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Drupady [299]
2 years ago
9

On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 103 N/C. Compare the

gravitational and electric forces on a small dust particle of mass 2.2 ✕ 10−15 g that carries a single electron charge. Fg FE = What is the acceleration (both magnitude and direction) of the dust particle? (Enter the magnitude in m/s2.) magnitude m/s2 direction ---Select---
Physics
1 answer:
Nina [5.8K]2 years ago
7 0

Answer:

a) FE = 0.764FG

b) a = 2.30 m/s^2

Explanation:

a) To compare the gravitational and electric force over the particle you calculate the following ratio:

\frac{F_E}{F_G}=\frac{qE}{mg}              (1)

FE: electric force

FG: gravitational force

q: charge of the particle = 1.6*10^-19 C

g: gravitational acceleration = 9.8 m/s^2

E: electric field = 103N/C

m: mass of the particle = 2.2*10^-15 g = 2.2*10^-18 kg

You replace the values of all parameters in the equation (1):

\frac{F_E}{F_G}=\frac{(1.6*10^{-19}C)(103N/C)}{(2.2*10^{-18}kg)(9.8m/s^2)}\\\\\frac{F_E}{F_G}=0.764

Then, the gravitational force is 0.764 times the electric force on the particle

b)

The acceleration of the particle is obtained by using the second Newton law:

F_E-F_G=ma\\\\a=\frac{qE-mg}{m}

you replace the values of all variables:

a=\frac{(1.6*10^{-19}C)(103N/C)-(2.2*10^{-18}kg)(9.8m/s^2)}{2.2*10^{-18}kg}\\\\a=-2.30\frac{m}{s^2}

hence, the acceleration of the particle is 2.30m/s^2, the minus sign means that the particle moves downward.

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A mass of 10 g of oxygen fill a weighted piston–cylinder device at 20 kPa and 110°C. The device is now cooled until the tempe
mezya [45]

Answer:

The change of the volume of the device during this cooling is 14.3\times10^{-3}\ m^3

Explanation:

Given that,

Mass of oxygen = 10 g

Pressure = 20 kPa

Initial temperature = 110°C

Final temperature = 0°C

We need to calculate the change of the volume of the device during this cooling

Using formula of change volume

\Delta V=V_{2}-V_{1}

\Delta V=\dfrac{mR}{P}(T_{2}-T_{1})

Put the value into the formula

\Delta V=\dfrac{0.3125\times0.0821}{2.0265\times10^{9}}(383-273)

\Delta V=14.297\ L

\Delta V=14.3\times10^{-3}\ m^3

Hence, The change of the volume of the device during this cooling is 14.3\times10^{-3}\ m^3

6 0
3 years ago
An object with mass 3.5 kg is attached to a spring with spring stiffness constant k = 270 N/m and is executing simple harmonic m
Elanso [62]

Answer:

Part a)

A = 0.066 m

Part b)

maximum speed = 0.58 m/s

Explanation:

As we know that angular frequency of spring block system is given as

\omega = \sqrt{\frac{k}{m}}

here we know

m = 3.5 kg

k = 270 N/m

now we have

\omega = \sqrt{\frac{270}{3.5}}

\omega = 8.78 rad/s

Part a)

Speed of SHM at distance x = 0.020 m from its equilibrium position is given as

v = \omega \sqrt{A^2 - x^2}

0.55 = 8.78 \sqrt{A^2 - 0.020^2}

A = 0.066 m

Part b)

Maximum speed of SHM at its mean position is given as

v_{max} = A\omega

v_{max} = 0.066(8.78) = 0.58 m/s

4 0
3 years ago
A certain resistance thermometer read 14.5 ohms in pure melting ice and 18.5 ohms in steam at standard atmospheric pressure what
Vadim26 [7]

The resistance of the thermometer at room temperature is 15.04 ohms.

<h3 />

<h3>What is a resistance thermometer?</h3>

A resistance thermometer is a type of thermometer that measures temperature through a change in resistance.

To calculate the resistance of the thermometer at room temperature, we use the formula below.

Formula:

  • 100/27 = 2/(x-14.5)..............Eqquation 1

Where:

  • x = Resistance of the thermometer at room temperature

Make x the subject of the equation

  • x = [(27×2)/100]+14.5
  • x = (54/100)+14.5
  • x = 0.54+14.5
  • x = 15.04 ohms.

Hence, The resistance of the thermometer at room temperature is 15.04 ohms.

Learn more about thermometers here: brainly.com/question/1531442

3 0
2 years ago
Write an essay on basketball history and how the game has changed over time
jarptica [38.1K]

Basket ball has gone through some changes and hence got into shape as what we play or watch nowadays. Overall rules have the same fundamental principles as set in 1891 by the founder of this game.

December 21st, in 1891, the introductory basketball game was played in Springfield, Massachusetts. It was first brought into shape by a Canadian-by birth, Dr. James Naismith. The basic idea of the new game was to keep the sports loving students in shape during the winters or in between the outdoor game seasons.

'The basket ball' which had a set of thirteen basic rules set up a strong foundation to the game and still played almost in the same style with few modifications.In the very year of 1891, after mixing and changing theme of several other played games of those days, basket ball was born.

Initially, basket ball with the 13 rules was played with 2 peach baskets setup as goals, which now a days are the baskets on a high poles although modified but with same basic idea. In the very first game played in Springfield, the set of players were able to score a single point only, in the whole game.

By comparing Naismith’s basketball and basketball of today we see a characteristic of the original game which had no dribbling, instead a player had to pass the ball to a another player of his team ,from his the very spot where he caught it. The second thing which has now changed is the fouls, in the original game if any team made 3 consecutive foul play ,the oponent received a goal in reward. Although this scoring technique doesn't exist in the modern basketball. Rather nowadays, if any team makes five fouls in one quarter, the offending team is in the penalty ,and shoot free throws are awarded against them.

All other rules are quiet intact in the modern play as well. As the game today,has extended to over 2 hundered countries and enjoyed by many on television screens, with modern umpiring and technologies being used in the game, but the original idea of the founder is still endured.

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