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kakasveta [241]
2 years ago
3

In the late 19th century, great interest was directed toward the study of electrical discharges in gases and the nature of so-ca

lled cathode rays. One remarkable series of experiments with cathode rays, conducted by J. J. Thomson around 1897, led to the discovery of the electron.
With the idea that cathode rays were charged particles, Thomson used a cathode-ray tube to measure the ratio of charge to mass, q/m, of these particles, repeating the measurements with different cathode materials and different residual gases in the tube.

Part A

What is the most significant conclusion that Thomson was able to draw from his measurements?

a. He found a different value of q/m for different cathode materials.
b. He found the same value of q/m for different cathode materials.
c. From measurements of q/m he was able to calculate the charge of an electron.
d. From measurements of q/m he was able to calculate the mass of an electron.

Part B

What is the distance Δy between the two points that you observe? Assume that the plates have length d, and use e and m for the charge and the mass of the electrons, respectively.

Express your answer in terms of e, m, d, v0, L, and E0.

Part C

Now imagine that you place your entire apparatus inside a region of magnetic field of magnitude B0 (Figure 2) . The magnetic field is perpendicular to E⃗ 0 and directed straight into the plane of the figure. You adjust the value of B0 so that no deflection is observed on the screen.

What is the speed v0 of the electrons in this case?

Express your answer in terms of E0 and B0.

Part D

In your experiment, you measure a total deflection of 4.12 cm when an electric field of 1.10×103V/m is established between the plates (with no magnetic field present). When you add the magnetic field as described in Part C, to what value do you have to adjust its magnitude B0 to observe no deflection?

Assume that the plates are 6.00 cm long and that the distance between them and the screen is 12.0 cm.

Express your answer numerically in tesla.
Physics
1 answer:
IRINA_888 [86]2 years ago
4 0

Answer:

b. He found the same value of q/m for different cathode materials

Explanation:

The correct answer is b. He found the same value of q/m for different cathode materials.

As stated in the question  Thomson thought the cathode rays were charged particle. From his experiments he was able to show that the particles in his cathode rays were negatively charged.

He went on to calculate the ratio q/m (charge to mass) by measuring amounts of delflection of the cathode beams.

By utilizing different cathode materials and different gases, he found that q/m was independent of both suggesting that the particles were fundamental and the same for all atoms, as   the electron is a an elementary fundamental particle.

As you can conclude a. is false since it is the opposite of the experimental result.

c, d are also false.

Thomson found the ratio q/m, and it was until  MIlikans oil dro experiment  experiment that the determination of the charge of the electron was made.

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Which of the following best describes the dependant variable?
weeeeeb [17]
Hello!

The independent variable is the variable deliberately changed.

The dependent variable is the variable that responds to change. So the answer is A.

Hope this helps. Any questions please just ask! Thank you!
5 0
2 years ago
A Honda Civic travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by
aleksklad [387]

a) Average velocity: 2.8 m/s

b) Average velocity: 5.2 m/s

c) Average velocity: 7.6 m/s

Explanation:

a)

The position of the car as a function of time t is given by

x(t)=\alpha t^2 - \beta t^3

where

\alpha = 1.50 m/s^2

\beta = 0.05 m/s^3

The average velocity is given by the ratio between the displacement and the time taken:

v=\frac{\Delta x}{\Delta t}

The position at t = 0 is:

x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0

The position at t = 2.00 s is:

x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m

So the displacement is

\Delta x = x(2)-x(0)=5.6-0=5.6 m

The time interval is

\Delta t = 2.0 s - 0 s = 2.0 s

And so, the average velocity in this interval is

v=\frac{5.6 m}{2.0 s}=2.8 m/s

b)

The position at t = 0 is:

x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0

While the position at t = 4.00 s is:

x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m

So the displacement is

\Delta x = x(4)-x(0)=20.8-0=20.8 m

The time interval is

\Delta t = 4.0 - 0 = 4.0 s

So the average velocity here is

v=\frac{20.8}{4.0}=5.2 m/s

c)

