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hodyreva [135]
3 years ago
8

A phonograph turntable rotating at 33 1 3 rev/min slows down and stops in 1.98 min. (Assume the turntable initially rotates in t

he negative (clockwise) direction.) (a) What is the turntable's angular acceleration (in rad/s2) assuming it is constant
Physics
1 answer:
enyata [817]3 years ago
6 0

Explanation:

photography presente in frame was of 11 meter thats why

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If a 0.08 kg cell phone falls off a table at 15 m/s, then what is its kinetic energy right before it hits the ground?
Mariana [72]

The kinetic energy of the phone right before it hits the ground is 9J.

<h3>Kinetic energy of the phone</h3>

The kinetic energy of the phone right before it hits the ground is calculated as follows;

K.E = ¹/₂mv²

where;

  • m is mass of the phone
  • v is velocity of the phone

K.E = ¹/₂(0.08)(15)²

K.E = 9 J

Thus, the kinetic energy of the phone right before it hits the ground is 9J.

Learn more about kinetic energy here: brainly.com/question/25959744

#SPJ1

7 0
2 years ago
79. An example of an electrical insulator is
vampirchik [111]

Answer:

B. Glass

Explanation:

An electrical insulator is a substance that does not conduct electricity.

Glass has tightly bounded electrons, that is why it is an insulator of electricity.

6 0
3 years ago
Read 2 more answers
A wire carries a steady current of 2.60 A. A straight section of the wire is 0.750 m long and lies along the x axis within a uni
ad-work [718]

Answer:

The magnetic force on the section of wire is -2.925\hat{j}\ N.

Explanation:

Given that,

Current I = 2.60\hat{i}\ A

Length = 0.750 m

Magnetic field B = 1.50\hat{k}\ T

We need to calculate the magnetic force on the section of wire

Using formula of magnetic force

\vec{F}=l\vec{I}\times\vec{B}

\vec{F}=0.750\times2.60\hat{i}\times1.50\hat{k}

Since, \hat{i}\times\hat{k}=-\hat{j}

\vec{F}=-2.925\hat{j}\ N

Hence, The magnetic force on the section of wire is -2.925\hat{j}\ N.

7 0
3 years ago
a horse moves a sleigh 1.00 kilometer by applying a horizontal 2,000 newton on its harness for 45 minutes.
sineoko [7]
Work:

1 kilometer = 1000 meters
45 × 60 =  2700

W = F × D
W = 2,000 N × 1,000 m
W = 2,000,000 J

P = W ÷ t
P = 2,000,000 J ÷ 2,700 s
P = 741 watts 


Answer:

741 watts of horse power.
5 0
3 years ago
A day on a distant planet observed orbiting a nearby star is 21.5 hr. Also, a year on the planet lasts 69.3 Earth days. In other
serg [7]

Answer:

Part A

The angular speed of rotation of the plane is 8.11781 × 10⁻⁵ rad/s

Part B

The angular speed of orbit of the planet is 1.04938 × 10⁻⁶ rad/s

Explanation:

The parameters of the planet are;

The duration of a day on the distant planet = 21.5 hr.

The duration of a year on the distant planet = 69.3 Earth days

Part A

The duration of a day = The time to make one complete revolution of 2·π radians

∴ The average angular speed about its axis, \omega_{rotation} = Angle turned/Time

∴ \omega_{rotation}  = 2·π/(21.5 × 60 × 60) s ≈ 8.11781 × 10⁻⁵ rad/s

The average angular speed of the planet about its own axis, \omega_{rotation}  = 8.11781 × 10⁻⁵ rad/s

The angular speed of rotation of the plane \omega_{rotation}  = 8.11781 × 10⁻⁵ rad/s

Part B

The time it takes the planet to revolve round the neighboring star once = 69.3 Earth days

Therefore, the average angular speed of the planet around its neighboring star, \omega _{Star}, is given as follows;

\omega _{Orbit}  = 2·π/((69.3 × 24 × 60 × 60) s) = 1.04938 × 10⁻⁶ rad/s

The average angular speed of orbit, \omega _{Orbit} = 1.04938 × 10⁻⁶ rad/s

The angular speed of orbit of the planet, \omega _{Orbit} = 1.04938 × 10⁻⁶ rad/s.

3 0
3 years ago
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