservation of energy states that:
K.E at A + P.E at A = K.E at B + P.E at B
(1/2)mvₐ² + mghₐ = (1/2)m(vb)² + mg(hb)
(1/2)vₙ² + ghₐ = (1/2)(vb)² + g(hb)
where,
vₙ = velocity of roller coaster at point a = 0 m/s
hₙ = height of roller coaster at point a = 25 m
g = 9.8 m/s²
vb = velocity of roller coaster at point B = ?
hb = Height of Point B = 0 m (since, point is the reference point)
Therefore,
(1/2)(0 m/s)² + (9.8 m/s²)(25 m) = (1/2)(vb)² + (9.8 m/s²)(0 m)
245 m²/s² * 2 = vb²
vb = √(490 m²/s²)
<u>vb = 22.13 m/s</u>
<u>So, the only thing that was measured here was the height of point A relative to point B. And the Law of Conservation of Energy was used.</u>
Answer: velocity
Explanation: it's the rate of change of the objects position/ consistent change
The calculated coefficient of kinetic friction is 0.33125.'
The rate of kinetic friction the friction force to normal force ratio experienced by a body moving on a dry, uneven surface is known as k. The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them. Typically, it is represented by the Greek letter mu (). In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.
given mass of the block=10 kg
spring constant k= 2250 Nm
now according to principal of conservation of energy we observe,
the energy possessed by the block initially is reduced by the friction between the points B and C and rest is used up in work done by the spring.
mgh= μ (mgl) +1/2 kx²
10 x 10 x 3= μ(600) +(1125) (0.09)
μ(600) =300 - 101.25
μ = 198.75÷600
μ =0.33125
The complete question is- A 10.0−kg block is released from rest at point A in Fig The track is frictionless except for the portion between point B and C, which has a length of 6.00m the block travels down the track, hits a spring of force constant 2250N/m, and compresses the spring 0.300m form its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between point Band (C)
Learn more about kinetic friction here-
brainly.com/question/13754413
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