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garri49 [273]
3 years ago
12

A small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal

circle so that the thread’s trajectory describes a cone. The acceleration of gravity is 9.8 m/s 2 . How long does it take for the ball to rotate once around the axis?
Physics
1 answer:
Gemiola [76]3 years ago
4 0

Answer:

Time taken, T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}

Explanation:

It is given that, a small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread’s trajectory describes a cone as shown in attached figure.

From the figure,

The sum of forces in y direction is :

T\ cos\theta-mg=0

T=\dfrac{mg}{cos\theta}

Sum of forces in x direction,

T\ sin\theta=\dfrac{mv^2}{r}

mg\ tan\theta=\dfrac{mv^2}{r}.............(1)

Also, r=l\ sin\theta

Equation (1) becomes :

mg\ tan\theta=\dfrac{mv^2}{l\ sin\theta}

v=\sqrt{gl\ tan\theta.sin\theta}...............(2)

Let t is the time taken for the ball to rotate once around the axis. It is given by :

T=\dfrac{2\pi r}{v}

Put the value of T from equation (2) to the above expression:

T=\dfrac{2\pi r}{\sqrt{gl\ tan\theta.sin\theta}}

T=\dfrac{2\pi l\ sin\theta}{\sqrt{gl\ tan\theta.sin\theta}}

On solving above equation :

T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}

Hence, this is the required solution.

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A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(
Natasha2012 [34]

Answer:

a) t=2s

b) h_{max}=1024ft

c) v_{y}=-256ft/s

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

0=(64ft/s)^2-2(32ft/s^2)(y-960ft)

y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

h=-16t^2+64t+960

1024=-16t^2+64t+960

0=-16t^2+64t-64

Solving the quadratic equation

t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

a=-16\\b=64\\c=-64

t=2s

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s

Since the projectile is going down the velocity at the instant it reaches the ground is:

v=-256ft/s

5 0
2 years ago
Calculate the force it would take to accelerate a 50 kg bike at a rate of 3 m/s2.
Ede4ka [16]

ummmm it might be 300... i used a calculator

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4 0
3 years ago
Which three temperature readings all mean the same thing?
Drupady [299]

Answer : The correct option is, (D) 273 Kelvin, 0 degrees Celsius, 32 degrees Fahrenheit

Explanation :

Conversion of degree Celsius to Kelvin :

K=^oC+273

Conversion of degree Celsius to degrees Fahrenheit :

^oF=(\frac{9}{5}\times ^oC)+32

By using these two conversion factors, we get the three temperature readings all mean the same thing.

For option A :

K=^oC+273=100+273=373K

^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 100)+32=212^oF

For option B :

K=^oC+273=100+273=373K

^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 100)+32=212^oF

For option C :

K=^oC+273=0+273=273K

^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 0)+32=32^oF

For option D :

K=^oC+273=0+273=273K

^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 0)+32=32^oF

From the given options, only option (D) is correct.

Hence, the correct option is, (D) 273 Kelvin, 0 degrees Celsius, 32 degrees Fahrenheit

7 0
3 years ago
Read 2 more answers
Asap help please for q3
Digiron [165]
Second option would be correct.
7 0
3 years ago
A helicopter’s velocity increases from 35 m/s to 70 m/s in 10 seconds. What is the acceleration of this helicopter?
Alenkinab [10]
3.5 meters per second second is the acceleration because we know that acceleration is change in velocity over time and the change is velocity here is 35 and the time is 10 so we can simply divide 35 by 10 which is 3.5 m/s squared
3 0
2 years ago
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