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zloy xaker [14]
4 years ago
13

Examples of field force

Physics
1 answer:
Sindrei [870]4 years ago
7 0
Here are some examples of field force:
1. <span>1) electrostatic force of attraction or repulsion on a charged particle in an electric field
2. </span><span>) magnetic force on charged particles moving at right angles with a magnetic field. Magnetic force also affects magnetic poles.


I hope this helps. </span>
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Follow these steps to solve this problem: Two identical loudspeakers, speaker 1 and speaker 2, are 2.0 m apart and are emitting
horsena [70]

Answer:

Explanation:

Wave length of sound from each of the  speakers = 340 / 1700 = .2 m = 20 cm

Distance between first speaker and the given point = 4 m.

Distance between second speaker and the given sound

= √ 4² + 2² = √16 +4 = √20 = 4.472 m

Path difference = 4.472 - 4 = .4722 m.

Path difference / wave length = 0.4772 / 0.2 = 2.386

This is a fractional integer which is neither an odd nor an  even multiple of half wavelength. Hence this point of neither a perfect constructive nor a perfect destructive interference.

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4 years ago
What is not true ozone?
Gnoma [55]
The second answer choice is not true (ozone is concentrated in the stratosphere, not the exosphere)
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3 years ago
the figure shows an initially stationary block of mass m on a floor. A force of magnitude 0.500mg is then applied at upward angl
Cerrena [4.2K]

Answer:

(a) 1.054 m/s²

(b) 1.404 m/s²

Explanation:

0.5·m·g·cos(θ) - μs·m·g·(1 - sin(θ))  - μk·m·g·(1 - sin(θ))  = m·a

Which gives;

0.5·g·cos(θ) - μ·g·(1 - sin(θ)   = a

Where:

m = Mass of the of the block

μ = Coefficient of friction

g = Acceleration due to gravity = 9.81 m/s²

a = Acceleration of the block

θ = Angle of elevation of the block = 20°

Therefore;

0.5×9.81·cos(20°) - μs×9.81×(1 - sin(20°)  - μk×9.81×(1 - sin(20°) = a

(a) When the static friction μs = 0.610  and the dynamic friction μk = 0.500, we have;

0.5×9.81·cos(20°) - 0.610×9.81×(1 - sin(20°)  - 0.500×9.81×(1 - sin(20°) = 1.054 m/s²

(b) When the static friction μs = 0.400  and the dynamic friction μk = 0.300, we have;

0.5×9.81·cos(20°) - 0.400×9.81×(1 - sin(20°)  - 0.300×9.81×(1 - sin(20°) = 1.404 m/s².

3 0
3 years ago
A proton is moved from a position where the electric potential is 125 V to a position where the electric potential is 275 V. The
liq [111]
We are asked to solve for the capacitance of a charged proton and the formula is shown below:
C=  q / ΔV  

The given values are the following:
ΔV = 275 volts - 125 volts
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q = C * ΔV
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6 0
4 years ago
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diamong [38]
ANSWER:

IV, Type of dish detergent. DV, height of foam. CV, type of container, amount of water in container, temperature of water, time the container is agitated.

Explanation:

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Constant variables(CV)- things that do not change in order to isolate the tested variables as much as possible.
4 0
3 years ago
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