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WITCHER [35]
4 years ago
6

What do we call any space that has no air?

Physics
1 answer:
user100 [1]4 years ago
5 0
This is known as a "Vaccum"
This is a common term when referring to Space. 
A Vaccum is the space that contains no air or any other particles. 

~Hope this helps!
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If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it will oscillate. If i
Blababa [14]

Explanation:

An object is attached to the spring and then released. It begins to oscillate. If it is displaced a distance 0.125 m from its equilibrium position and released with zero initial speed. The amplitude of a wave is the maximum displacement of the particle. So, its amplitude is 0.125 m.  

After 0.800 seconds, its displacement is found to be a distance 0.125 m on the opposite side. The time period will be, t=2\times 0.8=1.6\ s

We know that the relation between the time period and the time period is given by :

f=\dfrac{1}{t}

f=\dfrac{1}{1.6}

f = 0.625 Hz

So, the frequency of the object is 0.625 Hz. Hence, this is the required solution.

6 0
3 years ago
Read 2 more answers
When spray is applied to a car, the paint has a negative charge and the surface of the car has a positive charge. Some processes
RUDIKE [14]

Answer:

Based on the properties electrically charged particles, we have that unlike charges attract and like charges repel each other. In order for proper application of an electrostatically negatively charged paint to be properly applied on the metal body surface of a vehicle, require that for attraction, the surface of the vehicle should be grounded and positively charged so as to effectively attract the negatively charged paint particles as it exits the nozzle, to form a strong attachment with the positively charged surface of the vehicle

Explanation:

4 0
3 years ago
How does the work required to accelerate a particle from 10 m/s to 20 m/s compare to that required to accelerate it from 20 m/s
poizon [28]

To solve this problem we will apply the energy conservation theorem for which the work applied on a body must be equivalent to the kinetic energy of this (or vice versa) therefore

W = \Delta KE

\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2

Here,

m = mass

v_{f,i} = Velocity (Final and initial)

First case) When the particle goes from 10m/s to 20m/s

\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2

\Delta W = \frac{1}{2} (m)(20)^2 -\frac{1}{2} (m)(10)^2

W_1 = 150(m) J

Second case) When the particle goes from 20m/s to 30m/s

\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2

\Delta W = \frac{1}{2} (m)(30)^2 -\frac{1}{2} (m)(20)^2

W_1 = 250(m) J

As the mass of the particle is the same, we conclude that more energy is required in the second case than in the first, therefore the correct answer is A.

5 0
4 years ago
If R = 20 Ω, what is the equivalent resistance between points A and B in the figure?​
Paraphin [41]

Answer:

Option C. 70 Ω

Explanation:

Data obtained from the question include:

Resistor (R) = 20 Ω

From diagram given ABOVE, we observed the following

1. R and R are in parallel connections.

2. 2R and 2R are in parallel connections.

3. 4R and 4R are in parallel connections.

Next, we shall determine the equivalent resistance in each case.

This is illustrated below:

1. Determination of the equivalent resistance for R and R parallel connections.

R = 20 Ω

Equivalent R = (R×R) /(R+R)

Equivalent R = (20 × 20) /(20 + 20)

Equivalent R = 400/40

Equivalent R = 10 Ω

2. Determination of the equivalent resistance for 2R and 2R parallel connections.

R = 20 Ω

2R = 2 × 20 = 40 Ω

Equivalent 2R = (2R×2R) /(2R+2R)

Equivalent 2R = (40 × 40) /(40 + 40)

Equivalent 2R = 1600/80

Equivalent 2R = 20 Ω

3. Determination of the equivalent resistance for 4R and 4R parallel connections.

R = 20 Ω

4R = 4 × 20 = 80 Ω

Equivalent 4R = (4R×4R) /(4R+4R)

Equivalent 4R = (80 × 80) /(80 + 80)

Equivalent 4R = 6400/160

Equivalent 4R = 40 Ω

Thus, the equivalence of R, 2R and 4R are now in series connections. We can obtain the equivalent resistance in the circuit as follow:

Equivalent of R = 10 Ω

Equivalent of 2R = 20 Ω

Equivalent of 4R = 40 Ω

Equivalent =?

Equivalent = Equivalent of (R + 2R + 4R)

Equivalent = 10 + 20 + 40

Equivalent = 70 Ω

Therefore, the equivalent resistance between point A and B is 70 Ω.

6 0
3 years ago
Find the wavelength of light given off by a hydrogen atom when its electron drops from the n = 4 to n = 1 energy level
4vir4ik [10]

Answer: 1.55 x 10 -19J on top

Explanation:

4 0
3 years ago
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