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Anni [7]
3 years ago
7

) A stone initially moving at 8.0 m/s on a level surface comes to rest due to friction after it travels 11 m. What is the coeffi

cient of kinetic friction between the stone and the surface
Physics
1 answer:
natali 33 [55]3 years ago
4 0

Answer:

-0.3

Explanation:

F' = μmg ........... Equation 1

Where F' = Frictional force, μ = coefficient of kinetic friction, m = mass of the stone, g = acceleration due to gravity.

But,

F' = ma ............ Equation 2

Where a = acceleration of the stone.

Substitute equation 2 into equation 1

ma = μmg

dividing both side of the equation by m

a = μg

make μ the subject of the equation

μ = a/g............... Equation 3

From the equation of motion,

v² = u²+2as................. Equation 4

Where v and u are the final and the initial velocity respectively, s = distance.

Given: v = 0 m/s (to rest), u = 8.0 m/s, s = 11 m.

Substitute into equation 4

0² = 8² + 2×11×a

22a = -64

a = -64/22

a = -32/11 m/s² = -2.91 m/s²

substitute the values of a and g into equation 3

μ = -2.91/9.8

μ = -0.297

μ ≈ -0.3

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Read 2 more answers
A series circuit has a capacitor of 0.25 × 10−6 F, a resistor of 5 × 103 Ω, and an inductor of 1 H. The initial charge on the ca
svetoff [14.1K]

Answer:

q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Explanation:

Given that L=1H, R=5000\Omega, \ C=0.25\times10^{-6}F, \ \ E(t)=12V, we use Kirchhoff's 2nd Law to determine the sum of voltage drop as:

E(t)=\sum{Voltage \ Drop}\\\\L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\\\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+\frac{1}{0.25\times10^{-6}}q=12\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+4000000q=12\\\\m^2+5000m+4000000=0\\\\(m+4000)(m+1000)=0\\\\m=-4000  \ or \ m=-1000\\\\q_c=c_1e^{-4000t}+c_2e^{-1000t}

#To find the particular solution:

Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Hence the charge at any time, t is q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

6 0
3 years ago
The amplitude of a standing sound wave in a long pipe closed at the left end is sketched below. The vertical axis is the maximum
Rudiy27

Answer:

Check the explanation

Explanation:

A) 7th Harmonic. (Of an open ended pipe, odd harmonics are allowed (3rd overtone))

b) f = n v / 4 L

n = 7

f = 7 x 350 / 4 x 0.41 = 1493.9 Hz

c) Let level of water H, If reduces the effective length of pipe

Using, f = n v / 4 Leff

n = 1

251.8 = 1 x 350 / 4 ( 0.41 - H)

H = 0.0625m

H = 6.25 cm

5 0
3 years ago
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