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3 years ago

How would you feel if everyone hated you?

Physics

2 answers:

crimeas [40]3 years ago

8 0

Answer:

awful

Explanation:

can you help me with my question Which of the themes of Hawthorne's "Dr. Heidegger's Experiment" is illustrated by this passage? Paragraph 36: The most singular effect of their gayety was an impulse to mock the infirmity and decrepitude of which they had so lately been the victims. A. Gayety produces an impulse to mock. B. It is a great release to look back on our problems happily C It is human nature to mock or make light of problems we have been delivered from

Advocard [28]3 years ago

5 0

Really sad try not to let it get to you

You might be interested in

A loop circuit has a resistance of R1 and a current of 1.9 A. The current is reduced to 1.5 A when an additional 3.1 Ω resistor

Deffense [45]

Answer:

11.625 Ohm

Explanation:

Let V be the Voltage charge of the loop, as this is constant we know that before the resistor addition the current I is:

V/R1 = 1.9 or V = 1.9R1

After the resistor addition to series R = R1 + 3.1

I = V/R = V/(R1 + 3.1) = 1.5

We can substitute V = 1.9R1

1.9R1 = 1.5R1 + 1.5*3.1

0.4R1 = 4.65

R1 = 4.65/0.4 = 11.625 Ohm

4 0

4 years ago

A 54 kg man holding a 0.65 kg ball stands on a frozen pond next to a wall. He throws the ball at the wall with a speed of 12.1 m

Inessa [10]

Answer:

The velocity of the man is 0.144 m/s

Explanation:

This is a case of conservation of momentum.

The momentum of the moving ball before it was caught must equal the momentum of the man and the ball after he catches the ball.

Mass of ball = 0.65 kg

Mass of the man = 54 kg

Velocity of the ball = 12.1 m/s

Before collision, momentum of the ball = mass x velocity

= 0.65 x 12.1 = 7.865 kg-m/s

After collision the momentum of the man and ball system is

(0.65 + 54)Vf = 54.65Vf

Where Vf is their final common velocity.

Equating the initial and final momentum,

7.865 = 54.65Vf

Vf = 7.865/54.65 = 0.144 m/s

4 0

3 years ago

A snowball is rolling down a hill at 4.5 m/s and accumulating snow as it goes. Its diameter begins at 0.50 m and ends at the bot

Reil [10]

To find the change in centripetal acceleration, you should first look for the centripetal acceleration at the top of the hill and at the bottom of the hill.

The formula for centripetal acceleration is:
Centripetal Acceleration = v squared divided by r

where:
v = velocity, m/s
r= radium, m

assuming the velocity does not change:

at the top of the hill:
centripetal acceleration = (4.5 m/s^2) divided by 0.25 m
= 81 m/s^2

at the bottom of the hill:
centripetal acceleration = (4.5 m/s^2) divided by 1.25 m
= 16.2 m/s^2

to find the change in centripetal acceleration, take the difference of the two.
change in centripetal acceleration = centripetal acceleration at the top of the hill - centripetal acceleration at the bottom of the hill

= 81 m/s^2 - 16.2 m/s^2
= 64.8 m/s^2 or 65 m/s^2

6 0

4 years ago

Charge q1 = +2.00 μC is at -0.500 m along the x axis. Charge q2 = -2.00 μC is at 0.500 m along the x axis. Charge q3 = 2.00 μC i

Kobotan [32]

The magnitude of electrical force on charge $q_{3}$ due to the others is 0.102 newtons.

<h3>How to calculate the electrical force experimented on a particle</h3>

The vector position of each particle respect to origin are described below:

$\vec r_{1} = (-0.500, 0)\,[m]$

$\vec r_{2} = (+0.500, 0)\,[m]$

$\vec r_{3} = (0, +0.500)\,[m]$

Then, distances of the former two particles particles respect to the latter one are found now:

$\vec r_{13} = (+0.500, +0.500)\,[m]$

$r_{13} = \sqrt{\vec r_{13}\,\bullet\,\vec r_{13}} = \sqrt{(0.500\,m)^{2}+(0.500\,m)^{2}}$

$r_{13} =\frac{\sqrt{2}}{2}\,m$

$\vec r_{23} = (-0.500, +0.500)\,[m]$

$r_{23} = \sqrt{\vec r_{23}\,\bullet \,\vec r_{23}} = \sqrt{(-0.500\,m)^{2}+(0.500\,m)^{2}}$

$r_{23} =\frac{\sqrt{2}}{2}\,m$

The resultant force is found by Coulomb's law and principle of superposition:

$\vec R = \vec F_{13}+\vec F_{23}$ (1)

Please notice that particles with charges of same sign attract each other and particles with charges of opposite sign repeal each other.

$\vec R = \frac{k\cdot q_{1}\cdot q_{3}}{r_{13}^{2}}\cdot \vec u_{13} +\frac{k\cdot q_{2}\cdot q_{3}}{r_{23}^{2}}\cdot \vec u_{23}$ (2)

Where:

$k$ - Electrostatic constant, in newton-square meters per square Coulomb.
$q_{1}$ , $q_{2}$ , $q_{3}$ - Electric charges, in Coulombs.
$r_{13}$ , $r_{23}$ - Distances between particles, in meters.
$\vec u_{13}$ , $\vec u_{23}$ - Unit vectors, no unit.

If we know that $k = 8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}$ , $q_{1} = 2\times 10^{-6}\,C$ , $q_{2} = 2\times 10^{-6}\,C$ , $q_{3} = 2\times 10^{-6}\,C$ , $r_{13} =\frac{\sqrt{2}}{2}\,m$ , $r_{23} =\frac{\sqrt{2}}{2}\,m$ , $\vec u_{13} = \left(-\frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2} \right)$ and $\vec u_{23} = \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right)$ , then the vector force on charge $q_{3}$ is:

$\vec R = \frac{\left(8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot (2\times 10^{-6}\,C)\cdot (2\times 10^{-6}\,C)}{\left(\frac{\sqrt{2}}{2}\,m \right)^{2}} \cdot \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) + \frac{\left(8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot (2\times 10^{-6}\,C)\cdot (2\times 10^{-6}\,C)}{\left(\frac{\sqrt{2}}{2}\,m \right)^{2}} \cdot \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right)$