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HACTEHA [7]
2 years ago
10

A 150kg motorcycle starts from rest and accelerates at a constant rate along with a distance of 350m. The applied force is 250N

and the coefficient of kinetic friction is 0.03. What is the net force applied to the motorcycle?
Physics
1 answer:
mart [117]2 years ago
3 0

Answer:

205N

Explanation:

The net force (F) is the difference between the applied force(F_{A}) and the kinetic frictional force(F_{R}). i.e

F = F_{A} - F_{R}    -----------------(i)

Note that;

F_{R} = μmg

Where;

μ = coefficient of kinetic friction

m = mass of the body

g = acceleration due to gravity = 10m/s²

Equation (i) then becomes;

F = F_{A} - μmg        -------------------(ii)

<em>Given from question;</em>

m = mass of motorcycle = 150kg

μ = 0.03

F_{A} = 250N

Substitute these values into equation (ii) as follows;

F = 250 - (0.03 x 150 x 10)

F = 250 - (45)

F = 205N

Therefore, the net force applied to the motorcycle is 205N

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iris [78.8K]

Answer:

(a) 8.117 (b) 0.742m/sec^2

Explanation:

We have given distance s =40 m time t=7.5 sec

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So 2.55=u-7.5a --------------eqn 1

From the second equation of motion s=ut-\frac{1}{2}at^2 ( negative sign because there is retrdation as the truck speed is slowing down )

So 40=7.5u-0.5\times a\times 7.5^2

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7 0
3 years ago
Find the first three harmonics of a string of linear mass density 2. 00 g/m and length 0. 600 m when the tension in it is 50. 0
gavmur [86]

The first three harmonics of the string are 131.8 Hz, 263.6 Hz and 395.4 Hz.

<h3>Velocity of the wave</h3>

The velocity of the wave is calculated as follows;

v = √T/μ

where;

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v = √(50/0.002)

v = 158.1 m/s

<h3>First harmonic or fundamental frequency of the wave</h3>

f₀ = v/λ

where;

  • λ is the wavelength = 2L

f₀ = v/2L

f₀ = 158.1/(2 x 0.6)

f₀ = 131.8 Hz

<h3>Second harmonic of the wave</h3>

f₁ = 2f₀

f₁ = 2(131.8 Hz)

f₁ = 263.6 Hz

<h3>Third harmonic of the wave</h3>

f₂ = 3f₀

f₂ = 3(131.8 Hz)

f₂ = 395.4 Hz

Thus, the first three harmonics of the string are 131.8 Hz, 263.6 Hz and 395.4 Hz.

Learn more about harmonics here: brainly.com/question/4290297

#SPJ1

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