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andre [41]
3 years ago
11

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.16

kg and moves at v = 5.33 m/s. The circular path has a radius of R = 1.13 m. What is the minimum velocity so the string will not go slack as the ball moves around the circle?
Physics
2 answers:
ycow [4]3 years ago
4 0
We assign the variables: T as tension  and x the angle of the string
 The  <span>centripetal acceleration is expressed as v²/r=4.87²/0.9 and (0.163x4.87²)/0.9 = </span><span>T+0.163gcosx, giving T=(0.163x4.87²)/0.9 – 0.163x9.8cosx. 
</span>
<span>(1)At the bottom of the circle x=π and T=(0.163x4.87²)/0.9 – .163*9.8cosπ=5.893N. </span>
<span>(2)Here x=π/2 and T=(0.163x4.87²)/0.9 – 0.163x9.8cosπ/2=4.295N. </span>
<span>(3)Here x=0 and T=(0.163x4.87²)/0.9 – 0.163x9.8cos0=2.698N. </span>
<span>(4)We have T=(0.163v²)/0.9 – 0.163x9.8cosx.

</span><span>This minimum v is obtained when T=0 </span><span>and v verifies (0.163xv²)/0.9 – 0.163x9.8=0, resulting to v=2.970 m/s.</span>
kolezko [41]3 years ago
4 0

Answer:

v=3.57m/s

Explanation:

Hi, let's solve this!

The speed is minimal when the voltage T = 0N

This occurs when the balance is

T + m * g = m * v2 / r

Since the voltage is zero, then we cancel it and the equation is

m * g = m * v2 / r

We have the following data:

m = 0.16kg

g = 9.8m / s2

r = 1.13m

We cleared the speed and it is

v=\sqrt{g*r}

v=\sqrt{9.8m/s^{2}* 1.13m}

v=3.57m/s

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5 0
3 years ago
sara and tory are out fishing on the lake on a hot summer day when they both decide to go for a swim. sara dives off the front o
crimeas [40]

Here it is an application of Newton's III law

as we know by Newton's III law that every action has equal and opposite reaction

So here as we know that two boys jumps off the boat with different forces

from front side of the boat the boy jumps off with force 45 N which means as per Newton's III law if boy has a force of 45 N in forward direction then he must apply a reaction force on the boat in reverse direction of same magnitude

So boat must have an opposite force on front end with magnitude 45 N

Now similar way we can say

from back side of the boat the boy jumps off with force 60 N which means as per Newton's III law if boy has a force of 60 N in backward direction then he must apply a reaction force on the boat in reverse direction of same magnitude

So boat must have an opposite force on front end with magnitude 60 N

So here net force due to both jump on the boat is given by

F_{net} = F_1 - F_2

F_{net} = 60 - 45

F_{net} = 15 N

so boat will have net force F = 15 N in forward direction due to both jumps

3 0
3 years ago
If you stood on a planet with four times the mass of Earth, and twice Earth's radius, how much would you weigh?
nikdorinn [45]

Answer:

1/4 times your earth's weight

Explanation:

assuming the Mass of earth = M

Radius of earth = R

∴ the mass of the planet= 4M

the radius of the planet = 4R

gravitational force of earth is given as = \frac{GM}{R^{2} }

where G is the gravitational constant

Gravitational force of the planet = \frac{G4M}{(4R)^{2} }

                                                       =\frac{G4M}{16R^{2} }

                                                       =\frac{GM}{4R^{2} }

recall, gravitational force of earth is given as = \frac{GM}{R^{2} }

∴Gravitational force of planet = 1/4 times the gravitational force of the earth

you would weigh 1/4 times your earth's weight

3 0
3 years ago
In a television set, electrons are accelerated from rest through a potential difference of 20 kV. The electrons then pass throug
Svetradugi [14.3K]

Answer: Fmax = 5.54*10^-12 N

Explanation: From the question, we have the potential difference (V) =20kv = 20,000v and strength of magnetic field (B) =0.41 T.

The maximum force experienced by a charge of magnitude (q) is given as

Fmax = qvB

Where v = velocity of electron.

The velocity of the electron can be gotten by using the work energy theorem.

The kinetic energy of the electron (mv²/2) equals the work done needed to accelerate it.

mv²/2 = qV.

Where m = mass of an electronic charge = 9.11×10^-31 kg, q = magnitude of an electronic charge = 1.609×10^-19 c, v = velocity of electron, V = potential difference = 20,000v.

By substituting the parameters, we have that

(9.11×10^-31 × v²)/2 = 1.609×10^-19 × 20000

(9.11×10^-31 × v²) = 1.609×10^-19 × 20000 ×2

v² = (1.609×10^-19 × 20000 ×2)/9.11×10^-31

v² = 64.36*10^(-16)/9.11×10^-31

v² = 7.0647×10^15

v = √7.0647×10^15

v = 8.40×10^7 m/s

Fmax = 1.609×10^-19 × 8.40×10^7 × 0.41

Fmax = 5.54*10^-12 N

7 0
3 years ago
brainly a child on a swing set swings back and forth with a period of 3.3 s and an amplitude of 25°. what is the maximum speed o
beks73 [17]

Maximum speed of the child as she swings is  2.23 m/s.

<h3>Step by Step Calculation:</h3>

T=3.3 s is the oscillation's time period.

The swing's greatest angle is 25° (max).

The swing's bottom will have the following kinetic energy:

k=12mv2...........(1)

The mass in this situation is m, and the speed is v.

The potential energy change is expressed as,

∆U=mgL1-cosθmax...............(2)

Here, L is a string's length and g is the acceleration caused by gravity. L is given as,

L=gT24π2

Combine equation (1) with (2)

12mv2=mgL1-cos, maxv=2g, maxv=gT24, maxv=g2T22, maxv=9.8 m/s

22.33 m/s, 22.31 s22.31 cos25°

Therefore, the child's top speed is 2.23 m/s.

<h3>What is Oscillation ?</h3>
  • The process of any quantity or measure fluctuating repeatedly about its equilibrium value in time is known as oscillation.
  • A periodic change in a substance's value between two values or around its central value is another way to define oscillation.

To learn more about Oscillation refer to:

brainly.com/question/28312746

#SPJ1

6 0
1 year ago
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