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3 years ago

7

Under what circumstances does distance traveled equal magnitude of displacement? What is the only case in which magnitude of dis

placement and distance are exactly the same?

1 answer:

Sedaia [141]3 years ago

6 0

Answer: Please find the answer in the explanation

Explanation:

Under what circumstances does distance traveled equal magnitude of displacement?

When a body's motion is linear in one direction. Or a body moving in a straight line without turning back.

What is the only case in which magnitude of displacement and distance are exactly the same?

When the body is moving in a straight line with without changing direction or without turning back.

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All of the following are regulated by the medulla except __________.

Lelechka [254]

Answer:

i think its B sorry if its wrong

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3 years ago

Read 2 more answers

A uniform disk of mass M = 4.9 kg has a radius of 0.12 m and is pivoted so that it rotates freely about its axis. A string wrapp

AlekseyPX

Answer:

(A) 2.4 N-m

(B) $0.035kgm^2$

(C) 315.426 rad/sec

(D) 1741.13 J

(E) 725.481 rad

Explanation:

We have given mass of the disk m = 4.9 kg

Radius r = 0.12 m, that is distance = 0.12 m

Force F = 20 N

(a) Torque is equal to product of force and distance

So torque $\tau =Fr$ , here F is force and r is distance

So $\tau =20\times 0.12=2.4Nm$

(B) Moment of inertia is equal to $I=\frac{1}{2}mr^2$

So $I=\frac{1}{2}\times 4.9\times 0.12^2=0.035kgm^2$

Torque is equal to $\tau =I\alpha$

So angular acceleration $\alpha =\frac{\tau }{I}=\frac{2.4}{0.035}=68.571rad/sec^2$

(C) As the disk starts from rest

So initial angular speed $\omega _{0}=0rad/sec$

Time t = 4.6 sec

From first equation of motion we know that $\omega =\omega _0+\alpha t$

So $\omega =0+68.571\times 4.6=315.426rad/sec$

(D) Kinetic energy is equal to $KE=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 0.035\times 315.426^2=1741.13J$

(E) From second equation of motion

$\Theta =\omega _0t+\frac{1}{2}\alpha t^2=0\times 4.6+\frac{1}{2}\times 68.571\times 4.6^2=725.481rad$

3 0

3 years ago

One violin creates a sound whose intensity level is 60dB. Find the intensity level of 16 violins, each playing at this intensity

Yanka [14]

Answer:

72.04 dB.

Explanation:

The intensity level of 60dB corresponds to the sound intensity $I$ given by the equation

$60dB = 10log(\dfrac{I}{I_0} )$

where $I_0 = 1*10^{-12}W/m^2$

solving for $I$ we get:

$6 = log(\dfrac{I}{I_0} )$

$10^6 =\dfrac{I}{1*10^{-12}}$

$\boxed{I = 1*10^{-6} W/m^2}$

Now, when 16 violins are playing the intensity $I$ becomes

${I = 16(1*10^{-6} W/m^2)$

which on the decibel scale gives

$dB = 10log(\dfrac{16*10^{-6}}{1*10^{-12}} )$

$dB = 72.04\: dB$ .

Thus, playing 16 violins together gives the intensity level of 72 dB.

8 0

3 years ago

A 0.45 Caliber bullet (m = 0.162 kg) leaving the muzzle of a gun at 860

Charra [1.4K]

Answer:139.32kgm/s

Explanation:

Mass(m)=0.162kg

Velocity(v)=860m/s

Momentum=mass x velocity

Momentum=0.162 x 860

Momentum=139.32kgm/s

3 0

3 years ago

Yo-Yo man releases a yo-yo from rest and allows it to drop, as he keeps the top end of the string stationary. The mass of the yo

patriot [66]

To solve this problem it is necessary to use the conservation equations of both kinetic, rotational and potential energy.

By definition we know that

$KE + KR = PE$

Where,

KE =Kinetic Energy

KR = Rotational Kinetic Energy

PE = Potential Energy

In this way

$\frac{1}{2} mv^2 +\frac{1}{2} I\omega^2 = mgh$

Where,

m = mass

v= Velocity

I = Moment of Inertia

$\omega =$ Angular velocity

g = Gravity

h = Height

We know as well that $\omega = v/r$ for velocity (v) and Radius (r)

Therefore replacing we have

$\frac{1}{2} mv^2 +\frac{1}{2} I\omega^2 = mgh$

$[tex]h= \frac{1}{2} \frac{v^2}{g} +\frac{1}{2} \frac{I}{mg}(\frac{v}{r})^2$ [/tex]

$h= \frac{1}{2}v^2 ( \frac{1}{g} +\frac{I}{mg}\frac{1}{r^2} )$

$h= \frac{1}{2}0.75^2 ( \frac{1}{9.8} +\frac{2.9*10^{-5}}{(0.056)(9.8)}\frac{1}{(0.0064)^2} )$

$h = 0.3915m$

Therefore the height must be 0.3915 for the yo-yo fall has a linear speed of 0.75m/s

6 0

3 years ago