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sergeinik [125]
3 years ago
14

An object spins in a horizontal circle with a radius of 15.0 cm. The rotations are timed and the amount of time it takes for it

to go around once is 0.56 s. The centripetal force is measured to be 6.1 N.According to the experiment, the speed of the object is closest to:'
Physics
1 answer:
bezimeni [28]3 years ago
6 0

Answer:

1.7 m/s

Explanation:

Relevant Data provided as per the question below:-

Radius = 15.0 cm

Time = 0.56 s.

Based on the above information

The computation of the speed of the object is shown below:-

Velocity = \frac{2\times \pi \times Radius}{Time}

Velocity = \frac{2\times \frac{22}{7}  \times 0.15}{0.56}

= \frac{0.942857}{0.56}

= 1.683 m/s

or

= 1.7 m/s

Therefore for computing the speed of the object or velocity we simply applied the above formula by considering the pi and all other given data

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how long does it take sound to travel the distance between the two microphones? Given.:wave 1 of microphone 1 has T=2 sec and f=
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0.00583 seconds

Explanation:

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A golf ball is released from rest from the top of a very tall building. Choose a coordinate system whose origin is at the starti
Zigmanuir [339]

Answer:

Velocity of the ball after 3.04 (s) = 29.79 (m/s)

Explanation:

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The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at
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Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

C) f. None of the above (2149.24 m/s)

Explanation:

A)

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

T = Absolute Temperature

but,

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m  _____ eqn (1)

<u>FOR HYDROGEN:</u>

v = √(3KT)/m = 2000 m/s  _____ eqn (2)

<u>FOR OXYGEN:</u>

velocity of oxygen = √(3KT)/(mass of oxygen)  

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

<u>velocity of oxygen = (1/4)(2000 m/s) = 500 m/s</u>

B)

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

<u>K.E of Oxygen = 4000 J</u>

C)

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

<u>v = 2149.24 m/s</u>

<u></u>

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