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Anna35 [415]
3 years ago
11

A spring-mass system has a spring constant k = 383 N/m, length of the spring L = 0.5 m, and the mass attached to it is M = 3.8 k

g. What will be its frequency of oscillation if it is pulled by 5.0 cm from its equilibrium position and released?
Physics
1 answer:
sergiy2304 [10]3 years ago
4 0

Answer:

Frequency = f = 10.0394 (1/s)

Explanation:

The frequency of oscillation of the system is given by the action:

f= √(k/m)

f= system count

k = spring constant

m = mass connected to the spring

Therefore the frequency will be:

f= √(k/m) = √(383(N/m) / (3.8kg))= √( 100.7895 (kg×m/s²)/(kg ) =

= √( 100.7895 (1/s²) = 10.0394 (1/s)

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Answer:

Fnet = 0

Explanation:

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  • According Newton's 2nd Law, if the acceleration is zero, the net force on the sliding mass must be zero.
  • This means that there must be a friction force opposing to the horizontal component of the applied force, equal in magnitude to it:

       F_{appx} = F_{app} * cos \theta = 295 N * cos 35 = 242 N  (1)

  • In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:

      F_{appy} = F_{app} * cos \theta = 295 N * sin 35 = 169 N  (2)

⇒    169 N + Fn = Fg = 216 N  (3)

  • This means that there must be a normal force equal to the difference between Fappy and Fg, as follows:
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6 0
3 years ago
An Olympic-class sprinter starts a race with an acceleration of 5.10 m/s2. What is her speed 2.40 s later?
ivolga24 [154]

Answer:

12.24 m/s

Explanation:

Speed: This can be defined as the rate of change of distance with time. The S.I unit of speed is m/s.

Using the formula,

a = v/t................ Equation 1

Where a = acceleration of the sprinter, v = speed of the sprinter, t = time.

making v the subject of the equation,

v = at ................. Equation 2

Given: a = 5.1 m/s², t = 2.4 s.

Substitute into equation 2

v = 5.1(2.4)

v = 12.24 m/s.

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Answer:

Chemical Energy

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Answer:

Y component = 32.37

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Angle of projection of the rocket is, \theta=33.6

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If a vector 'A' makes angle \theta with the horizontal, then the horizontal and vertical components are given as:

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Here, as the velocity is a vector quantity and makes an angle of 33.6 with the horizontal, its Y component is given as:

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Therefore, the Y component of initial velocity is 32.37.

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