A wave is a disturbance that transmits energy from one location to another without transmitting matter with it.

Un lector de DVD, la velocidad de giro es de 5400 rpm. determina el valor velocidad angular en rad/s,la frecuencia y el periodo

zubka84 [21]

Responder:

A) ω = 565.56 rad / seg

B) f = 90Hz

C) 0.011111s

Explicación:

Dado que:

Velocidad = 5400 rpm (revolución por minuto)

La velocidad angular (ω) = 2πf

Donde f = frecuencia

ω = 5400 rev / minuto

1 minuto = 60 segundos

2πrad = I revolución

Por lo tanto,

ω = 5400 * (rev / min) * (1 min / 60s) * (2πrad / 1 rev)

ω = (5400 * 2πrad) / 60 s

ω = 10800πrad / 60 s

ω = 180πrad / seg

ω = 565.56 rad / seg

SI)

Dado que :

ω = 2πf

donde f = frecuencia, ω = velocidad angular en rad / s

f = ω / 2π

f = 565.56 / 2π

f = 90.011669

f = 90 Hz

C) Periodo (T)

Recordar T = 1 / f

Por lo tanto,

T = 1/90

T = 0.0111111s

3 0

2 years ago

A physicist is constructing a solenoid. She has a roll of insulated copper wire and a power supply. She winds a single layer of

Leni [432]

Answer:

P =105.44 W

Explanation:

Given that

D= 10 cm ,L= 60 cm

d= 0.1 cm ,B= 6.4 mT

ρ= 1.7 x 10⁻⁸ Ω · m

The number of turns N

N= L/d

N= 60/0.1 = 600 turns

Length of wire

Lc= πDN

Lc= 3.14 x 0.1 x 600

Lc=188.4 m

The magnetic filed given as

$B=\dfrac{\mu_oNI}{L}$

$I=\dfrac{BL}{\mu_oN}$

Now by putting the values

$I=\dfrac{0.0064\times 0.6}{4\pi \times 10^{-7}\times 600}$

I=5.09 A

The resistance R given as

$R=\dfrac{\rho L_c}{A}$

$A=\dfrac{\pi}{4}\times d^2$

$A=\dfrac{\pi}{4}\times (0.1\times 10^{-2})^2$

$R=\dfrac{1.7\times 10^{-8} \times 188.4}{\dfrac{\pi}{4}\times (0.1\times 10^{-2})^2}$

R=4.07 Ω

Power P

p =I²R

P= 5.09² x 4.07 W

P =105.44 W

5 0

2 years ago

A force f = bx 3 acts in the x direction, where the value of b is 3.7 n/m3. how much work is done by this force in moving an obj

Elodia [21]

The answer would be 21.6 but rounded up it would be 22J.

8 0

2 years ago

A uniform sphere with mass M and radius R is rotating with angular speed ω1 about a frictionless axle along a diameter of the sp

liq [111]

Answer:

$W_2=\sqrt{\frac{3}{5} }W_1$

Explanation:

For the first ball, the moment of inertia and the kinetic energy is:

$I_1 =\frac{2}{5}MR^2$

$K_1 = \frac{1}{2}IW_1^2$

So, replacing, we get that:

$K_1 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2$

At the same way, the moment of inertia and kinetic energy for second ball is:

$I_2 =\frac{2}{3}MR^2$

$K_2 = \frac{1}{2}IW_2^2$

So:

$K_2 = \frac{1}{2}(\frac{2}{3}MR^2)W_2^2$

Then, $K_2$ is equal to $K_1$ , so:

$K_2 = K_1$

$\frac{1}{2}(\frac{2}{3}MR^2)W_2^2 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2$

$\frac{1}{3}MR^2W_2^2 = \frac{1}{5}MR^2W_1^2$

$\frac{1}{3}W_2^2 = \frac{1}{5}W_1^2$

Finally, solving for $W_2$ , we get:

$W_2=\sqrt{\frac{3}{5} }W_1$

5 0

2 years ago

A cart, which has a mass of 2.30 kg is sitting at the top of an inclined plane, which is 4.50 meters long and meets the horizont

expeople1 [14]

