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omeli [17]
3 years ago
12

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
Physics
2 answers:
Semenov [28]3 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]3 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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2 ways weak nuclear forces and gravity are alike <br><br> PLEASE HELP ):
shtirl [24]
They are attractive
They don’t depend on charge
7 0
3 years ago
What the unit of work?​
ra1l [238]

Answer:

yes

Explanation:

5 0
2 years ago
A 10 g particle undergoes SHM with an amplitude of 2.0 mm and a maximum acceleration of magnitude 8.0 multiplied by 103 m/s2, an
Nat2105 [25]

Answer:

a)T=0.0031416s

b)v_{max}=6.283\frac{m}{s}

c) E=0.1974J

d)F=80N

e)F=40N

Explanation:

1) Important concepts

Simple harmonic motion is defined as "the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's Law (F=-Kx). The motion experimented by the particle is sinusoidal in time and demonstrates a single resonant frequency".

2) Part a

The equation that describes the simple armonic motion is given by X=Acos(\omega t +\phi)    (1)

And taking the first and second derivate of the equation (1) we obtain the velocity and acceleration function respectively.

For the velocity:

\frac{dX}{dt}=v(t)=-A\omega sin(\omega t +\phi)   (2)

For the acceleration

\frac{d^2 X}{dt}=a(t)=-A\omega^2 cos(\omega t+\phi)   (3)

As we can see in equation (3) the acceleration would be maximum when the cosine term would be -1 and on this case:

A\omega^2=8x10^{3}\frac{m}{s^2}

Since we know the amplitude A=0.002m  we can solve for \omega like this:

\omega =\sqrt{\frac{8000\frac{m}{s^2}}{0.002m}}=2000\frac{rad}{s}

And we with this value we can find the period with the following formula

T=\frac{2\pi}{\omega}=\frac{2 \pi}{2000\frac{rad}{s}}=0.0031416s

3) Part b

From equation (2) we see that the maximum velocity occurs when the sine function is euqal to -1 and on this case we have that:

v_{max}=A\omega =0.002mx2000\frac{rad}{s}=4\frac{m rad}{s}=4\frac{m}{s}

4) Part c

In order to find the total mechanical energy of the oscillator we can use this formula:

E=\frac{1}{2}mv^2_{max}=\frac{1}{2}(0.01kg)(6.283\frac{m}{s})^2=0.1974J

5) Part d

When we want to find the force from the 2nd Law of Newton we know that F=ma.

At the maximum displacement we know that X=A, and in order to that happens cos(\omega t +\phi)=1, and we also know that the maximum acceleration is given by::

|\frac{d^2X}{dt^2}|=A\omega^2

So then we have that:

F=ma=mA\omega^2

And since we have everything we can find the force

F=ma=0.01Kg(0.002m)(2000\frac{rad}{s})^2 =80N

6) Part e

When the mass it's at the half of it's maximum displacement the term cos(\omega t +\phi)=1/2 and on this case the acceleration would be given by;

|\frac{d^2X}{dt^2}|=A\omega^2 cos(\omega t +\phi)=A\omega^2 \frac{1}{2}

And the force would be given by:

F=ma=\frac{1}{2}mA\omega^2

And replacing we have:

F=\frac{1}{2}(0.01Kg)(0.002m)(2000\frac{rad}{s})^2 =40N

8 0
3 years ago
A low C (f=65Hz) is sounded on a piano. If the length of the piano wire is 2.0 m and
WITCHER [35]

Answer:

T = 676 N

Explanation:

Given that: f = 65 Hz, L = 2.0 m, and ρ = 5.0 g/m^{2} = 0.005 kg

A stationary wave that is set up in the string has a frequency of;

f = \frac{1}{2L}\sqrt{\frac{T}{M} }

⇒      T = 4L^{2}f^{2}M

Where: t is the tension in the wire, L is the length of the wire, f is the frequency of the waves produced by the wire and M is the mass per unit length of the wire.

But M = L × ρ = (2 × 0.005) = 0.01 kg/m

T = 4 × 2^{2} ×65^{2} × 0.01

   = 4 × 4 ×4225 × 0.01

   = 676 N

Tension of the wire is 676 N.

4 0
3 years ago
What are the units for mass and weight?
Luda [366]

The kilogram is the SI unit of mass and it is the almost universally used standard mass unit. The associated SI unit of force and weight is the Newton, with 1 kilogram weighing 9.8 Newtons under standard conditions on the Earth's surface.

6 0
3 years ago
Read 2 more answers
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