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2 years ago

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A cathode ray tube is made of glass with a small amount of some kind of gas in it. It has metal electrodes at each end to pick u

p an electric current. The electrodes are named "positive” and "negative,” which were words used by Benjamin Franklin in the 1700s to describe electricity. A bright ray forms in the gas when an electric current is applied to metal electrodes. In the 1800s, an important scientist suspected that the negatively charged particles in a cathode ray were present in all atoms. What procedure would allow that scientist to investigate this idea

2 answers:

Semenov [28]2 years ago

7 0

Answer:

Option B.

Making cathode ray tubes out of different materials to see if the ray is the same. (edg2020)

LuckyWell [14K]2 years ago

6 0

Scientists could investigate this idea by making cathode ray tubes out of different materials to see if the ray was the same.

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The variation in the pressure of helium gas, measured from its equilibrium value, is given by ΔP = 2.9 × 10−5 cos (6.20x − 3 000

nadya68 [22]

Answer:

The wavelength of this wave is 1.01 meters.

Explanation:

The variation in the pressure of helium gas, measured from its equilibrium value, is given by :

$\Delta P=2.9\times 10^{-5}\ cos(6.2x-3000t)$ ..............(1)

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$\Delat P=P_o\ cos(kx-\omega t)$ ...........(2)

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3 years ago

A torque of 36.5 N · m is applied to an initially motionless wheel which rotates around a fixed axis. This torque is the result

vivado [14]

Answer:

$21.6\ \text{kg m}^2$

$3.672\ \text{Nm}$

$54.66\ \text{revolutions}$

Explanation:

$\tau$ = Torque = 36.5 Nm

$\omega_i$ = Initial angular velocity = 0

$\omega_f$ = Final angular velocity = 10.3 rad/s

t = Time = 6.1 s

I = Moment of inertia

From the kinematic equations of linear motion we have

$\omega_f=\omega_i+\alpha_1 t\\\Rightarrow \alpha_1=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha_1=\dfrac{10.3-0}{6.1}\\\Rightarrow \alpha_1=1.69\ \text{rad/s}^2$

Torque is given by

$\tau=I\alpha_1\\\Rightarrow I=\dfrac{\tau}{\alpha_1}\\\Rightarrow I=\dfrac{36.5}{1.69}\\\Rightarrow I=21.6\ \text{kg m}^2$

The wheel's moment of inertia is $21.6\ \text{kg m}^2$

t = 60.6 s

$\omega_i$ = 10.3 rad/s

$\omega_f$ = 0

$\alpha_2=\dfrac{0-10.3}{60.6}\\\Rightarrow \alpha_1=-0.17\ \text{rad/s}^2$

Frictional torque is given by

$\tau_f=I\alpha_2\\\Rightarrow \tau_f=21.6\times -0.17\\\Rightarrow \tau=-3.672\ \text{Nm}$

The magnitude of the torque caused by friction is $3.672\ \text{Nm}$

Speeding up

$\theta_1=0\times t+\dfrac{1}{2}\times 1.69\times 6.1^2\\\Rightarrow \theta_1=31.44\ \text{rad}$

Slowing down

$\theta_2=10.3\times 60.6+\dfrac{1}{2}\times (-0.17)\times 60.6^2\\\Rightarrow \theta_2=312.03\ \text{rad}$

Total number of revolutions

$\theta=\theta_1+\theta_2\\\Rightarrow \theta=31.44+312.03=343.47\ \text{rad}$

$\dfrac{343.47}{2\pi}=54.66\ \text{revolutions}$

The total number of revolutions the wheel goes through is $54.66\ \text{revolutions}$ .

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2 years ago