Answer:

a) The gravitational potential energy before the cart rolls down the incline is 24.6 J.

b) The magnitude of the force that causes the cart to roll down is 5.47 N.

c) The acceleration of the cart is 2.38 m/s²

d) It takes the cart 1.94 s to reach the bottom of the incline.

e) The velocity of the cart at the bottom of the inclined plane is 4.62 m/s.

f) The kinetic energy of the cart as it reaches the bottom of the incline is 24.6 J.

g) The work done by the gravitational force is 24.6 J.

Explanation:

Hi there!

a) The gravitational potential energy is calculated using the following equation:

EP = m · g · h

Where:

EP = gravitational potential energy.

m = mass of the object.

g = acceleration due to gravity.

h = height at which the object is located.

The height of the inclined plane can be calculated using trigonomoetry:

sin 14.0° = height / lenght

sin 14.0° = height / 4.50 m

4.50 m · sin 14.0° = height

height = 1.09 m

Then, the gravitational potential energy will be:

EP = m · g · h

EP = 2.30 kg · 9.81 m/s² · 1.09 m = 24.6 J

The gravitational potential energy before the cart rolls down the incline is 24.6 J.

b) Please, see the attached figure for a graphical description of the problem and the forces acting on the cart. The force that causes the cart to accelerate down the incline is the horizontal component of the weight (Fwx in the figure). The magnitude of this force can be obtained using trigonometry:

sin 14° = Fwx / Fw

The weight of the cart (Fw) is calculated as follows:

Fw = m · g

Fw = 2.30 kg · 9.81 m/s²

Fw = 22.6 N

Then, the x-component of the weight will be:

FW · sin 14° = Fwx

22.6 N · sin 14° = Fwx

Fwx = 5.47 N

The magnitude of the force that causes the cart to roll down is 5.47 N.

c)Using the equation of Fwx we can calculate the acceleration of the cart:

Fwx = m · a

Where "m" is the mass of the cart and "a" is the acceleration.

Fwx / m = a

5.47 N / 2.30 kg = a

a = 2.38 m/s²

The acceleration of the cart is 2.38 m/s²

d) To calculate the time it takes the cart to reach the bottom of the incline, let´s use the equation of position of the cart:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the cart at time t.

x0 = initial position.

v0 = initial velocity.

a = acceleration.

t = time.

Considering the initial position as the point at which the cart starts rolling (x0 = 0) and knowing that the cart starts from rest (v0 = 0), let´s find the time it takes the cart to travel the 4.50 m of the inclined plane:

x = 1/2 · a · t²

4.50 m = 1/2 · 2.38 m/s² · t²

2 · 4.50 m / 2.38 m/s² = t²

t = 1.94 s

It takes the cart 1.94 s to reach the bottom of the incline.

e) The velocity of the cart at the bottom of the inclined plane can be obtained using the following equation:

v = v0 + a · t

v = 0 m/s + 2.38 m/s² · 1.94 s

v = 4.62 m/s

The velocity of the cart at the bottom of the inclined plane is 4.62 m/s.

f) The kinetic energy can be calculated using the following equation:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass of the cart.

v = velocity of the cart.

KE = 1/2 · 2.30 kg · (4.62 m/s)²

KE = 24.6 J

The kinetic energy of the cart as it reaches the bottom of the incline is 24.6 J.

The gain of kinetic energy is equal to the loss of gravitational potential energy.

g) The work done by the gravitational force can be calculated using the work-energy theorem: the work done by the gravitational force is equal to the negative change in the gravitational potential energy:

W = -ΔPE

W = -(final potential energy - initial potential energy)

W = -(0 - 24.6 J)

W = 24.6 J

This can also be calculated using the definition of work:

W = Fw · d

Where "d" is the distance traveled in the direction of the force, that is the height of the inclined plane:

W = 22.6 N · 1.09 m = 24.6 J.

The work done by the gravitational force is 24.6 J.

4 0

2 years ago