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Lapatulllka [165]

3 years ago

8

Two solutions, Solutions A and Solution B, are tested and found to have pH values of 3.0 and 7.0 respectively. Which statement b

elow correctly compares the two solutions?
A. Solution B has an equal hydroxyl ion concentration to Solution A.
B. Solution A has a higher hydroxyl ion concentration than Solution B.
C. Solution A has a higher hydronium ion concentration than Solution B.
D. Solution B has a higher hydronium ion concentration than Solution A.

1 answer:

aleksklad [387]3 years ago

7 0

Answer:

i love your profile picture :3

Explanation:

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An electric motor spins at 1000 rpm and is slowing down at a rate of 10 t rad/s2 ; where t is measured in seconds. (a) If the mo

REY [17]

Answer:

a) The tangential component of acceleration at the edge of the motor at $t = 1.5\,s$ is -1.075 meters per square second.

b) The electric motor will take approximately 3.963 seconds to decrease its angular velocity by 75 %.

Explanation:

The angular aceleration of the electric motor ( $\alpha$ ), measured in radians per square second, as a function of time ( $t$ ), measured in seconds, is determined by the following formula:

$\alpha = -10\cdot t\,\left[\frac{rad}{s^{2}} \right]$ (1)

The function for the angular velocity of the electric motor ( $\omega$ ), measured in radians per second, is found by integration:

$\omega = \omega_{o} - 5\cdot t^{2}\,\left[\frac{rad}{s} \right]$ (2)

Where $\omega_{o}$ is the initial angular velocity, measured in radians per second.

a) The tangential component of aceleration ( $a_{t}$ ), measured in meters per square second, is defined by the following formula:

$a_{t} = R\cdot \alpha$ (3)

Where $R$ is the radius of the electric motor, measured in meters.

If we know that $R = 7.165\times 10^{-2}\,m$ , $\alpha = 10\cdot t$ and $t = 1.5\,s$ , then the tangential component of the acceleration at the edge of the motor is:

$a_{t} = (7.165\times 10^{-2}\,m)\cdot (-10)\cdot (1.5\,s)$

$a_{t} = -1.075\, \frac{m}{s^{2}}$

The tangential component of acceleration at the edge of the motor at $t = 1.5\,s$ is -1.075 meters per square second.

b) If we know that $\omega_{o} = 104.720\,\frac{rad}{s}$ and $\omega = 26.180\,\frac{rad}{s}$ , then the time needed is:

$26.180\,\frac{rad}{s} = 104.720\,\frac{rad}{s}-5\cdot t^{2}$

$5\cdot t^{2} = 104.720\,\frac{rad}{s}-26.180\,\frac{rad}{s}$

$t^{2} = \frac{104.720\,\frac{rad}{s}-26.180\,\frac{rad}{s} }{5}$

$t = \sqrt{\frac{104.720\,\frac{rad}{s}-26.180\,\frac{rad}{s} }{5} }$

$t \approx 3.963\,s$

The electric motor will take approximately 3.963 seconds to decrease its angular velocity by 75 %.

8 0

3 years ago

Einhorn and Finkle are in a convenience store together. For a while they talk about stealing hot dogs and deodorant, for they ar

dangina [55]

We can conclude that Einhorn decides not to steal according to the pre-conventional level where as Finkle decides per the conventional level according to Kohlberg's theory.

Explanation:

Kohlberg's theory is explained in three stages:

The pre-conventional stage where sense of moral is controlled externally by rules of authority figures such as parents and teachers.An action is viewed according to the results obtained after the act.
Conventional stage is when a person's morals are controlled by authority figures and their positive relation to the society norms.
Post-conventional level relies on a person's principles and values where some laws are seen as unjust.

In this case, Einhorn recognizes that if you steal, you might end up being caught and punished because it is not accepted by authority figures and could led to punishment. This is stage one. Finkle recognizes that stealing is unaccepted by the law and people in a society should avoid it.This is stage two.

We can conclude that Einhorn's decision not to steal is influenced by the pre-conventional level where as Finkle's by the conventional level according to Kohlberg's theory of moral development.

