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erastovalidia [21]

2 years ago

15

What is the tangential velocity at the edge of a disk of radius 10cm when it spins with a frequency of 10Hz? Give your answer wi

thout a unit, in cm⋅s−1, and correct to two significant figures

1 answer:

Nina [5.8K]2 years ago

8 0

Answer:

630cm/s

Explanation:

In simple harmonic motion, the tangential velocity is expressed mathematically as v = ὦr

ὦ is the angular velocity = 2πf

r is the radius of the disk

f is the frequency

Given the radius of disk = 10cm

frequency = 10Hz

v = 2πfr

v = 2π×10×10

v = 200π

v = 628.32 cm/s

The tangential velocity = 630cm/s ( to 2 significant figures)

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A 1.15-kg mass oscillates according to the equation x = 0.650 cos(8.40t) where x is in meters and t in seconds. Determine (a) th

zheka24 [161]

Answer:

(a) A = 0.650 m

(b) f = 1.3368 Hz

(c) E = 17.1416 J

(d) K = 11.8835 J

U = 5.2581 J

Explanation:

Given

m = 1.15 kg

x = 0.650 cos (8.40t)

(a) the amplitude,

A = 0.650 m

(b) the frequency,

if we know that

ω = 2πf = 8.40 ⇒ f = 8.40 / (2π)

⇒ f = 1.3368 Hz

(c) the total energy,

we use the formula

E = m*ω²*A² / 2

⇒ E = (1.15)(8.40)²(0.650)² / 2

⇒ E = 17.1416 J

(d) the kinetic energy and potential energy when x = 0.360 m.

We use the formulas

K = (1/2)*m*ω²*(A² - x²) (the kinetic energy)

and

U = (1/2)*m*ω²*x² (the potential energy)

then

K = (1/2)*(1.15)*(8.40)²*((0.650)² - (0.360)²)

⇒ K = 11.8835 J

U = (1/2)*(1.15)*(8.40)²*(0.360)²

⇒ U = 5.2581 J

4 0

2 years ago

A cylinder rotating about its axis with a constant angular acceleration of 1.6 rad/s2 starts from rest at t = 0. At the instant

OverLord2011 [107]

Answer:

The magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²

Explanation:

The total linear acceleration is the vector sum of the tangential acceleration and radial acceleration.

The radial acceleration is given by;

$a_t = ar$

where;

a is the angular acceleration and

r is the radius of the circular path

$a_t = ar\\\\a_t = 1.6 *0.13\\\\a_t = 0.208 \ m/s^2$

Determine time of the rotation;

$\theta = \frac{1}{2} at^2\\\\0.4 = \frac{1}{2} (1.6)t^2\\\\t^2 = 0.5\\\\t = \sqrt{0.5} \\\\t = 0.707 \ s\\\\$

Determine angular velocity

ω = at

ω = 1.6 x 0.707

ω = 1.131 rad/s

Now, determine the radial acceleration

$a_r = \omega ^2r\\\\a_r = 1.131^2 (0.13)\\\\a_r = 0.166 \ m/s^2$

The magnitude of total linear acceleration is given by;

$a = \sqrt{a_t^2 + a_r^2} \\\\a = \sqrt{0.208^2 + 0.166^2} \\\\a = 0.266 \ m/s^2\\\\a = 0.27 \ m/s^2$

Therefore, the magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²

5 0

2 years ago

When light energy hits the retina, the retinal changes from a _____ to a _____ configuration.

gayaneshka [121]

Answer:

Cis, Trans.

Explanation:

Rhodopsin also known as visual purple, pigment which contains sensory protein that helps to convert light into an electrical signal. Rhodopsin present in wide range of organisms from bacteria to vertebrates.

Rhodopsin is composed of opsin, and 11-cis-retinaldehyde which is derived from vitamin A. When the eye contact with light the 11-cis component converted to all trans-retinal, which results in the changes in configuration fundamental in the rhodopsin molecule.

5 0

2 years ago

A 50.-kilogram rock rolls off the edge of a cliff. if it is traveling at a speed of 24.2 m/s when it hits the ground, what is th

ElenaW [278]

The correct answer to the question is : 29.88 m.

EXPLANATION :

As per the question, the mass of the rock m = 50 Kg.

The rock is rolling off the edges of the cliff.

The final velocity of the rock when it hits the ground v = 24 .2 m/s.

Let the height of the cliff is h.

The potential energy gained by the rock at the top of the cliff = mgh.

Here, g is known as acceleration due to gravity, and g = $9.8\ m/s^2$

When the rock rolls off the edge of the cliff, the potential energy is converted into kinetic energy.

When the rock hits the ground, whole of its potential energy is converted into its kinetic energy.

The kinetic energy of the rock when it touches the ground is given as -

Kinetic energy K.E = $\frac{1}{2}mv^2$ .

From above we know that -

Kinetic energy at the bottom of the cliff = potential energy at a height h

$\frac{1}{2}mv^2=\ mgh$

⇒ $v^2=\ 2gh$

⇒ $h=\ \frac{v^2}{2g}$

⇒ $h=\ \frac{(24.2)^2}{2\times 9.8}$

⇒ $h=\ 29.88\ m$

Hence, the height of the cliff is 29.88 m

5 0

2 years ago

Kiting during a storm. The legend that Benjamin Franklin flew a kite as a storm approached is only a legend — he was neither stu

Dvinal [7]

Answer:

The current is $I = 1.1434*10^{-5}}\ A$

Explanation:

From the question we are told that

The radius of the kite string is $R = 2.02 mm = 0.00202 \ m$

The distance it extended upward is $D = 0.823 km = 823 \ m$

The thickness of the water layer is $d = 0.506 mm = 0.000506 \ m$

The resistivity is $\rho = 159\ \Omega \cdot m$

The potential difference is $V = 186 MV = 186 *10^{6} \ V$

Generally the cross sectional area of the water layer is mathematically represented as

$A = \pi r^2$

Here r is mathematically represented as

$r = [(R + d ) - R]$

=> $r = [(0.00202 + 0.000506 ) - 0.00202]$

=> $r = 0.000506$

=> $A = 3.142 * [0.000506]^2$

=> $A = 8.0447*10^{-7}\ m^2$

Generally the resistance of the water is mathematically represented as

$R = \frac{\rho * D }{A}$

=> $R = \frac{159 *823 }{8.0447*10^{-7}}$

=> $R = 1.62662 * 10^{11} \ \Omega$

Generally the current is mathematically represented as

$I = \frac{V}{R}$

=> $I = \frac{186 *10^{6} }{1.62662 * 10^{11}}$

=> $I = 1.1434*10^{-5}}\ A$

8 0

2 years ago