Is stress a individual matter

Plasma is the most abundant state of matter in the universe

Darina [25.2K]

Plasmas are the most common state of matter in the universe. <span>A plasma is a gas that has been energized to the point that some of the electrons break free from, but travel with, their nucleus.</span>

6 0

4 years ago

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A flywheel with a radius of 0.200m starts from rest and accelerates with a constant angular acceleration of 0.900rad/s2.

kobusy [5.1K]

Answer:

The magnitude of the radial acceleration is 0.754 rad/s²

Explanation:

Given;

radius of the flywheel, r = 0.2 m

initial angular velocity of the flywheel, $\omega_i = 0$

angular acceleration of the flywheel, a = 0.900 rad/s².

angular distance, θ = 120⁰

the angular distance in radian = $\frac{120}{180} \pi = \frac{2 \pi}{3} \ rad$

Apply the following kinematic equation to determine the final angular velocity;

$\omega _f^2 = \omega _i ^2 + 2\alpha \theta\\\\\omega _f^2 = 0 + 2(0.9)(\frac{2\pi}{3} )\\\\\omega _f^2 = 3.7699\\\\\omega _f = \sqrt{ 3.7699} \\\\ \omega _f = 1.942 \ rad/s$

The magnitude of the radial acceleration is calculated as;

$\alpha _r = \omega ^2r\\\\\alpha _r = (1.942)^2 (0.2)\\\\\alpha _r =0.754 \ rad/s^2$

Therefore, the magnitude of the radial acceleration is 0.754 rad/s²

4 0

3 years ago

A current loop ABCDA consists of a metal rod. a resistor R, and a pair of conducting rails separated by a distance d. The rod ha

Lilit [14]

Answer:

a) $\vec{d} =B-A$

b) $I=\frac{B*v*d*\sin 90 \textdegree}{R}$

c) $v \approx 0.6m/s$

Explanation:

From the question we are told that

Magnetic field strength $B=3.6T$

Distance traveled $d=7m$

Mass $m=4.7 kg$

Resistance $r=8.2 ohm$

Gravitational acceleration $g=9.8m/s^2$

$\theta =90 \textdegree$ Because of perpendicularity

a)

Generally the direction of the current will be given as

$\vec{d} =B-A$

Because it opposes increases of magnetic flux

b)

Generally the equation for induced EMF $E$ is mathematically given as

$E=B*v*d*\sin \theta$

$E=B*v*d*\sin 90 \textdegree$

Generally the equation for induced current $I$ is mathematically given as

$I=E/R$

$I=\frac{B*v*d*\sin 90 \textdegree}{R}$

c)

Generally the the equation for force F at terminal speed is mathematically given as

$F=mg$

$mg=B*I*d*\sin 90 \textdegree$

$mg=B*(\frac{B * v * d }{R}) *d$

$v=\frac{m*g*R}{B^2*D^2}$

$v=\frac{4.7*9.8*8.2}{3.6^2*7^2}$

$v=0.59475m/s$

$v \approx 0.6m/s$

7 0

3 years ago

9. A bicycle tire is spinning clockwise at 3.80 rad/s. During a time period of t = 2.35 s, the tire is stopped and spun in the o

Dima020 [189]

Answer:

option E

Explanation:

given,

angular speed of bicycle tire ()= 3.80 rad/s

time period = 2.35 s

tire stops and spun in opposite direction with angular velocity = 3.80 rad/s

average angular acceleration

initial angular speed = ω₀ = - 3.80 rad/s

final angular speed = ω₁ = 3.80 rad/s

change in angular velocity

$\Delta \omega = \omega - \omega_0$

$\Delta \omega = 3.80 - (-3.80)$

$\Delta \omega = 7.6\ rad/s$

average angular acceleration

$\alpha = \dfrac{\Delta \omega}{t}$

$\alpha = \dfrac{7.6}{2.35}$

$\alpha = 3.23\ rad/s^2$

approximately equal to 3.4 rad/s²

the correct answer is option E

4 0

3 years ago

A 1kg box is pushed on a flat surface that is 250m long. The box is initially at rest and then pushed with a constant Net force

sasho [114]

Answer:

C) 50 m/s

Explanation:

With the given information we can calculate the acceleration using the force and mass of the box.

Newton's 2nd Law: F = ma

5 N = 1 kg * a
a = 5 m/s²

List out known variables:

v₀ = 0 m/s
a = 5 m/s²
v = ?
Δx = 250 m

Looking at the constant acceleration kinematic equations, we see that this one contains all four variables:

v² = v₀² + 2aΔx

Substitute known values into the equation and solve for v.

v² = (0)² + 2(5)(250)
v² = 2500
v = 50 m/s

The final velocity of the box is C) 50 m/s.

7 0

3 years ago