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3 years ago

Your roommate is working on his bicycle and has the bike upside down. He spins the 60 cm

1 answer:

5 0

Diameter = 60 cm, Radius = 60/2 = 30 cm = 30/100 = 0.3 m.

The pebble in the tread goes by 3 times every second.

This is the same as 3 times per second.

Recall the unit of frequency is Hertz or per second, s⁻¹

So 3 times per second, Frequency, f = 3s⁻¹ or 3 Hertz

For angular motion:

Angular speed, ω = 2πf

   = 2*π*3

   = 6π   rad/s

Linear speed, v = ωr = 6π * 0.3 = 1.8π m/s

Linear acceleration, a = v² / r

   a = 1.8π * 1.8π / 0.3 = 10.8π² m/s²

Angular acceleration α = a/r = 10.8π² / 0.3 = 36π² rad/s²

Angular speed = 6π rad/s ≈ 18.840 rad/s

The linear speed of the pebble = 1.8π m/s ≈ 5.655 m/s

The angular acceleration = 36π² rad/s² ≈ 355.306 rad/s²

The linear acceleration of the pebble = 10.8π² m/s ≈ 106.592 m/s²

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I need it due in 10 mins

ExtremeBDS [4]

Answer:

B. 14.4 N

Rotational speed (Angular Velocity) = 2

The Radius of the circle = 1.2 m

Velocity = Angular velocity × radius = 2×1.2 = 2.4 m/s

Centripetal force= mv²/r = 3 × 2.4×2.4/1.2 = 3 × 2.4 × 2

= 14.4 N

7 0

3 years ago

Calculate the speed of a proton after it accelerates from rest through a potential difference of 350 V.

AVprozaik [17]

The speed of a proton after it accelerates from rest through a potential difference of 350 V is $25.86 \times 10^4 ~m/s$ .

Initial velocity of the proton $u = 0$

Given potential difference $\Delta V = 350V$

let's assume that the speed of the proton is $v$ ,

Since the proton is accelerating through a potential difference, proton's potential energy will change with time. The potential energy of a particle of charge $q$ when accelerated with a potential difference $\Delta V$ is,

$U = q \Delta V$

Due to Work-Energy Theorem and Conservation of Energy - <em>If there is no non-conservative force acting on a particle then loss in Potential energy P.E must be equal to gain in Kinetic Energy K.E</em> i.e

$\Delta K = \Delta V$

If the initial and final velocity of the proton is $u$ and $v$ respectively then,

change in Kinetic Energy $\implies \Delta K = \frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}mv^2 - 0$

change in Potential Energy $\implies \Delta U = q\Delta V$

from conservation of energy,

$v= \sqrt{\frac{2q\Delta V}{m}}$

so, $v = \sqrt{\frac{2\times 350 \times 1.6\times 10^{-19}}{1.67 \times 10^{-27}}$

$= 25.86 \times 10^4 ~m/s$

To read more about the conservation of energy, please go to brainly.com/question/14668053

7 0

1 year ago

A 2 kg ball is thrown upward with a velocity of 15 m/s. What is the kinetic energy of the ball as it is being thrown?

marusya05 [52]

KE=1/2 m v^2
KE= .5 x 2kg x 15m/s to the 2nd power
KE=225 km/s

7 0

4 years ago

Read 2 more answers

A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the magnitude of the dri

nordsb [41]

Complete Question

A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm2.) mm/s

Answer:

The drift velocity is $v = 0.0002808 \ m/s$