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Anika [276]
3 years ago
10

Cathode ray particles have what type of charge

Physics
1 answer:
zlopas [31]3 years ago
3 0

Answer:

negative charge

Explanation:

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The table below shows two types of electromagnetic waves and three random applications of electromagnetic waves.
Musya8 [376]

Answer:

a. Microwaves—3 and infrared waves—1

Explanation:

Microwaves and infrared waves are both part of the electromagnetic spectrum, but they have different frequency and wavelength.

In particular:

- Microwaves are long-wavelength electromagnetic waves, with wavelength between 1 mm and 1 m. Their wavelength is longer than visible light

- Infrared waves are also long-wavelength electromagnetic waves, but their wavelength is shorter than microwaves: between 700 nm and 1 mm. Their wavelength is also longer than visible light.

The two types of waves are also used for different purposes. In particular:

- Infrared waves are emitted by any hot object, and their intensity depends on the temperature of the object. Therefore, they are used in astronomy to show the heat released by astronomical objects (option 1)

- Microwaves are used to study the Cosmic Microwave Background (CMB). This is electromagnetic radiation that permeates the whole universe, and its wavelength depends inversely on the local temperature. Therefore, areas with longer wavelength have lower temperature, and viceversa. Therefore, microwaves are used to measure temperature differences in space (option 3).

7 0
3 years ago
If a body accelerates from a state of relative rest in a gravity field, where does the energy come from to transfer as kinetic e
Delvig [45]

The energy that transforms into kinetic energy is the Potential Energy. It happens that objects can store energy as a result of its position. Image for example a slingshot. When you stretch the slingshot, it stores energy, this energy would be the energy you used to stretch the slingshot, the material aborbs it and then release to throw the projectile.

Now, on earth and everywhere in the universe where you are close to an object with mass, it exists a force called gravity that attracts you towards that object. Every object that has mass exercises gravitational attration towards the other objects. It just happens that Earth is has so much mass that its gravitational pull is way stronger that the gravitational pull of another object on its surface. This means things will tend to be as close as earth as possible, and in order to move something away from earth, you will have to perform a force in the opposite direction to Earth and, therefore, consume energy. This energy will be store as potential energy, and when you drop the object, the potential energy will be the energy that will transform to kinetic energy.

5 0
3 years ago
A horizontal force of 750 N is needed to overcome the force of static friction between a level floor and a 250-kg crate. If g-9.
seraphim [82]

Answer:

0.31

Explanation:

horizontal force, F = 750 N

mass of crate, m = 250 kg

g = 9.8 m/s^2

The friction force becomes applied force = 750 N

According to the laws of friction,

Friction force = μ x Normal reaction of the surface

here, μ be the coefficient of friction

750 = μ x m g

750 = μ x 250 x 9.8

μ = 0.31

Thus, the coefficient of static friction is 0.31.

7 0
3 years ago
A factory has 1200 workers of which 720 are male and the rast are female what percent of workers are female​
umka21 [38]

Answer:

Percent of Female Workers = 40%

Explanation:

The percentage of the female workers in the given group of workers can be easily found by the following formula:

Percent\ of\ Female\ Workers = \frac{No.\ of\ Female\ Workers}{Total\ No.\ of\ Workers}\ x\ 100\%

where,

Total No. of Workers = 1200

No. of Female Workers = Total Workers - No. of Male Workers

No. of Female Workers = 1200 - 720 = 480

Therefore,

Percent\ of\ Female\ Workers = \frac{480}{1200}\ x\ 100\%

<u>Percent of Female Workers = 40%</u>

7 0
3 years ago
Hey stob it.<br> Please help me.<br> Cmon help me.<br> Plz.
Anna [14]

Answer:

3) D: 31 m/s

4) D: 84.84 metres

Explanation:

3) Initial velocity along the x-axis is;

v_x = v_o•cos θ

Initial velocity along the y-axis is;

v_y = v_o•sin θ

Plugging in the relevant values, we have;

v_x = 31 cos 60

v_x = 31 × 0.5

v_x = 15.5 m/s

Similarly,

v_y = 31 sin 60

v_y = 31 × 0.8660

v_y = 26.85 m/s

Thus, magnitude of the initial velocity is;

v = √(15.5² + 26.85²)

v ≈ 31 m/s

4) Formula for horizontal range is;

R = (v² sin 2θ)/g

R = (31² × sin (2 × 60))/9.81

R = 84.84 m

6 0
3 years ago
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