What forms when acid reacts with a base? please help

El motor de una licuadora gira a 3600 rpm, disminuye su velocidad angular hasta 2000 rpm realizando 120 vueltas. Calcular: a) La

allsm [11]

Answer:

a) α = -65,2 rad/s².

b) t = 2,57 s.

Explanation:

a) La aceleración angular se puede calcular usando la siguiente ecuación:

$\omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \theta$

En donde:

$\omega_{f}$ : es la velocidad angular final = 2000 rpm = 209,4 rad/s

$\omega_{0}$ : es la velocidad angular inicial = 3600 rpm = 377,0 rad/s

α: es la aceleración angular=?

θ: es el desplazamiento o número de vueltas = 120 rev = 754,0 rad

Las conversiones de unidades se hicieron sabiendo que 1 revolución = 2π radianes y que 1 minuto = 60 segundos.

Resolviendo la ecuación (1) para α, tenemos:

$\alpha = \frac{\omega_{f}^{2} - \omega_{0}^{2}}{2\theta} = \frac{(209,4 rad/s)^{2} - (377,0 rad/s)^{2}}{2*754,0 rad} = -65,2 rad/s^{2}$

Entonces, la aceleración angular es -65,2 rad/s². El signo negativo se debe a que el motor está desacelerando.

b) El tiempo transcurrido se puede encontrar como sigue:

$\omega_{f} = \omega_{0} + \alpha t$

Resolviendo para t, tenemos:

$t = \frac{\omega_{f} - \omega_{0}}{\alpha} = \frac{209,4 rad/s - 377,0 rad/s}{-65,3 rad/s^{2}} = 2,57 s$

Por lo tanto, el tiempo transcurrido fue 2,57 s.

Espero que te sea de utilidad!

3 0

3 years ago

A gold wire has a cross-sectional area of 1.0 cm^2 and a resistivity of 2.8 × 10^-8 Ω ∙ m at 20°C. How long would it have to be

Karo-lina-s [1.5K]

Answer:

Length, l = 3.57 meters

Explanation:

It is given that,

Area of cross-section of gold wire, $A=1\ cm^2=0.0001\ m^2$

Resistivity of gold wire, $\rho=2.8\times 10^{-8}\ \Omega-m$

Resistance, R = 0.001 ohms

Resistance in terms of length and area is given by :

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

$l=\dfrac{RA}{\rho}$

$l=\dfrac{0.001\times 0.0001}{2.8\times 10^{-8}}$

l = 3.57 meters

So, the length of the wire is 3.57 meters. Hence, this is the required solution.

4 0

3 years ago

If a honeybee is flying 7.0 m/s, what is the kinetic energy (Joules) if its mass is 0.1 g?

svetlana [45]

Answer:

KINETIC ENERGY=K.E=0.00245 Joules ≈ 0.003 Joules

Explanation:

mass=m=0.1g=0.0001kg

velocity=v=7.0m/s

kinetic energy=K.E.=?

as we know that

$kinetic energy=\frac{1}{2}mv^2\\ putting values\\kinetic energy=\frac{1}{2}(0.0001kg)(7.0m/s)^2\\ kinetic energy=\frac{1}{2} 0.0049joules\\kinetic energy= 0.00245Newton meter or joules$

hope it will help ^_^

5 0

3 years ago

What is the electromagnetic spectrum?

Pavel [41]

What is the electromagnetic spectrum?
The electromagnetic spectrum is the range of frequencies (the spectrum) of electromagnetic radiation and their respective wavelengths and photon energies. The electromagnetic spectrum covers electromagnetic waves with frequencies ranging from below one hertz to above 10²⁵ hertz, corresponding to wavelengths from thousands of kilometers down to a fraction of the size of an atomic nucleus.

How the light affect the color we see?
All of the colors we see are a byproduct of spectrum light, as it is reflected off or absorbed into an object. An object that reflects back all of the rays of light will appear white; an object that absorbs all of the rays, black. All of the millions of other colors are produced by a combination of light rays being absorbed and reflected.

8 0

3 years ago

an object is acted on by a drag force with a magnitude that is proportional to the speed. the object accelerates downward at 3.0

Nutka1998 [239]

The terminal speed of the object falling down is 66.67 m/s.

The terminal speed acquired by the body when,

Weight of the body = Drag force of the body

It is given,

Drag force is directly proportional to the speed,

So,

F = CV

Where F is drag force,

V is the speed,

C is the constant,

So, it can be written as C = F/V.

The weight of the body = mg

The weight of the body = 10m

M is the mass and g is the acceleration due to gravity,

The drag force when the speed is 20m/s.

Drag force = ma

a is the acceleration during the drag force which is given to be 3m/s²,

Drag force = 3m

Now we can write,

F₁/V₁ = F₂/V₂

F₁ is the drag force at 20m/s speed.

F₂ is the weight of the body and V₂ is the terminal speed,

Now, it can be written,

3m/20 = 10m/V₂

V₂ = 66.67 m/s.

So, the terminal speed is 66.67m/s.

To know more about terminal speed, visit,

brainly.com/question/14605362

#SPJ4

7 0

1 year ago