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4 years ago

Why should you never use the coarse adjustment on high power

Physics

1 answer:

umka21 [38]4 years ago

3 0

Answer:

First, the image moves in and out of focus too quickly, so that it is difficult to precisely adjust the focus. Second, you run the risk of crashing the objective into the slide. Use the coarse focus only with the 4x low power objective. You can use the fine focus knob with all objectives.

Explanation:

You might be interested in

What is the momentum of an 18-kg object moving at 0.1 m/s ?

Anit [1.1K]

Answer:

1.8 m/s

Explanation:

here's the solution : -

momentum = mass × velocity

=》18 × 0.1

=》1.8 m/s

6 0

3 years ago

What is the wavelength in nanometers of light when the energy is 1. 91 × 10^6 j for a mole of photons?.

JulijaS [17]

The wavelength in nanometers of light when the energy is 1. 91 × 10^6 j for a mole of photons is <u>62. 8 nm.</u>

Wavelength is the distance among the same points (adjacent crests) within the adjoining cycles of a waveform signal propagated in space or along a cord. In wi-fi structures, this period is typically specified in meters (m), centimeters (cm), or millimeters (mm).

The wavelength is the distance between wave crests, and it is going to be the same for troughs. The frequency is the variety of vibrations that skip over a given spot in one 2nd, and it's far measured in cycles consistent with the second (Hz) (Hertz).

Frequency is the ratio of pace and wavelength in relation to speed. In comparison, wavelength refers to the ratio of pace and frequency. Audible sound waves are characterized by way of a frequency range of 20 to 20 kHz. In contrast, the variety of wavelengths of visible light is from four hundred to seven hundred nm.

<u>calculation:-</u>

*E=hc/λ

1.91 × 10^6 J = (6.62610⁻³⁴) (3.00*10⁸) / λ

λ= (6.62610⁻³⁴) (3.00*10⁸) / 1.91 × 10⁶ J

λ= 1.0410⁻³¹× 10⁻⁹ × 6.022*10²³

=<u> 62. 8 nm </u>

Learn more about wavelength here:-brainly.com/question/10728818

#SPJ4

6 0

1 year ago

Calculate the electric field on the surface of a charged sphere with a radius of 5cm and charge 10μC. Thanks

victus00 [196]

As you approach the surface of the sphere very closely, the electric field should resemble more and more the electric field from an infinite plane of charge.

If you check Gauss's law (recalling that the field in the conductor is zero) you will see that if the surface charge density is σ=Q/4πR2, then indeed the field at the surface is σ/ϵ0 as in the infinite charge of plane case.

Such a field is constant, the field lines are parallel and non-diverging, and the infinities associated with the field due to point charge do not arise.

Explanation:

4 0

2 years ago

a ball of mass 100g moving at a velocity of 100m/s collides with another ball of mass 400g moving at 50m/s in same direction, if

klio [65]

Answer:

Velocity of the two balls after collision: $60\; \rm m \cdot s^{-1}$ .

$100\; \rm J$ of kinetic energy would be lost.

Explanation:

<h3>Velocity</h3>

Because the question asked about energy, convert all units to standard units to keep the calculation simple:

Mass of the first ball: $100\; \rm g = 0.1\; \rm kg$ .
Mass of the second ball: $400\; \rm g = 0.4 \; \rm kg$ .

The two balls stick to each other after the collision. In other words, this collision is a perfectly inelastic collision. Kinetic energy will not be conserved. The velocity of the two balls after the collision can only be found using the conservation of momentum.

Assume that the system of the two balls is isolated. Thus, the sum of the momentum of the two balls will stay the same before and after the collision.

The momentum of an object of mass $m$ and velocity $v$ is: $p = m \cdot v$ .

Momentum of the two balls before collision:

First ball: $p = m \cdot v = 0.1\; \rm kg \times 100\; \rm m \cdot s^{-1} = 10\; \rm kg \cdot m \cdot s^{-1}$ .
Second ball: $p = m \cdot v = 0.4\; \rm kg \times 50\; \rm m \cdot s^{-1} = 20\; \rm kg \cdot m \cdot s^{-1}$ .
Sum: $10\; \rm kg \cdot m \cdot s^{-1} + 20 \; \rm kg \cdot m \cdot s^{-1} = 30 \; \rm kg \cdot m \cdot s^{-1}$ given that the two balls are moving in the same direction.

Based on the assumptions, the sum of the momentum of the two balls after collision should also be $30\; \rm kg \cdot m \cdot s^{-1}$ . The mass of the two balls, combined, is $0.1\; \rm kg + 0.4\; \rm kg = 0.5\; \rm kg$ . Let the velocity of the two balls after the collision $v\; \rm m \cdot s^{-1}$ . (There's only one velocity because the collision had sticked the two balls to each other.)

Momentum after the collision from $p = m \cdot v$ : $(0.5\, v)\; \rm kg \cdot m \cdot s^{-1$ .
Momentum after the collision from the conservation of momentum: $30\; \rm kg \cdot m \cdot s^{-1}$ .

These two values are supposed to describe the same quantity: the sum of the momentum of the two balls after the collision. They should be equal to each other. That gives the equation about $v$ :

$0.5\, v = 30$ .

$v = 60$ .

In other words, the velocity of the two balls right after the collision should be $60\; \rm m \cdot s^{-1}$ .

<h3>Kinetic Energy</h3>

The kinetic energy of an object of mass $m$ and velocity $v$ is $\displaystyle \frac{1}{2}\, m \cdot v^{2}$ .

Kinetic energy before the collision:

First ball: $\displaystyle \frac{1}{2} \, m \cdot v^2 = \frac{1}{2}\times 0.1\; \rm kg \times \left(100\; \rm m \cdot s^{-1}\right)^{2} = 500\; \rm J$ .
Second ball: $\displaystyle \frac{1}{2} \, m \cdot v^2 = \frac{1}{2}\times 0.4\; \rm kg \times \left(50\; \rm m \cdot s^{-1}\right)^{2} = 500\; \rm J$ .
Sum: $500\; \rm J + 500\; \rm J = 1000\; \rm J$ .

The two balls stick to each other after the collision. Therefore, consider them as a single object when calculating the sum of their kinetic energies.

Mass of the two balls, combined: $0.5\; \rm kg$ .
Velocity of the two balls right after the collision: $60\; \rm m\cdot s^{-1}$ .

Sum of the kinetic energies of the two balls right after the collision:

$\displaystyle \frac{1}{2} \, m \cdot v^{2} = \frac{1}{2}\times 0.5\; \rm kg \times \left(60\; \rm m \cdot s^{-1}\right)^2 = 900\; \rm J$ .

Therefore, $1000\; \rm J - 900\; \rm J = 100\; \rm J$ of kinetic energy would be lost during this collision.

7 0

4 years ago

PLZ HELP!!!!

irinina [24]

<span>First blank: B) Cutting down trees instead of growing trees
Cutting down trees means less plants that are photosynthesizing and consuming CO2 from the atmosphere.

Second blank: </span><span>D) Using clean energy sources instead of fossil fuels
Clean energy sources have little to no CO2 emissions. On the other hand, fossil fuels, when burned/combusted release large amounts of CO2 into the atmosphere.</span>

4 0

4 years ago

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