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Scorpion4ik [409]

2 years ago

6

In Thomson experiment, why was the glowing beam repelled by a negatively charged plate

2 answers:

algol132 years ago

6 0

<span>In Thomson experiment, why was the glowing beam repelled by a negatively charged plate, because the glowing beam was negatively charged. The glowing beam particles were attracted to the positive plate.

</span><span>J.JThomson proved that the cathode rays produced a stream of negatively charged particles called electrons. </span>

Julli [10]2 years ago

4 0

Answer:

Because the beam contained negatively charged particles

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In an experiment designed to measure the Earth's magnetic field using the Hall effect, a copper bar 0.550 cm thick is positioned

motikmotik

Answer:

$B= 4.197*10^-5T$

Explanation:

We know that the magnetic Field of the Earth is 50\mu T, so the equation for

Hall effect voltage und magnetic field are defined as,

$\Delta V_H = \frac{IB}{nqt}$

$B=\frac{nqt\Delta V_h}{I}$

So,

$B=\frac{(8.48*10^{28})(1.6*10^{-19})(0.0055)(4.5*10^-{12})}{8}$

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2 years ago

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5) A person holds a 1.7 kg bucket and lets it move down at

Elena L [17]

Answer:

12.5J

Explanation:

Given parameters:

Mass of bucket = 1.7kg

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To solve this problem, we use the work done equation;

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2 years ago

Find the fundamental frequency and the next three frequencies that could cause standing-wave patterns on a string that is 30.0 m

maksim [4K]

Answer:

0.786 Hz, 1.572 Hz, 2.358 Hz, 3.144 Hz

Explanation:

The fundamental frequency of a standing wave on a string is given by

$f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}$

where

L is the length of the string

T is the tension in the string

$\mu$ is the mass per unit length

For the string in the problem,

L = 30.0 m

$\mu=9.00\cdot 10^{-3} kg/m$

T = 20.0 N

Substituting into the equation, we find the fundamental frequency:

$f=\frac{1}{2(30.0)}\sqrt{\frac{20.0}{(9.00\cdot 10^{-3}}}=0.786 Hz$

The next frequencies (harmonics) are given by

$f_n = nf$

with n being an integer number and f being the fundamental frequency.

So we get:

$f_2 = 2 (0.786 Hz)=1.572 Hz$

$f_3 = 3 (0.786 Hz)=2.358 Hz$

$f_4 = 4 (0.786 Hz)=3.144 Hz$

6 0

3 years ago

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pantera1 [17]

Answer:

It is the 3 one

Explanation:

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2 years ago