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ycow [4]
3 years ago
15

THE LENGTH OF A PENDULUM IS (1.5±0.01)m AND THE ACCELERATION DUE TO GRAVITY IS TAKEN AS (9.8±0.1)ms-² calculate the time period

of the pendulum with uncertainty in it
Physics
1 answer:
tiny-mole [99]3 years ago
5 0

Answer:

  2.4583 ± 0.0207 seconds

Explanation:

The time period of a pendulum is approximately given by the formula ...

  T = 2π√(L/g)

The maximum period will be achieved when length is longest and gravity is smallest:

  Tmax = 2π√(1.51/9.7) ≈ 2.47903 . . . seconds

The minimum period will be achieved for the opposite conditions: minimum length and maximum gravity:

  Tmin = 2π√(1.49/9.9) ≈ 2.43756 . . . seconds

If we want to express the uncertainty using a symmetrical range, we need to find half their sum and half their difference.

  T = (2.47903 +2.43756)/2 ± (2.47903 -2.43756)/2

  T ≈ 2.4583 ± 0.0207 . . . seconds

__

We have about 2+ significant digits in the given parameters, so the time might be rounded to 2.46±0.02 seconds.

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the mass of one water drop is 0.0008kg and the gravitational field strength is 10N/kg what is its weight
djyliett [7]
Weight = (mass) x (gravity)

Weight = (8 x 10⁻⁴ kg) x (10 N/kg) = 0.008 Newton
8 0
3 years ago
Can a goalkeeper at his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance will
zmey [24]

The goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground

Explanation:

Consider the vertical motion of ball,

We have equation of motion v = u + at

     Initial velocity, u  = u sin θ

     Final velocity, v =  0 m/s    

     Acceleration = -g

     Substituting

                      v = u + at  

                      0 = u sin θ - g t

                      t=\frac{usin\theta }{g}

This is the time of flight.

Consider the horizontal motion of ball,

        Initial velocity, u =  u cos θ

        Acceleration, a =0 m/s²  

        Time, t=\frac{usin\theta }{g}  

     Substituting

                      s = ut + 0.5 at²

                      s=ucos\theta \times \frac{usin\theta }{g}+0.5\times 0\times (\frac{usin\theta }{g})^2\\\\s=\frac{u^2sin\theta cos\theta}{g}\\\\s=\frac{u^2sin2\theta}{2g}

This is the range.

In this problem

              u = 30 m/s

              g = 9.81 m/s²

              θ = 45° - For maximum range

Substituting

               s=\frac{30^2\times sin(2\times 45)}{2\times 9.81}=45.87m

Maximum horizontal distance traveled by ball without touching ground is 45.87 m, which is less than 95 m.

So the goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground

6 0
3 years ago
70.5
ra1l [238]

Answer:5000000000hertz

Explanation:

Wavelength=6cm=6/100 m=0.06m

Frequency=velocity/wavelength

Frequency=(3×10^8)÷0.06

Frequency=5000000000 hertz

5 0
3 years ago
A dog jumps 0.80 m to catch a treat. The dog's displacement vector is shown below.
Brrunno [24]

Answer:D

Explanation:

It was right o khan academy

6 0
2 years ago
Student 1 lifts a box with a force of 500 N and sets it on a tabletop 1.2 m high. Student 2 pushes an identical box up a 5 m ram
Troyanec [42]

The student who did the most work is student 2 with 2500 Joules.

<u>Given the following data:</u>

  • Force 1 = 500 Newton
  • Distance 1 = 1.2 meter
  • Force 2 = 500 Newton
  • Distance 2 = 5 meter

To determine which of the students did the most work:

Mathematically, the work done by an object is given by the formula;

Work\;done = Force \times distance

<u>For </u><u>student 1</u><u>:</u>

Work\;done = 500 \times 1.2

Work done = 600 Joules

<u>For </u><u>student 2</u><u>:</u>

Work\;done = 500 \times 5

Work done = 2500 Joules.

Therefore, the student who did the most work is student 2 with 2500 Joules.

Read more: Read more: brainly.com/question/13818347

7 0
3 years ago
Read 2 more answers
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