Learn More

Kohlberg's theory of moral development : brainly.com/question/8885213

Keywords: store, stealing,afraid, caught, punished, wrong, act, comparing, moral development, motivation, Kohlberg's theory

#LearnwithBrainly

8 0

3 years ago

(b) How far from the surface will a particle go if it leaves the asteroid's surface with a radial speed of 1510 m/s? (c) With wh

makvit [3.9K]

Answers:

(a) $2509.98 m/s$

(b) 397042.215 m

(c) 1917.76 m/s

Explanation:

The question is incomplete, please remember to write the whole question :) However, part (a) is written below:

(a) What is the escape speed on a spherical asteroid whose radius is $700 km$ and whose gravitational acceleration at the surface is $a_{g}=4.5 m/s^{2}$

Knowing this, let's begin:

a) In this part we need to find the escape speed $V_{e}$ on the asteroid:

$V_{e}=\sqrt{\frac{2GM}{R}}$ (1)

Where:

$G$ is the universal gravitational constant

$M$ is the mass of the asteroid

$R=700 km=700(10)^{3} m$ is the radius of the asteroid

On the other hand we know the gravitational acceleration is $a_{g}=4.5 m/s^{2}$ , which is given by:

$a_{g}=\frac{GM}{R^{2}}$ (2)

Isolating $GM$ :

$GM=a_{g}R^{2}$ (3)

Substituting (3) in (1):

$V_{e}=\sqrt{\frac{2a_{g}R^{2}}{R}}=\sqrt{2a_{g}R}$ (4)

$V_{e}=\sqrt{2(4.5 m/s^{2})(700(10)^{3} m)}$ (5)

$V_{e}=2509.98 m/s$ (6) This is the escape velocity

b) In this part we will use the Conservation of mechanical energy principle:

$E_{o}=E_{f}$ (7)

Being:

$E_{o}=K_{o}+U_{o}=\frac{1}{2}m V^{2} - \frac{GMm}{R}$ (8)

$E_{f}=K_{f}+U_{f}=0 - \frac{GMm}{R+h}$ (9)

Where:

$E_{o}$ is the initial mechanical energy

$E_{f}$ is the final mechanical energy

$K_{o}$ is the initial kinetic energy

$K_{f}=0$ is the final kinetic energy

$U_{o}$ is the initial gravitational potential energy

$U_{f}$ is the final gravitational potential energy

$m$ is the mass of the object

$V=1510 m/s$ is the radial speed of the object

$h$ is the distance above the surface of the object

Then:

$\frac{1}{2}m V^{2} - \frac{GMm}{R}=- \frac{GMm}{R+h}$ (10)

Isolating $h$ :

$h=\frac{2 a_{g} R^{2}}{2a_{g}R-V^{2}}-R$ (11)

$h=\frac{2 (4.5 m/s^{2}) (700(10)^{3} m)^{2}}{2(4.5 m/s^{2})(700(10)^{3} m)-(1510 m/s)^{2}}-700(10)^{3} m$ (11)

$h=397042.215 m$ (12) This is the distance above the asteroid's surface

c) We will use the Conservation of mechanical energy principle again, but now the condition is that the object is dropped at a distance $h=981.8 km=981.8(10)^{3} m$ . This means that at the begining the object only has gravitational potential energy and then it has kinetic energy and gravitational potential energy:

$\frac{-GMm}{R+h}=\frac{-GMm}{R}+\frac{1}{2}mV^{2}$ (13)

Isolating $V$ :

$V=\sqrt{2a_{g} R(1-\frac{R}{R+h})}$ (14)

$V=\sqrt{2(4.5 m/s^{2}) (700(10)^{3} m)(1-\frac{700(10)^{3} m}{700(10)^{3} m+981.8(10)^{3} m})}$ (15)

Finally:

$V=1917.76 m/s$

7 0

3 years ago

What are some INDIRECT applications of centripetal force in real life?

Eddi Din [679]

Answer:

The circular turning of roads

Explanation:

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The answer is magnet away from the coil

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