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uranmaximum [27]
1 year ago
12

A ball on a string travels once around a circle with a circumference of 2. 0 m. The tension in the string is 5. 0 n.

Physics
1 answer:
Schach [20]1 year ago
4 0

The work done by the ball on a string is 0J if a ball is travelling around as circle with a circumference of 2.0m

We know that circumference or perimeter of circle=2πr where r is the radius of the circle.

So, we have circumference=2m

Therefore,2πr=2

=>r=2/2π

=>r=(1/π) m

Now, we know very well that when a body moves a displacement dx under the action of force, some amount of work is done by the force on the body and that force is given by

W=\int\limits {F}. \, dx

where F is defined as the force acting on the body

and dx is the displacement of the body.

On expanding the above formula, we get

=>W=Fdxcosθ where cosθ is the angle between the force and displacement of the body.

Since Tension is actually centripetal force which is acting along the center of the circle where ball displacement is in perpendicular direction to the tension.

It means angle between Tension and ball's displacement is 90°. So,cos90°=0

Therefore, W=Fdxcos90

=>W=0J.

Hence, amount of work done is 0J.

To know more about work, visit here:

brainly.com/question/18094932

#SPJ4

(Complete question) is:

A ball on a string travels once around a circle with a circumference of 2. 0 m. The tension in the string is 5. 0 N. What amount of work done by the ball on the string?

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A certain spring has a spring constant k1 = 660 N/m as the spring is stretched from x = 0 to x1 = 35 cm. The spring constant the
pantera1 [17]

(a) The equation for the work done in stretching the spring from x1 to x2 is ¹/₂K₂Δx².

(b) The work done, in stretching the spring from x1 to x2 is 11.25 J.

(c) The work, necessary to stretch the spring from x = 0 to x3 is 64.28 J.

<h3>Work done in the spring</h3>

The work done in stretching the spring is calculated as follows;

W = ¹/₂kx²

W(1 to 2) = ¹/₂K₂Δx²

W(1 to 2)  =  ¹/₂(250)(0.65 - 0.35)²

W(1 to 2)  = 11.25 J

W(0  to 3) = ¹/₂k₁x₁² + ¹/₂k₂x₂² + ¹/₂F₃x₃

W(0  to 3) = ¹/₂(660)(0.35)² + ¹/₂(250)(0.65 - 0.35)² + ¹/₂(105)(0.89 - 0.65)

W(0  to 3) = 64.28 J

Learn more about work done here: brainly.com/question/25573309

#SPJ1

6 0
1 year ago
A flute is designed so that it plays a frequency of 261.6 Hz, middle C, when all the holes are covered and the temperature is 20
irinina [24]

Answer:

0.655 m

13.468°C

Explanation:

v = Speed of sound at 20.0°C = 343 m/s (general value)

For one both end open we have the expression

L=\dfrac{\lambda_1}{2}\\\Rightarrow L=\dfrac{\dfrac{v}{f_1}}{2}\\\Rightarrow L=\dfrac{\dfrac{343}{261.6}}{2}\\\Rightarrow L=0.655581039755\ m

The length of the flute is 0.655 m

Beat frequency is given by

\Delta f=f_1-f_2\\\Rightarrow 3=261.6-f_2\\\Rightarrow f_2=261.6-3\\\Rightarrow f_2=258.6\ Hz

Velocity of the wave is

v=f_2\lambda_1\\\Rightarrow v=258.6\times \dfrac{343}{261.6}\\\Rightarrow v=339.066513761\ m/s

The temperature is given by

T=273(\dfrac{v}{331})^2\\\Rightarrow T=273(\dfrac{339.066513761}{331})^2\\\Rightarrow T=286.468227799\ K=286.468227799-273=13.468227799^{\circ}C

The temperature of the room is 13.468°C

4 0
3 years ago
Based on The MOHS hardness Scale, which mineral could be scratched by a penny but not by a fingernail
melomori [17]

The correct answer is B. Calcite

Explanation:

Mohs hardness scale indicates the hardness of minerals using a scale from 1 to 10 as well as defining the objects or tools that can be used to scratch the minerals. These two features of minerals are shown in the table of the image. About this, it is shown gypsum and talc can be scratched by just a fingernail, considering minerals with a hardness of 2.5 or below can be scratched by a fingernail. In the case of calcite that has a hardness of 3, this cannot be scratched by a fingernail, but it can be scratched by a penny, which works for minerals with a hardness of 3.5 or below. Thus, the correct answer is Calcite.

6 0
2 years ago
A particle (mass = 2.0 mg, charge = −6.0 μC) moves in the positive direction along the x axis with a velocity of 3.0 km/s. It en
Virty [35]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The acceleration would be  a = 0.003* 6 =0.018\  m/s^2    

Explanation:

The objective of this solution is to obtain the acceleration of the particle

       Now looking at Newton law which is mathematically represented as

                       F = ma

  Where F is the force experience by a particle

              m is the mass of the particle

              a is the acceleration of the particle

  And also this force is equivalent to magnetic force in a magnetic field which is mathematically represented as

                    F = qvB

Where q is the charge of the particle

             v is the velocity of the  charge

             B is the magnetic field the charge is under it influence

  Now equating this two formulas

                   ma = qvB

 Making a the subject we have

                  a = \frac{qvB}{m}

In the question the direction of the is in the positive x-axis which is i hence the direction would be in the i direction

      So substituting  (2.0i +3.0j+4.0k)mT = (2.0i +3.0j+4.0k)*10^{-3}T for B

                    a = \frac{q}{m} * v (2.0i +3.0j +4.0k)*10^{-3}

      Substituting       3.0 Km /s = 3.0*10^{3}\ m/s  for v  and -6.0 \muC = -6.0*10^{-6} C for q

                     a = \frac{-6.0*10^{-6}}{2.0*10^{-3}} * 3.0*10^{3} *(2i+3j+4k) *10^{-3}

                       a = 0.003 * 3i(2i+3j+4k)

                      a = 0.003 *((3*2)i \ \cdot i \ +(3*3) i \ \cdot \ j  \ + (3*4)i \ \cdot \ k)

According to vector multiplication

                                             i \cdot i = j \cdot j = k\cdot k = 1\\\\and \ i\cdot j = i\cdot k  = 0

     So

               a = 0.003* 6 =0.018\  m/s^2          

     

8 0
3 years ago
When positively charged particles were radiated onto a gold atom, a few of the particles bounced back. Which of the following ca
ratelena [41]

Positively charged protons in the nucleus of the gold atom .... rutherford scattering ???


3 0
3 years ago
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