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Ksju [112]
3 years ago
9

A 500-N box is pushed across a horizontal floor with a force of 250 N. The coefficient of kinetic friction is 0.4 Find the work

done in pushing the box 20 m. How much work went into overcoming friction and how much into accelerating the box?
Physics
1 answer:
lozanna [386]3 years ago
4 0

Answer:

Explanation:

W = Fd

work input to the box

W = 250(20) = 5000 J

work of friction

Wf = μ(mg)d = 0.4(500)(-20) = -4000 J

Kinetic energy change (accelerating the box)

5000 - 4000 = 1000 J

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A uniform string of length 10.0 m and weight 0.32 N is attached to the ceiling. A weight of 1.00 kN hangs from its lower end. Th
Juliette [100K]

Answer: 0.0180701 s

Explanation:

Given the following :

Length of string (L) = 10 m

Weight of string (W) = 0.32 N

Weight attached to lower end = 1kN = 1×10^3

Using the relation:

Time (t) = √ (weight of string * Length) / weight attached to lower end * acceleration due to gravity

g = acceleration due to gravity = 9.8m/s^2

Weight of string = 0.32N

Time(t) = √ (0.32 * 10) / [(1*10^3) * (9.8)]

Time = √3.2 / 9800

= √0.0003265

= 0.0180701s

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Why is it warmer near the equator and cooler near the poles?
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How is the radiating electric field (or electromagnetic signal) produced when radio stations broadcast
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On the sonometer shown below, a horizontal cord of length 5 m has a mass of 1.45 g. When the cord was plucked the wave produced
Korolek [52]

Answer:

(a) T = 0.015 N

(b) M = 1.53 x 10⁻³ kg = 1.53 g

Explanation:

(a) T = 0.015 N

First, we will find the speed of waves:

v =f\lambda

where,

v = speed of wave = ?

f = frequency = 120 Hz

λ = wavelength = 6 cm = 0.06 m

Therefore,

v = (120 Hz)(0.06 m)

v = 7.2 m/s

Now, we will find the linear mass density of the coil:

\mu = \frac{m}{l}

where,

μ = linear mass density = ?

m = mass = 1.45 g = 1.45 x 10⁻³ kg

l = length = 5 m

Thereforre,

\mu = \frac{1.45\ x\ 10^{-3}\ kg}{5\ m}\\\\\mu = 2.9\ x\ 10^{-4}\ kg/m

Now, for the tension we use the formula:

v = \sqrt{\frac{T}{\mu}}\\\\7.2\ m/s = \sqrt{\frac{T}{2.9\ x\ 10^{-4}\ kg/m}}\\\\(51.84\ m^2/s^2)(2.9\ x\ 10^{-4}\ kg/m) = T

<u>T = 0.015 N</u>

<u></u>

(b)

The mass to be hung is:

T = Mg\\\\M = \frac{T}{g}\\\\M = \frac{0.015\ N}{9.8\ m/s^2}\\\\

<u>M = 1.53 x 10⁻³ kg = 1.53 g</u>

4 0
3 years ago
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