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3 years ago

Difference between relating velocity and angular velocity?

Physics

1 answer:

bazaltina [42]3 years ago

4 0

Answer:

Linear velocity is speed in a straight line (measured in m/s) whileangular velocity is the change in angle over time (measured in rad/s, which can be converted into degrees as well).

Explanation:

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(4%) Problem 9: A mass is connected to a spring and is allowed to move horizontally. The mass is at a position L when the spring

skad [1K]

Answer:

a) Acceleration is zero , c) Speed is cero

Explanation:

a) the equation that governs the simple harmonic motion is

x = A cos (wt +φφ)

Where A is the amplitude of the movement, w is the angular velocity and φ the initial phase determined by the initial condition

Body acceleration is

a = d²x / dt²

Let's look for the derivatives

dx / dt = - A w sin (wt + φ)

a = d²x / dt² = - A w² cos (wt + φ)

In the instant when it is not stretched x = 0

As the spring is released at maximum elongation, φ = 0

0 = A cos wt

Cos wt = 0 wt = π / 2

Acceleration is valid for this angle

a = -A w² cos π/2 = 0

Acceleration is zero

c) When the spring is compressed x = A

Speed is

v = dx / dt

v = - A w sin wt

We look for time

A = A cos wt

cos wt = 1 wt = 0, π

For this time the speedy vouchers

v = -A w sin 0 = 0

Speed is cero

5 0

3 years ago

An accelerator produces a beam of protons with a circular cross section that is 2.0 mm in diameter and has a current of 1.0 mA.

rewona [7]

Answer:

the number density of the protons in the beam is 3.2 × 10¹³ m⁻³

Explanation:

Given that;

diameter D = 2.0 mm

current I = 1.0 mA

K.E of each proton is 20 MeV

the number density of the protons in the beam = ?

Now, we make use of the relation between current and drift velocity

I = MeAv ⇒ 1 / eAv

The kinetic energy of protons is given by;

K = $\frac{1}{2}$ $m_{p}$ v²

v = √( 2K / $m_{p}$ )

lets relate the cross-sectional area A of the beam to its diameter D;

A = $\frac{1}{4}$ πD²

now, we substitute for v and A

n = I / $\frac{1}{4}$ πeD² ×√( 2K / $m_{p}$ )

n = 4I/π eD² × √( $m_{p}$ / 2K )

so we plug in our values;

n = ((4×1.0 mA)/(π(1.602×10⁻¹⁹C)(2mm)²) × √(1.673×10⁻²⁷kg / 2×( 20 MeV)(1.602×10⁻¹⁹ J/ev )

n = 1.98695 × 10¹⁸ × 1.6157967 × 10⁻⁵

n = 3.2 × 10¹³ m⁻³

Therefore, the number density of the protons in the beam is 3.2 × 10¹³ m⁻³

8 0

3 years ago

Initial velocity vector vA has a magnitude of 3.00 meters per second and points 20.0o north of east, while final velocity vector

garri49 [273]

Answer:

$5.2\ \text{m/s}$

$70^{\circ}$ south of east

Explanation:

$v_a$ = 3 m/s

$\theta_a$ = $20^{\circ}$ north of east

$v_b$ = 6 m/s

$\theta_b$ = $40^{\circ}$ south of east = $360-40=320^{\circ}$ north of east

x and y component of $v_a$

$v_{ax}=v_a\cos \theta\\\Rightarrow v_{ax}=3\times \cos 20^{\circ}\\\Rightarrow v_{ax}=2.82\ \text{m/s}$

$v_{ay}=v_a\sin\theta\\\Rightarrow v_{ay}=3\times \sin20^{\circ}\\\Rightarrow v_{ay}=1.03\ \text{m/s}$

x and y component of $v_b$

$v_{bx}=v_b\cos \theta\\\Rightarrow v_{bx}=6\times \cos 320^{\circ}\\\Rightarrow v_{bx}=4.6\ \text{m/s}$

$v_{by}=v_b\sin\theta\\\Rightarrow v_{by}=6\times \sin320^{\circ}\\\Rightarrow v_{by}=-3.86\ \text{m/s}$

$\Delta v=v_b-v_a\\\Rightarrow \Delta v=(4.6-2.82)\hat{i}+(-3.86-1.03)\hat{j}\\\Rightarrow \Delta v=1.78\hat[i}-4.89\hat{j}$

Magnitude

$|\Delta v|=\sqrt{(-4.89)^2+1.78^2}\\\Rightarrow \Delta v=5.2\ \text{m/s}$

Direction

$\theta=\tan{-1}|\dfrac{-4.89}{1.78}|\\\Rightarrow \theta=70^{\circ}$

The magnitude of the change in velocity vector is $5.2\ \text{m/s}$ and the direction is $70^{\circ}$ south of east.

6 0

3 years ago

A person hears the echo of his own sound from a distance hill after 2 seconds . How far is the person from the hill , if the spe

Scorpion4ik [409]

Answer:

the distance between the person and the hill is 330 m.

Explanation:

Given;

speed of sound in air, v = 330m/s

time of sound travel, t = 2 s

The distance between the person and the hill is calculated as;

$Distance = Speed \ \times \ Time\\\\Time \ of \ echo = \frac{2\ \times \ distance}{speed} \\\\distance = \frac{Time \ of \ echo \ \times \ speed}{2} \\\\distance = \frac{2 \ \times \ 330}{2} \\\\distance = 330 \ m$

Therefore, the distance between the person and the hill is 330 m.

3 0

3 years ago

Conclusion 1. why do engineers place tolerances on dimensions? 2. what are the three types of tolerances that appear on dimensio

Butoxors [25]

Tolerance enables the engineer to be informed when somethings requires replacement or if there is a drawback with too much war.

The three types of tolerances that appear on dimensioned drawings are limit, bilateral, and unilateral.

<span>General tolerances are normally found in the information blocks of the blueprint while a specific tolerance is noted for certain areas of the blueprint.</span>

7 0

3 years ago

Difference between relating velocity and angular velocity?​

Difference between relating velocity and angular velocity?