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balandron [24]
3 years ago
10

Irene was investigating the behavior of gases at certain conditions. Which would most likely cause an increase in the rate of co

llisions between gas molecules in a container?
Physics
1 answer:
kipiarov [429]3 years ago
6 0

Answer:

Increase in temperature

Increase in pressure

Explanation:

For the rate of collisions to increase, It can happen in various ways

1. The gasses can be made to move faster: This can be done by increasing their temperature. An increase in the temperature of the gases directly relates to a corresponding increase in their kinetic energy. This means that the molecules of the gases can move with a greater velocity, which will increase collision rates.

2. The volume of the container can be reduced: Consider this; If the container is made smaller, the molecules of the gases will have less space to move about. This means that they will be bumping into the walls of the container more frequently. This reduction in volume leads to an increase in the pressure of the gas. This is exactly what happens in a piston-cylinder assembly

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Q1 is located at the origin, Q2 is located at x = 2.50 cm and Q3 is located at x = 3.50 cm. Q1 has a charge of +4.92μC and Q3 ha
Inessa05 [86]

Answer:

+1.11\mu C

Explanation:

A charge located at a point will experience a zero electrostatic force if the resultant electric field on it due to any other charge(s) is zero.

Q_1 is located at the origin. The net force on it will only be zero if the resultant electric field intensity due to Q_2 and Q_3 at the origin is equal to zero. Therefore we can perform this solution without necessarily needing the value of Q_1.

Let the electric field intensity due to Q_2 be +E_2 and that due to Q_3 be -E_3 since the charge is negative. Hence at the origin;

+E_2-E_3=0..................(1)

From equation (1) above, we obtain the following;

E_2=E_3.................(2)

From Coulomb's law the following relationship holds;

+E_2=\frac{kQ_2}{r_2^2}\\  

-E_3=\frac{kQ_3}{r_3^2}

where r_2 is the distance of Q_2 from the origin, r_3 is the distance of Q_3 from the origin and k is the electrostatic constant.

It therefore means that from equation (2) we can write the following;

\frac{kQ_2}{r_2^2}=\frac{kQ_3}{r_3^2}.................(3)

k can cancel out from both side of equation (3), so that we finally obtain the following;

\frac{Q_2}{r_2^2}=\frac{Q_3}{r_3^2}................(4)

Given;

Q_2=?\\r_2=2.5cm=0.025m\\Q_3=-2.18\mu C=-2.18* 10^{-6}C\\r_3=3.5cm=0.035m

Substituting these values into equation (4); we obtain the following;

\frac{Q_2}{0.025^2}=\frac{2.18*10^{-6}}{0.035^2}\\\\hence;\\\\Q_2=\frac{0.025^2*2.18*10^{-6}}{0.035^2}\\

Q_2=\frac{0.00136*10^{-6}}{0.00123}=1.11*10^{-6}C\\\\Q_3=+1.11\mu C

6 0
3 years ago
A hot air balloon is traveling vertically upward at a constant speed of 4.5 m/s. When it is 28 m above the ground, a package is
ella [17]

Answer:

1.97 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

s=ut+\frac{1}{2}at^2\\\Rightarrow 21=4.5t+\frac{1}{2}\times -9.8\times t^2\\\Rightarrow 21=-4.5t-4.9t^2\\\Rightarrow 4.9t^2+4.5t-28=0\\\Rightarrow 49t^2+45t-280=0

Solving the above equation we get

t=\frac{-45+\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}, \frac{-45-\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}\\\Rightarrow t=1.97, -2.89

So, the time the package was in the air is 1.97 seconds

3 0
3 years ago
Compare and contrast climate and weather
Levart [38]
It’s 49% of its original gig hope this helps!
8 0
3 years ago
Read 2 more answers
Approximately how much air is in a column 1-cm2 in cross section that extends from sea level to the top of the atmosphere?
laila [671]

Approximately 101 N air is in a column 1-cm2 in cross-section that extends from sea level to the top of the atmosphere

The basic level for determining height and depth on Earth is the sea level. The ocean's surface tends to seek the same level since it is one continuous body of water. However, the sea level is never fully level due to winds, currents, river discharges, and changes in gravity and temperature.

At the equator, the radius of the Earth at sea level is 6378.137 km (3963.191 mi). At the poles, it is 6,356.752 km (3,949.903 km), and on average, it is 6,371.001 km (3,958.756 mi). The elevation of the shoreline—the boundary between the ocean and the land—is referred to as sea level. Land that is higher than this altitude is above sea level, and land that is lower is below sea level.

To learn more about sea level please visit -
brainly.com/question/2113249
#SPJ4

8 0
2 years ago
A 0.2 kg block sliding on a horizontal table slows down from 25 m/s to 20 m/s. How much energy does the block lose due to fricti
Papessa [141]

Answer:

the kinetic energy lost due to friction is 22.5 J

Explanation:

Given;

mass of the block, m = 0.2 kg

initial velocity of the block, u = 25 m/s

final velocity of the block, v = 20 m/s

The kinetic energy lost due to friction is calculated as;

\Delta K.E= K.E_f - K.E_i\\\\\Delta K.E= \frac{1}{2}mv^2 -  \frac{1}{2}mu^2\\\\\Delta K.E= \frac{1}{2}m(v^2 -u^2)\\\\\Delta K.E= \frac{1}{2} \times 0.2 (20^2 - 25^2)\\\\\Delta K.E= -22.5 \ J

Therefore, the kinetic energy lost due to friction is 22.5 J

7 0
3 years ago
Read 2 more answers
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