The position at t = 2 s is:

x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m

While the position at t = 4 s is:

x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m

So the displacement is

\Delta x = 20.8 - 5.6 = 15.2 m

While the time interval is

\Delta t = 4.0 - 2.0 = 2.0 s

So the average velocity is

v=\frac{15.2}{2.0}=7.6 m/s

Learn more about average velocity:

brainly.com/question/8893949

brainly.com/question/5063905

#LearnwithBrainly

6 0
2 years ago
An ant is crawling along a yardstick that is pointed with the 0-inch mark to the east and the 36-inch mark to the west. It start
FrozenT [24]

Answer:

  • The total distance traveled is 28 inches.
  • The displacement is 2 inches to the east.

Explanation:

Lets put a frame of reference in the problem. Starting the frame of reference at the point with the 0-inch mark, and making the unit vector \hat{i} pointing in the west direction, the ant start at position

\vec{r}_0 = 16 \ inch \ \hat{i}

Then, moves to

\vec{r}_1 = 29 \ inch \ \hat{i}

so, the distance traveled here is

d_1 = |\vec{r}_1 - \vec{r}_0  | = | 29 \ inch   \ \hat{i} - 16 \ inch   \ \hat{i}  |

d_1 =  | 13 \ inch   \ \hat{i}  |

d_1 =  13 \ inch

after this, the ant travels to

\vec{r}_2 = 14 \ inch \ \hat{i}

so, the distance traveled here is

d_2 = |\vec{r}_2 - \vec{r}_1  | = | 14 \ inch   \ \hat{i} - 29 \ inch   \ \hat{i}  |

d_2 =  | - 15 \ inch   \ \hat{i}  |

d_2 =  15 \ inch

The total distance traveled will be:

d_1 + d_2 = 13 \ inch + 15 \ inch = 28 \ inch

The displacement is the final position vector minus the initial position vector:

\vec{D}=\vec{r}_2 - \vec{r}_1

\vec{D}= 14 \ inch   \ \hat{i} - 16 \ inch \ \hat{i}

\vec{D}= - 2 \ inch \ \hat{i}

This is 2 inches to the east.

6 0
2 years ago
A shuttle on Earth has a mass of 4.5 E 5 kg. Compare its weight on Earth to its weight while in orbit at a height of 6.3 E 5 met
faltersainse [42]

Answer:

83%

Explanation:

On the surface, the weight is:

W = GMm / R²

where G is the gravitational constant, M is the mass of the Earth, m is the mass of the shuttle, and R is the radius of the Earth.

In orbit, the weight is:

w = GMm / (R+h)²

where h is the height of the shuttle above the surface of the Earth.

The ratio is:

w/W = R² / (R+h)²

w/W = (R / (R+h))²

Given that R = 6.4×10⁶ m and h = 6.3×10⁵ m:

w/W = (6.4×10⁶ / 7.03×10⁶)²

w/W = 0.83

The shuttle in orbit retains 83% of its weight on Earth.

4 0
2 years ago
A sample of water is heated at a constant pressure of one atmosphere. Initially, the sample is ice at 260 K, and at the end the
USPshnik [31]

In <u>370 K to 375 K </u>temperature intervals of 5 K, would be the greatest increase in the entropy of the sample.

Option: C

<u>Explanation</u>:

Because the largest difference in molar entropy occurs when a condensed phase (solid/liquid) transforms to the gas phase. Then change in entropy is equal to heat transfer divided by temperature: \Delta \mathrm{S}=\frac{\Delta Q}{\mathrm{m} T}.

According to given ice sample at 260 K, when this solid sample start converting into liquid sample it will gain positive temperature and steam will take place near 373 K (273 K ice temperature + 100^{\circ} \mathrm{C} temperature of boiling water). Therefore it’s very obvious that greatest increase in entropy will occur during 370 K – 375 K.

5 0
2 years ago
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