A surfactant with a Hydrophile-

Saatlemati

vodomira [7]

Answer:

what are u asking there isnt a question

3 0

2 years ago

Please list them from top to bottom, for exp like in your response, a,g,q,d. Giving quite some points for it.

Aleks04 [339]

D,f,g,h,i,a,e,c,j. I’m sure that it

7 0

2 years ago

Can anyone help me with 11 and 12 please

dexar [7]

The 1st one is basically B because it will stay in motion with the same speed and in the same direction unless acted on by an unbalanced force and the 2cd one is A because most of is transformed into thermos energy. hope this helps!

7 0

3 years ago

A 49.0 kg wheel, essentially a thin hoop with radius 0.730 m, is rotating at 114 rev/min. It must be brought to a stop in 22.0 s

belka [17]

Explanation:

Mass of the wheel, m = 49 kg

Radius of the hoop, r = 0.73 m

Initial angular speed of the wheel, $\omega_i=114\ rev/min = 11.93\ rad/s$

Final angular speed of the wheel, $\omega_f=0$

Time, t = 22 s

(a) If I is the moment of inertia of the hoop. It is equal to,

$I=mr^2$

$I=49\times (0.73)^2$

$I=26.11\ kg-m^2$

We know that the work done is equal to change in kinetic energy.

$W=\Delta E$

$W=\dfrac{1}{2}I(\omega_f^2-\omega_i^2)$

$W=-\dfrac{1}{2}\times 26.11\times (11.93^2)$

W = -1858.05 Joules

(b) Let P is the average power. It is given by :

$P=\dfrac{W}{t}$

$P=\dfrac{1858.05\ J}{22\ s}$

P =84.45 watts

Hence, this is the required solution.

4 0

3 years ago

A 0.30-kg object connected to a light spring with a force constant of 22.6 N/m oscillates on a frictionless horizontal surface.

gtnhenbr [62]

Answer:

(a) vmax = 0.34m/s

(b) v = 0.13m/s

(c) v = 0.31m/s

(d) x = 0.039m

Explanation:

Given information about the spring-mass system:

m: mass of the object = 0.30kg

k: spring constant = 22.6 N/m

A: amplitude of the motion = 4.0cm = 0.04m

(a) The maximum speed of the object is given by the following formula:

$v_{max}=\omega A$ (1)

w: angular frequency of the motion.

The angular frequency is calculated with the following relation:

$\omega=\sqrt{\frac{k}{m}}$ (2)

You replace the expression (2) into the equation (1) and replace the values of the parameters:

$v_{max}=\sqrt{\frac{k}{m}}A=\sqrt{\frac{22.6N/m}{0.30kg}}(0.04m)=0.34\frac{m}{s}$

The maximum speed of the object is 0.34 m/s

(b) If the object is compressed 1.5cm the amplitude of its motion is A = 0.015m, and the maximum speed is:

$v_{max}=\sqrt{\frac{22.6N/m}{0.30kg}}(0.015m)=0.13\frac{m}{s}$

The speed is 0.13m/s

(c) To find the speed of the object when it passes the point x=1.5cm, you first take into account the equation of motion:

$x=Acos(\omega t)$

You solve the previous equation for t:

$t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\\omega=\sqrt{\frac{22.6N/m}{0.30kg}}=8.67\frac{rad}{s}\\\\t=\frac{1}{8.67}cos^{-1}(\frac{1.5cm}{4.0cm})=0.13s$

With this value of t, you can calculate the speed of the object with the following formula:

$v=\omega Asin(\omega t)\\\\v=(8.67rad/s)(0.04m)sin((8.67rad/s)(0.13s))=0.31\frac{m}{s}$

The speed of the object for x = 1.5cm is v = 0.31 m/s

(d) To calculate the values of x on which v is one-half the maximum speed, you first calculate the time t:

$\frac{v_{max}}{2}=\omega A sin(\omega t)\\\\t=\frac{1}{\omega}sin^{-1}(\frac{v_{max}}{2\omega A})\\\\t=\frac{1}{8.67rad/s}sin^{-1}(\frac{0.13m/s}{2(8.67rad/s)(0.04m)})=0.021s$

The position will be:

$x=Acos(\omega t)=0.04mcos((8.67rad/s)(0.021s))=0.039m$

The position of the object on which its speed is one-half its maximum velocity is 0.039

5 0

3 years ago