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horsena [70]
3 years ago
14

a mass spring system makes 50 complete osillations in 10 seconds. what is the period and frequency of the oscillations

Physics
1 answer:
Oxana [17]3 years ago
5 0

Answer: Period = 0.2 seconds; frequency = 5Hz

Explanation:

Number of oscillations = 50

Time required = 10 seconds

Period (T) = ?

Frequency of the oscillations (F) = ?

A) Recall that frequency is the number of oscillations that the mass spring system completes in one second.

i.e Frequency = (Number of oscillations / time taken)

F = 50/10 = 5Hz

B) Period, T is inversely proportional to frequency. i.e Period = 1/Frequency

T = 1/5Hz

T = 0.2 seconds

Thus, the the period and frequency of the oscillations are 0.2 seconds and 5Hz respectively.

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3 years ago
A loaded ore car has a mass of 950 kg. and rolls on rails ofnegligible friction. It starts from rest ans is pulled up a mineshaf
stiks02 [169]

(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

F - mg sin \theta = 0

where

F is the force applied to pull the car

m = 950 kg is the mass of the car

g=9.8 m/s^2 is the acceleration of gravity

\theta=30.0^{\circ} is the angle of the incline

Solving for F,

F=mg sin \theta = (950)(9.8)(sin 30.0^{\circ})=4655 N

Now we know that the car is moving at constant velocity of

v = 2.20 m/s

So we can find the power done by the motor during the constant speed phase as

P=Fv = (4655)(2.20)=10241 W

(b) 10624 W

The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

v=u+at

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2

So now the equation of the forces along the direction parallel to the incline is

F - mg sin \theta = ma

And solving for F, we find the maximum force applied by the motor:

F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N

The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:

P=Fv=(4829)(2.20)=10624 W

(c) 5.82\cdot 10^6 J

Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.

The change in potential energy of the car is:

\Delta U = mg \Delta h

where

m = 950 kg is the mass

g=9.8 m/s^2 is the acceleration of gravity

\Delta h is the change in height, which is

\Delta h = L sin 30^{\circ}

where L = 1250 m is the total distance covered.

Substituting, we find the energy transferred:

\Delta U = mg L sin \theta = (950)(9.8)(1250)(sin 30^{\circ})=5.82\cdot 10^6 J

8 0
3 years ago
HELP IM TIMED. ILL MARK BRAINLIEST PLSSSSSS.
Alex17521 [72]

Answer:

According to Newton's Second Law of Motion :    

Where,

F = Force Applied

m = Mass of the object

a = Acceleration

Now, we will use this law to solve this question.

Given :

Acceleration or a = 15.3 m/s²

Force = 44 N

Mass = ?

Substitute, the given values in the formula.

F = ma

⇒ m = F/a

   m = 44/15.3

<u>m = 2.9 kg</u>

5 0
3 years ago
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