1. What force is needed to accelerate the stone of

Why did thomson’s results from experimenting with cathode rays cause a big change in scientific thought about atoms?

natima [27]

Answer:

Explanation:

Before Thomson's discovery, atoms were believed according to the "Dalton's atomic theory" to be the smallest indivisible particle of any matter. This makes atoms the smallest unit of a matter.

Thomson in 1897, used the discharge tube to discover cathode rays which are today called electrons.

The discovery of electrons provided more light into the structure and nature of atoms. Atoms were now being seen in a different light as particles that are made up of other smaller sized particles.

Thomson through his experiment was able determine perfectly well the nature of the rays he saw emanating from the cathode. One of his findings shows that the rays are negatively charged and are repelled by negative charges.

The discovery of electrons further led to more works on the atom and other particles were discovered. Atoms were no longer seen as indivisible or the smallest particles of matter.

3 0

3 years ago

Question 11

notka56 [123]

The volume of the gas once it reaches the surface of water is 2 liters.

The volume of the air in balloon at depth of 100ft (30m) is 500ml.

The pressure at this point is 4 atm.

Assuming that the balloon have no compression by rubber of balloon the volume of air at the water surface is V.

The pressure at the surface of water is 1 atms.

As we know, from the ideal gas equation,

PV = nRT

Where,

P is the Pressure of gas,

V is the volume of the gas,

n is the number of moles,

R is the gas constant whose value is 0.082057 L atm mol-1 K-1,

T is the temperature.

Assuming that the temperature is constant,

We know,

PV = nRT

All quantities on the right side are constants,

So, we can write,

P₁V₁ = P₂V₂

Putting all the the values,

4(0.5) = 1V₂

V₂ = 2 Liters.

The volume of the air at the surface is 2 liters.

1 liter = 0.001 m

Hence,

2 liters = 0.002 m³

So the volume of air at the surface is 0.002m³.

To know more about Ideal gas Equation, visit,

brainly.com/question/20348074

#SPJ9

3 0

1 year ago

A ball is thrown up into the air with an initial velocity of 18 m/s. A) How high does the ball go? B) Calculate the time needed

kaheart [24]

Answer:

B) t = 1.83 [s]

A) y = 16.51 [m]

Explanation:

To solve this problem we must use the following equation of kinematics.

$v_{f} =v_{o} -g*t$

where:

Vf = final velocity = 0

Vo = initial velocity = 18 [m/s]

g = gravity acceleration = 9.81 [m/s²]

t = time [s]

Note: the negative sign in the above equation means that the acceleration of gravity is acting in the opposite direction to the motion.

A) The maximum height is reached when the final velocity of the ball is zero.

0 = 18 - (9.81*t)

9.81*t = 18

t = 18/9.81

t = 1.83 [s], we found the answer for B.

Now using the following equation.

$y = y_{o} + v_{o}*t - 0.5*g*t^{2}\\$

where:

y = elevation [m]

Yo = initial elevation = 0

y = 18*(1.83) - 0.5*9.81*(1.83)²

y = 16.51 [m]

7 0

3 years ago

When measuring an unknown voltage with an analog VOM, you should first

Soloha48 [4]

If you have no idea what the voltage is that you're about to measure,
then you should set the meter to the highest range before you connect
it to the two points in the circuit.

Analog meters indicate the measurement by moving a physical needle
across a physical card with physical numbers printed on it. If the unknown
voltage happens to be 100 times the full range to which the meter is set,
then the needle may find itself trying to move to a position that's 100 times
past the highest number on the meter's face. You'll hear a soft 'twang',
followed by a louder 'CLICK'. Then you'll wonder why the meter has no
needle on it, and then you'll walk over to the other side of the room and
pick up the needle off the floor, and then you'll probably put the needle
in your pocket. That will end your voltage measurements for that day,
and certainly for that meter.

Been there.
Done that.

8 0

3 years ago

Read 2 more answers

A child has an ear canal that is 1.3 cm long. Assume the speed of sound is v = 344 m/s.

kap26 [50]

Answer:

The frequencies are $(f, f_1) = (6615.4 \ Hz , 19846.2\ Hz)$

Explanation:

From the question we are told that

The length of the ear canal is $l = 1.3 \ cm =\frac{1.3}{100} = 0.013 \ m$

The speed of sound is assumed to be $v_s = 344 \ m/s$

Now taking look at a typical ear canal we see that we assume it is a closed pipe

Now the fundamental harmonics for the pipe(ear canal) is mathematically represented as

$f = \frac{v_s}{4 * l }$

substituting values

$f = \frac{344}{4 * 0.013 }$

$f = 6615.4 \ Hz$

Also the the second harmonic for the pipe (ear canal) is mathematically represented as

$f_1 = \frac{3v_s}{4 * l}$

substituting values

$f_1 = \frac{3 * 344}{4 * 0.013}$

$f_1 = 19846.2 \ Hz$

Given that sound would be loudest in the pipe at the frequency, it implies that the child will have an increased audible sensitivity at this frequencies

6 0

3 years ago