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mestny [16]
3 years ago
6

A pendulum oscillates 50 times in 6 seconds. Find its time period and frequency? ​

Physics
1 answer:
iren [92.7K]3 years ago
4 0

Explanation:

time taken fir 50 oscillations is 6 seconds

time taken for 1 oscillation is 6/50

convert it into a decimal

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A mass MM uniform solid cylinder of radius RR and a mass MM thin uniform spherical shell of radius RR roll without slipping. If
vampirchik [111]

Answer:

vcyl / vsph = 1.05

Explanation:

  • The kinetic energy of a rolling object can be expressed as the sum of a translational kinetic energy plus a rotational kinetic energy.
  • The traslational part can be written as follows:

       K_{trans} = \frac{1}{2}* M* v_{cm} ^{2}  (1)

  • The rotational part can be expressed as follows:

       K_{rot} = \frac{1}{2}* I* \omega ^{2}  (2)

  • where I = moment of Inertia regarding the axis of rotation.
  • ω = angular speed of the rotating object.
  • If the object has a radius R, and it rolls without slipping, there is a fixed relationship between the linear and angular speed, as follows:

       v = \omega * R (3)

  • For a solid cylinder, I = M*R²/2 (4)
  • Replacing (3) and (4)  in (2), we get:

       K_{rot} = \frac{1}{2}* \frac{1}{2} M*R^{2} * \frac{v_{cmc} ^{2}}{R^{2}} = \frac{1}{4}* M* v_{cmc}^{2}  (5)

  • Adding (5) and (1), we get the total kinetic energy for the solid cylinder, as follows:

       K_{cyl} = \frac{1}{2}* M* v_{cmc} ^{2}  +\frac{1}{4}* M* v_{cmc}^{2}  =  \frac{3}{4}* M* v_{cmc} ^{2} (6)

  • Repeating the same steps for the spherical shell:

        I_{sph} = \frac{2}{3} * M* R^{2} (7)  

       K_{rot} = \frac{1}{2}* \frac{2}{3} M*R^{2} * \frac{v_{cms} ^{2}}{R^{2}} = \frac{1}{3}* M* v_{cms}^{2}  (8)

      K_{sph} = \frac{1}{2}* M* v_{cms} ^{2}  +\frac{1}{3}* M* v_{cms}^{2}  =  \frac{5}{6}* M* v_{cms} ^{2} (9)

  • Since we know that both masses are equal each other, we can simplify (6) and (9), cancelling both masses out.
  • And since we also know that both objects have the same kinetic energy, this means that (6) are (9) are equal each other.
  • Rearranging, and taking square roots on both sides, we get:

       \frac{v_{cmc}}{v_{cms}} =\sqrt{\frac{10}{9} } = 1.05 (10)

  • This means that the solid cylinder is 5% faster than the spherical shell, which is due to the larger moment of inertia for the shell.
3 0
2 years ago
Julian and Joshua each maintain a constant speed as they run laps around a 400-meter track. In the time it takes Julian to compl
Marysya12 [62]

Answer:

the time Joshua travels 1 mile is 12.5 min

Explanation:

Let's start by finding the distance traveled on each lap,

Let's reduce everything to the SI system

    R = 400 m

    d = 1 mile (1609 m / 1 mile) = 1609 m

    L = 2 pi R

    L = 2 pi 400

    L = 2513 m

Let us form a rule of proportions if 2 turns of Julian is 3 turns Joshua, for 1 turn of Joshua how many turns Julian took

    lap Julian = 2/3 turn Joshua

Let's calculate what distance is the same for both of them since they are on the same track

    1 lap = 2513 m

    d. Julian = 2/3 2513 m

    d Julian = 1675 m distance Joshua

Let us form the last rule of three or proportions if 1609 m you travel in 12 min how long it takes to travel 1675 m

    t Julian = 1675/1609 12

    t = 12.5 s

Since this is the distance Joshua travels, this is the time Joshua travels 1 mile

5 0
3 years ago
Read 2 more answers
How are balanced & unbalanced forces related to net force?
Levart [38]

Answer:

An unbalanced force (net force) acting on an object changes its speed and/or direction of motion. ... A net force = unbalanced force. If however, the forces are balanced (in equilibrium) and there is no net force, the object will not accelerate and the velocity will remain constant.

Explanation:

4 0
2 years ago
30cm³ of brine of relative density 1.15 and 42cm³ of water are mixed. What is the density of the final solution​
Aleks04 [339]

Answer:

I think it's the most important part in this

7 0
3 years ago
What force is required to accelerate to 10 kg object to 5.9 m/s/s?
g100num [7]

Force required to accelerate 10 kg object to 5.9 m/s/s ?

Mass = 10 kg

Acceleration = 5.9 m/s^2

Force = Mass * Acceleration

Force = 10 kg * 5.9 m/s^2

Force = 59 kg m /s^2 = 59 N

3 0
3 years ago
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