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Vanyuwa [196]
3 years ago
5

Does anyone know how to solve this?

Physics
2 answers:
kompoz [17]3 years ago
8 0

Answer:

110 m

Explanation:

Draw a free body diagram of the car.  The car has three forces acting on it: normal force up, weight down, and friction to the left.

Sum of the forces in the y direction:

∑F = ma

N − mg = 0

N = mg

Sum of the forces in the x direction:

∑F = ma

-F = ma

-Nμ = ma

Substitute:

-mgμ = ma

-gμ = a

Given μ = 0.40:

a = -(9.8 m/s²) (0.40)

a = -3.92 m/s²

Given that v₀ = 30 m/s and v = 0 m/s:

v² = v₀² + 2aΔx

(0 m/s)² = (30 m/s)² + 2 (-3.9s m/s²) Δx

Δx ≈ 110 m

s2008m [1.1K]3 years ago
7 0
110
Might be wrong but always I helped
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An unknown source plays a pitch of middle C (262 Hz). How fast would the sound wave from this source have to travel to raise
Viefleur [7K]

Answer:

The sound travels at v_{s}=11.4 \mathrm{m} / \mathrm{s}

Option: c

Explanation:

Unknown source plays of middle C (fs) = 262 Hz

The sound wave from this source have to travel to raise the pitch to C sharp is (fd) = 272 Hz

\begin{array}{l}{velocity of sound in air(v)=343 \mathrm{m} / \mathrm{s}, f_{\mathrm{s}}=262 \mathrm{Hz}, f_{\mathrm{d}}=271 \mathrm{Hz}} \\ {velocity of receiver(v_{\mathrm{d}})=0 \mathrm{m} / \mathrm{s},velocity of source( v_{\mathrm{s}}) \text { is unknown }}\end{array}

\text { Speed of sound } \mathrm{V}_{\mathrm{S}}=343 \mathrm{m} / \mathrm{s}

f_{\mathrm{d}}=f_{\mathrm{s}}\left(\frac{v-v_{\mathrm{d}}}{v-v_{\mathrm{s}}}\right)

\frac{f_{d}}{f_{s}}=\left(\frac{v-v_{d}}{v-v_{s}}\right)

\left(v-v_{s}\right)=\frac{f_{s}}{f_{d}}\left(v-v_{d}\right)

v_{s}=v-\frac{f_{s}}{f_{d}}\left(v-v_{d}\right)

Substitute the given values in the formula,

v_{s}=343+\frac{262}{271}(343-0)

v_{s}=343+0.966(343)

v_{s}=343-331.33

v_{s}=11.4 \mathrm{m} / \mathrm{s}

Therefore, The sound travels at v_{s}=11.4 \mathrm{m} / \mathrm{s}

4 0
3 years ago
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