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Scorpion4ik [409]
3 years ago
6

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

Physics
2 answers:
algol133 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]3 years ago
4 0

Answer:

Because the beam contained negatively charged particles

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A force of 150 N is applied to a 25 cm- piston in a hydraulic machine. What is the area
maxonik [38]

Answer:

8.3 cm^2

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6 0
3 years ago
Object 1 of mass m moves with speed v in the positive direction. Object 2 of mass 3 m moves with speed 4 v in the negative x-dir
Klio2033 [76]

Answer:

This means that the kinetic energy of second object is 48times that of the first object

Explanation:

Kinetic energy is the energy possessed by a body by virtue of its motion e.g motion of an accelerating car. Mathematically,

Kinetic energy = 1/2mv² where;

m is the mass of the object

v is the velocity of the object

If Object 1 of mass m moves with speed v in the positive direction, its kinetic energy will be expressed as;

K1 = 1/2mv²

For Object 2 of mass 3m moving with speed 4v in the negative x-direction, its kinetic energy can be expressed as;

K2 = 1/2(3m)(4v)²

K2 = 1/2(3m)(16v²)

K2 = (3m)(8v²)

K2 = 24mv²

To compare the kinetic energy of both bodies, we will take the ratio of K2:K1 to have;

K2/K1 = 24mv²/(1/2)mv²

K2/K1 = 24/(1/2)

K2/K1 = 48

K2 = 48K1

This means that the kinetic energy of second object is 48times that of the first object and moving in the negative x direction since the body of mass 3m initially moves in the negative x direction.

3 0
3 years ago
Explain the origin of the magnitude designation for determining the brightness of stars. Why does it seem to go backward, with s
Mashcka [7]

Answer:

Hipparchus was an ancient Greek who classified stars based on the brightness in 129 B.C. He grouped the brightest stars and ranked them 1 (first magnitude) and dimmest stars as 6 (sixth magnitude). Thus, the smaller numbers indicated brighter stars. Now, the scale extends in negative axis as well. More the negative number, brighter is the star. For example, Sun has magnitude -26.74.

This the apparent magnitude which means the classification is based on the brightness of the star as it appears from the Earth.

5 0
3 years ago
A beam of monochromatic light with a wavelength of 400 nm in air travels into water. what is the wavelength of the light in wate
slava [35]
The refractive index of water is n=1.33. This means that the speed of the light in the water is:
v= \frac{c}{n}= \frac{3 \cdot 10^8 m/s}{1.33 }=2.26 \cdot 10^8 m/s

The relationship between frequency f and wavelength \lambda of a wave is given by:
\lambda= \frac{v}{f}
where v is the speed of the wave in the medium. The frequency of the light does not change when it moves from one medium to the other one, so we can compute the ratio between the wavelength of the light in water \lambda_w to that in air \lambda as
\frac{\lambda_w}{\lambda}= \frac{ \frac{v}{f} }{ \frac{c}{f} } = \frac{v}{c}
where v is the speed of light in water and c is the speed of light in air. Re-arranging this formula and by using \lambda=400 nm, we find
\lambda_w = \lambda \frac{v}{c}=(400 nm) \frac{2.26 \cdot 10^8 m/s}{3 \cdot 10^8 m/s}=301 nm
which is the wavelength of light in water.
5 0
2 years ago
Add these measurements, using significant digit rules:
tia_tia [17]

Here we have to add the two measurements given in the question

The measurement values are given as 1.0090 cm and 0.02 cm.we have to  add them on the basis of significant figure rules.

As per the addition rule in terms of significant figures

1-First we have to select the number of significant digits after the decimal point of each quantity.

2-Now we have to remember that during the addition ,the resultant of two quantities will follow the quantity having least number of significant figures after the decimal point.

3-Here we are considering the minimum number of significant figures after the decimal points not the minimum number of significant figures in case of multiplication and division

Now we have to add these two quantities as per the above rule-

         1.0090 cm +0.02 cm

         =1.0290 cm

Here the result  will follow 0.02 which has minimum number of significant figures after the decimal points.

Hence we have to round off the number from 9 of 1.0290

As 9 is  greater than 5 ,so he actual result will be 1.03 cm

       

3 0
3 years ago
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