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makkiz [27]
3 years ago
7

When Emma pushes a bag with a force of 27 newtons, the coefficient of kinetic friction between the bag and the floor is 0.23. Wh

at is the normal force acting on the bag by the floor?. A. 0.085 newtons B. 27 newtons C. 2.7 × 10^2 newtons . D. 1.2 × 10^2 newtons
Physics
2 answers:
vichka [17]3 years ago
6 0
<span>The normal force acting on the bag by the floor is calculated using th data on the force used to push the bag and the coefficient of friction between the bag and the floor. The normal force is determined through the formula Force / kin.coeff. Substituting, 27 N / 0.23 is equal to 117. 39 N. The answer is D.</span>
klemol [59]3 years ago
5 0

Answer:

the answer is D, hope this helps

Explanation:

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A 1.30-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v
atroni [7]

Answer:

1. t = 0.0819s

2. W = 0.25N

3. n = 36

4. y(x , t)= Acos[172x + 2730t]

Explanation:

1) The given equation is

y(x, t) = Acos(kx -wt)

The relationship between velocity and propagation constant is

v = \frac{\omega}{k}=\frac{2730rad/sec}{172rad/m}\\\\

v = 15.87m/s

Time taken, t = \frac{\lambda}{v}

= \frac{1.3}{15.87}\\\\=0.0819 sec

t = 0.0819s

2)

The velocity of transverse wave is given by

v = \sqrt{\frac{T}{\mu}}

v = \sqrt{\frac{W}{\frac{m}{\lambda}}}

mass of string is calculated thus

mg = 0.0125N

m = \frac{0.0125N}{9.8N/s}

m = 0.00128kg

\omega = \frac{v^2m}{\lambda}

\omega = \frac{(15.87^2)(0.00128)}{1.30}

\omega = 0.25N

3)

The propagation constant k is

k=\frac{2\pi}{\lambda}

hence

\lambda = \frac{2\pi}{k}\\\\\lambda = \frac{2 \times 3.142}{172}

\lambda = 0.036 m

No of wavelengths, n is

n = \frac{L}{\lambda}\\\\n = \frac{1.30m}{0.036m}\\

n = 36

4)

The equation of wave travelling down the string is

y(x, t)=Acos[kx -wt]\\\\becomes\\\\y(x , t)= Acos[(172 rad.m)x + (2730 rad.s)t]

without, unit\\\\y(x , t)= Acos[172x + 2730t]

7 0
3 years ago
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