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grigory [225]
3 years ago
11

A 10 kg ball is swung in a horizontal circle at the end of a 2 - m rope over a persons head. The ball makes 30 revolutions per m

inute. How much time does it take for the ball to go around once and what is the speed of the ball?
Physics
1 answer:
steposvetlana [31]3 years ago
8 0

Answer : T = 2 s and Speed = 6.28 m/s.

Explanation :

It is given that,

Mass of the ball, m = 10 kg

Radius of the circle, r = 2 m

The ball makes 30 revolutions per minute. So, frequency is given by :

f=\dfrac{30}{60\ s}

f=\dfrac{1}{2}\ Hz

We know that time period is defined as the reciprocal of frequency i.e.

T=\dfrac{1}{f}

T=2\ s

So, the time period for the ball to go around once is 2 s.

Now, speed is defined as distance travelled per unit time. Distance travelled by ball in circular path is equal to the circumference of the circle.

Speed =\dfrac{distance}{time}

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

S=\dfrac{2\times 3.14\times 2\ m}{2\ s}

S=6.28\ m/s

Speed of the ball is 6.28 m/s.

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How many earth years does it take mars to orbit the sun
Colt1911 [192]

This is a question that would have literally have taken two seconds to look up on google but the answer is 1.88 years.

4 0
3 years ago
An airplane is being pushed West by its engines with a force of 350 N as it starts to take off the runway at the Atlanta airport
Marizza181 [45]

Answer:

option C

Explanation:

given,

Force by the engine on plane in West direction = 350 N

Frictional force on the runway = 100 N in east

force exerted by the wind = 100 N in east

net force and direction = ?

consider west to be positive and east be negative.

when airplane will be moving there will be frictional as well as wind resistance will be acting in opposite direction of airplane

Net force = 350 N - 100 N - 100 N

                 = 150 N

as our answer comes out to be positive so the airplane will be moving in West

hence, the correct answer is option C

8 0
3 years ago
In a series RLC resonance circuit, the resonance frequency f0 = 700 kHz. The resistor R = 10 Ohm. The specified bandwidth (BW) s
sladkih [1.3K]

Answer:

  • quality factor (Q) = 69.99
  • inductor = 1.591 x 10⁻⁴ H
  • capacitor = 3.248 x 10⁻¹⁰ F

Explanation:

Given;

resonance frequency (F₀) = 700 kHz

resistor, R =  10 Ohm

bandwidth (BW) = 10 kHz

bandwidth (BW)  = \frac{R}{2\pi L}

BW = \frac{R}{2\pi L}

make L (inductor) the subject of the formula

L = \frac{R}{2\pi *BW}  =  \frac{10}{2\pi *10,000} =1.591 *10^{-4} \ H = \ 0.1591\ mH

F_o =\frac{1}{2\pi\sqrt{LC} } \\\\\sqrt{LC} = \frac{1}{2\pi F_o} \\\\LC = \frac{1}{4\pi ^2F_o^2}= \frac{1}{4\pi ^2(700,000)^2} = 5.168*10^{-14}

make C (capacitor)  the subject of the formula

C = \frac{5.168*10^{-14}}{1.591*10^{-4}} = 3.248*10^{-10} \ F = \ 3.248*10^{-4} \ \mu F

quality factor (Q) = \frac{1}{R} \sqrt{\frac{L}{C}} \ = \frac{1}{10} \sqrt{\frac{1.591*10^{-4}}{3.248*10^{-10}}}=69.99

quality factor (Q) =  69.99

5 0
4 years ago
Question: Is a flame matter?<br> .<br> Claim: (Write a statement that answers the question)
Alenkinab [10]

Answer:

Matter is anything that has mass and occupies space. The flame itself is a mixture of gases (vaporized fuel, oxygen, carbon dioxide, carbon monoxide, water vapor, and many other things) and so is matter. <em><u>The light produced by the flame is energy, not matter.</u></em>

<em><u></u></em>

6 0
3 years ago
g A thin-walled hollow cylinder and a solid cylinder, both have same mass 2.0 kg and radius 20 cm, start rolling down from rest
ArbitrLikvidat [17]

Answer:

a. i. 3.43 m/s ii. 2.8 m/s

b. The thin-walled cylinder

Explanation:

a. Find translational speed of each cylinder upon reaching the bottom

The potential energy change of each mass = total kinetic energy gain = translational kinetic energy + rotational kinetic energy

So, mgh = 1/2mv² + 1/2Iω² where m = mass of object = 2.0 kg, g =acceleration due to gravity = 9.8 m/s², h = height of incline = 1.2 m, v = translational velocity of object, I = moment of inertia of object and ω = angular speed = v/r where r = radius of object.

i. translational speed of thin-walled cylinder upon reaching the bottom

So, For the thin-walled cylinder, I = mr², we find its translational velocity, v

So, mgh = 1/2mv² + 1/2Iω²

mgh = 1/2mv² + 1/2(mr²)(v/r)²  

mgh = 1/2mv² + 1/2mv²

mgh = mv²

v² = gh

v = √gh

v = √(9.8 m/s² × 1.2 m)

v = √(11.76 m²/s²)

v = 3.43 m/s

ii. translational speed of solid cylinder upon reaching the bottom

So, For the solid cylinder, I = mr²/2, we find its translational velocity, v'

So, mgh = 1/2mv'² + 1/2Iω²

mgh = 1/2mv² + 1/2(mr²/2)(v'/r)²  

mgh = 1/2mv'² + mv'²

mgh = 3mv'²/2

v'² = 2gh/3

v' = √(2gh/3)

v' = √(2 × 9.8 m/s² × 1.2 m/3)

v' = √(23.52 m²/s²/3)

v' = √(7.84 m²/s²)

v' = 2.8 m/s

b. Determine which cylinder has the greatest translational speed upon reaching the bottom.

Since v = 3.43 m/s > v'= 2.8 m/s,

the thin-walled cylinder has the greatest translational speed upon reaching the bottom.

3 0
3 years ago
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