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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 1000 kg elevator accelerates upward at 1.0 m/s2 for 10 m, starting from rest. a. How much work does gravity do on the elevator
cricket20 [7]

Answer:

a)= 98kJ

b)=108kJ

c) = 10kJ

Explanation:

a. The work that is done by gravity on the elevator is:

Work = force * distance  

= mass * gravity * distance

= 1000 * 9.81 * 10  

= 98,000 J

= 98kJ

b)The net force equation in the cable

T - mg = ma

T = m(g+a)

T = 1000(9.8 + 10)

T = 10800N

The work done by the cable is

W = T × d

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=108kJ

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108,100 J = KE + 98,100 J  

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7 0
3 years ago
Read 2 more answers
4 A magnet can exert a force of attraction or a force of repulsion on another magnet.
GenaCL600 [577]

Answer:

Push -repulsion

Pull - attraction

Explanation:

When two magnets are brought together, a push happens when a force of repulsion is experienced where the magnets move away from each other. This means their polarity is the same and this will cause the magnet to push away from each other.

When two magnets are brought together , a pull happens when a force of attraction is experienced where the magnets move close to each other. This means their polarity is different and thus causes the magnets to pull closer to each other.

7 0
3 years ago
ENERGY SAVERS RACE, BRAIN BURNER. This question is about solar cars at the Chuck Norris Institute of Technology, CNIT, in Ocala,
Hunter-Best [27]

Answer:

a) d = 6.0 m

Explanation:

Since car is accelerating at uniform rate then here we can say that the distance moved by the car with uniform acceleration is given as

d = \frac{(v_f + v_i)}{2} \times t

here we know that

v_f = 10 m/s

v_i = 0

t = 1.2 s

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d = \frac{(10 + 0)}{2}\times 1.2

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d = 6.0 m

4 0
3 years ago
The maximum speed with which you can throw a stone is about 20 m/s. Can you hit a window 45 m away horizontally and 10 m up from
Allushta [10]

Answer:

 y = 17 m

Explanation:

For this projectile launch exercise, let's write the equation of position

          x = v₀ₓ t

          y = v_{oy} t - ½ g t²

let's substitute

          45 = v₀ cos θ t

          10 = v₀ sin θ t - ½ 9.8 t²

the maximum height the ball can reach where the vertical velocity is zero

 

           v_{y} = v_{oy} - gt

           0 = v₀ sin θ - gt

           0 = v₀ sin θ - 9.8 t

Let's write our system of equations

         45 = v₀ cos θ  t

         10 = v₀ sin θ t - ½ 9.8 t²

         0 = v₀ sin θ - 9.8 t

We have a system of three equations with three unknowns for which it can be solved.

Let's use the last two

        v₀ sin θ = 9.8 t

we substitute

        10 = (9.8 t) t - ½ 9.8 t2

        10 = ½ 9.8 t2

        10 = 4.9 t2

        t = √ (10 / 4.9)

        t = 1,429 s

Now let's use the first equation and the last one

         45 = v₀ cos θ t

         0  = v₀ sin θ - 9.8 t

         9.8 t = v₀  sin θ

         45 / t = v₀ cos θ

we divide

         9.8t / (45 / t) = tan θ

          tan θ = 9.8 t² / 45

          θ = tan⁻¹ ( 9.8 t² / 45 )

          θ = tan⁻¹ (0.4447)

          θ = 24º

Now we can calculate the maximum height

         v_y² = v_{oy}^2 - 2 g y

         vy = 0

          y = v_{oy}^2 / 2g

          y = (20 sin 24)²/2 9.8

          y = 3,376 m

the other angle that gives the same result is

       θ‘= 90 - θ

       θ' = 90 -24

       θ'= 66'

for this angle the maximum height is

 

          y = v_{oy}^2 / 2g

          y = (20 sin 66)²/2 9.8

           y = 17 m

thisis the correct

6 0
3 years ago
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