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jeka57 [31]
4 years ago
5

illustrates a method for solving this problem. A "swing" ride at a carnival consists of chairs that are swung in a circle by 19.

6 m cables attached to a vertical rotating pole, as the drawing shows. Suppose the total mass of a chair and its occupant is 250 kg. (a) Determine the tension in the cable attached to the chair. (b) Find the speed of the chair.
Physics
1 answer:
dlinn [17]4 years ago
4 0

Explanation:

Given that,

Length of the cable is 19.6 m, l = 19.6 m

Let us assume that the angle with vertical rotating pole is 63 degrees.

The total mass of a chair and its occupant is 250 kg.

(a) Let T is the tension in the cable attached to the chair. So,

T\cos\theta=mg\\\\T=\dfrac{mg}{\cos\theta}\\\\T=\dfrac{250\times 9.8}{\cos(63)}\\\\T=5396.58\ N

(b) The centripetal acceleration acts on it such that,

\dfrac{v^2}{r}=g\tan\theta\\\\v=\sqrt{Rg\tan\theta} \\\\v=\sqrt{l\sin\theta g\tan\theta}\\\\v=\sqrt{19.6\times \sin(63)9.8\times \tan(63)}\\\\v=18.32\ m/s

Hence, this is the required solution.

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Please answer!
Elden [556K]
I can confirm, your answer is 800J (Or 800 Joules)!

I just took the test and this was the correct answer :)
4 0
3 years ago
Read 2 more answers
A small steel wire of diameter 1.0mm is connected to an oscillator and is under a tension of 5.7N . The frequency of the oscilla
kotegsom [21]

Answer:

The power output of the oscillator is 0.350 watt.

Explanation:

Given that,

Diameter = 1.0 mm

Tension = 5.7 N

Frequency = 57.0 Hz

Amplitude = 0.54 cm

We need to calculate the power output of the oscillator

Using formula of the power

P=\dfrac{1}{2}\times\mu\times\omega^2\times a^2\times v

Put the value into the formula

P=\dfrac{1}{2}\times A\times\rho\times\omega^2\times a^2\times\dfrac{\sqrt{T}}{\mu}

P=\dfrac{1}{2}\times3.14\times(0.0005)^2\times7850\times(2\times\pi\times57.0)^2\times(0.54\times10^{-2})^2\times\sqrt{\dfrac{5.7}{3.14\times(0.0005)^2\times7850}}

P=0.350\ Watt

Hence, The power output of the oscillator is 0.350 watt.

3 0
3 years ago
A billiard ball is moving in the x-direction at 30.0 cm/s and strikes another billiard ball moving in the y-direction at 40.0 cm
PSYCHO15rus [73]

Explanation:

Given that,

Initial speed of the billiard ball 1, u = 30i cm/s

Initial speed of another billiard ball 2, u' = 40j cm/s

After the collision,

Final speed of first ball, v = 50 cm/s

Final speed of second ball, v' = 0 (as it stops)

Let us consider that both balls have same mass i.e. m

Initial kinetic energy of the system is :

K_i=\dfrac{1}{2}mu^2+\dfrac{1}{2}mu'^2\\\\K_i=\dfrac{1}{2}m(u^2+u'^2)\\\\K_i=\dfrac{1}{2}m((30)^2+(40)^2)\\\\K_i=1250m\ J

Final kinetic energy of the system is :

K_f=\dfrac{1}{2}mv^2+\dfrac{1}{2}mv'^2\\\\K_f=\dfrac{1}{2}m(v^2+v'^2)\\\\K_f=\dfrac{1}{2}m((50)^2+(0)^2)\\\\K_f=1250m\ J

The change in kinetic energy of the system is equal to the difference of final and initial kinetic energy as :

\Delta K=K_f-K_i\\\\\Delta K=1250m-1250m\\\\\Delta K=0

So, the change in kinetic energy of the system as a result of the collision is equal to 0.    

7 0
4 years ago
An nfl linebacker can go from 0 m/s to 2.5 m/s in 2.5s. What is his acceleration?
dybincka [34]

Answer:

a= 1 m/s^2

Explanation:

a= (Vf-Vi)/t

a=(2.5m/s - 0m/s)/2.5 s

a=2.5 m/s / 2.5 s

a= 1 m/s^2

5 0
3 years ago
Spaceship 1 and spaceship 2 have equal masses of 200 kg. Spaceship 1 has a speed of 0 m/s, and spaceship 2 has a speed of 10 m/s
m_a_m_a [10]

Answer:

2000 kg m/s

Explanation:

The momentum of an object is a vector quantity whose magnitude is given by

p=mv

where

m is the mass of the object

v is the velocity of the object

and its direction is the same as the velocity.

In this problem, we have:

- Spaceship 1 has

m = 200 kg (mass)

v = 0 m/s (zero velocity)

So its momentum is

p_1 =(200)(0)=0

- Spaceship 2 has

m = 200 kg (mass)

v = 10 m/s (velocity)

So its momentum is

p_2=(200)(10)=2000 kg m/s

Therefore, the combined momentum of the two spaceships is

p=p_1+p_2=0+2000=2000 kg m/s

3 0
3 years ago
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