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Alecsey [184]
3 years ago
12

A phonograph record 0.15 m in its radius rotates 18 times per 90 seconds what is the frequency?

Physics
1 answer:
ioda3 years ago
6 0

Answer:

The frequency of the phonograph record is 0.2 Hz

Explanation:

The frequency of an object moving in uniform circular motion is the number of completed cycles the object makes in a specified time period

The given parameters of the phonograph record are;

The radius of the record = 0.15 m

The number of times the phonograph record rotates, n = 18 times

The time it takes the phonograph record to rotate the 18 times, t = 90 seconds

The frequency of the phonograph record, f = (The number of times the phonograph record rotates) ÷ (The time it takes the phonograph record to rotate the 18 times)

∴ The frequency of the phonograph record, f = n/t = 18/(90 s) = 0.2 Hz

The frequency of the phonograph record = 0.2 Hz.

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You have been hired as a technical consultant for an early-morning cartoon series for children to make sure that the science is
katen-ka-za [31]

The initial potential energy of the wagon containing gold boxes will enable

it roll down the hill when cut loose.

The Lone Ranger and Tonto have approximately <u>5.1 seconds</u>.

Reasons:

Mass wagon and gold = 166 kg

Location of the wagon = 77 meters up the hill

Slope of the hill = 8°

Location of the rangers = 41 meters from the canyon

Mass of Lone Ranger, m₁ = 65 kg

Mass of Tonto m₂ = 66 kg

Solution;

Height of the wagon above the level ground, h = 77 m × sin(8°) ≈ 10.72 m

Potential energy = m·g·h

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Potential energy of wagon, P.E. ≈ 166 × 9.81 × 10.72 = 17457.0912

Potential energy of wagon, P.E. ≈ 17457.0912 J

By energy conservation, P.E. = K.E.

K.E. = \mathbf{\dfrac{1}{2} \cdot m \cdot v^2}

Where;

v = The velocity of the wagon a the bottom of the cliff

Therefore;

\dfrac{1}{2} \times 166 \times v^2 = 17457.0912

v = \sqrt{\dfrac{17457.0912}{\dfrac{1}{2} \times 166} } \approx 14.5

Velocity of the wagon, v ≈ 14.5 m/s

Momentum = Mass, m × Velocity, v

Initial momentum of wagon = m·v

Final momentum of wagon and ranger = (m + m₁ + m₂)·v'

By conservation of momentum, we have;

m·v = (m + m₁ + m₂)·v'

\therefore v' = \mathbf{ \dfrac{m \cdot v}{(m + m_1 + m_2)  }}

Which gives;

\therefore v' = \dfrac{166 \times 14.5}{(166 + 65 + 66)  } \approx 8.1

The velocity of the wagon after the Ranger and Tonto drop in, v' ≈ 8.1 m/s

Time = \dfrac{Distance}{Velocity}

\mathrm{The \ time \ the\ Lone \  Ranger \  and  \ Tonto \  have,  \ t} = \dfrac{41 \, m}{8.1 \, m/s} \approx 5.1 \, s

The Lone Range and Tonto have approximately <u>5.1 seconds</u> to grab the

gold and jump out of the wagon before the wagon heads over the cliff.

Learn more here:

brainly.com/question/11888124

brainly.com/question/16492221

5 0
3 years ago
What are the middle part of the ear
ella [17]

The middle ear transmits sound from the outer ear to the inner ear. The middle ear consists of three bones: the hammer (malleus), the anvil (incus) and the stirrup (stapes), the oval window, the round window and the Eustrachian tube. its called tympanic cavity

hope this helps :)

7 0
4 years ago
Read 2 more answers
a ball is thrown in the air at 20 m/s. ignoring redistance, what is the speed when it returns to the level at which it was throw
inysia [295]

Ball is projected upwards with speed 20 m/s

now using kinematics

v_f^2 - v_i^2 = 2 a\delta x

here when ball returns back to same position.

total displacement of ball will be zero.

v_f^2 - v_i^2 = 0

vf = vi

So ball will return to initial position with same speed in opposite direction.

6 0
3 years ago
When the net force acts at right angles to an object’s motion, what does it do to the object’s motion?
denpristay [2]

Answer:

B. It changes the objects direction

8 0
3 years ago
A body of mass m, = 0.050 kg and initial velocity v₁ = 9 m / s turned to the right collides with a body of mass m₂ = 0.030 g wit
Andrej [43]

Answer:

1 m/s

Explanation:

Po = Pf

(0.050kg)(9 m/s) + (0.030kg)(v2) = (0.05kg + 0.030)(6 m/s)

0.45 kgm/s + (0.030kg)(v2) = 0.48 kgm/s

(0.030kg)(v2) = 0.03 kgm/s

v2 = 1 m/s

4 0
2 years ago
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