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dsp73
3 years ago
12

What is the parameter of motion?

Physics
2 answers:
zvonat [6]3 years ago
4 0
Distance, speed, time, mass, work, energy, density, volume, area, pressure, etc. are some examples.
docker41 [41]3 years ago
3 0

Answer:

Scalars

Explanation:

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Is it true that playing badmenton help you to become a better person?
Anuta_ua [19.1K]

Answer:

There is no scientific evidence that playing specifically l badminton makes you a better person, but sport and exercise in general release hormones which can make you feel more happy therefore making you nicer to the people around you and 'a better person'.

8 0
3 years ago
Read 2 more answers
Particle A of charge 2.79 10-4 C is at the origin, particle B of charge -5.64 10-4 C is at (4.00 m, 0), and particle C of charge
kirill [66]

Answer:

a) 0 b) 29.9 N c) 21.7 N d) -17.4 N e) -13.0 N f) -17.4 N  g) 16.9 N

h) 24.3 N θ = 44.2º

Explanation:

a) As the electric force is exerted along the line that joins the charges, due to any of the charges A or C has non-zero x-coordinates, the force has no x components either.

So, Fcax = 0

b) Similarly, as Fx = 0, the entire force is directed along the y-axis, and is going upward, due both charges repel each other.

Fyca = k*qa*qc / rac² = (9.10⁹ N*m²/C²*(2.79)*(1.07)*10⁻⁸ C²) / 9.00 m²

Fyca = 29. 9 N

c) In order to get the magnitude of the force exerted by B on C, we need to know first the distance between both charges:

rbc² = (3.00 m)² + (4.00m)² = 25.0 m²

⇒ Fbc = k*qb*qc / rbc² = (9.10⁹ N*m²/C²*(5.64)*(1.07)*10⁻⁸ C²) / 25.0 m²

⇒ Fbc = 21.7 N

d) In order to get the x component of Fbc, we need to get the projection of Fcb over the x axis, taking into account that the force on particle C is attractive, as follows:

Fbcₓ = Fbc * cos (-θ) where θ, is the angle that makes the line of action of the force, with the x-axis, so we can write:

cos θ = x/r = 4.00 / 5.00 m =

Fcbx = 21.7*(-0.8) = -17.4 N

e) The  y component can be calculated in the same way, projecting the force over the y-axis, as follows:

Fcby = Fcb* sin (-θ) = 21.7* (-3.00/5.00) = -13.0 N

f) The sum of both x components gives :

Fcx = 0 + (-17.4 N) = -17.4 N

g) The sum of both y components gives :

Fcy = 29.9 N + (-13.0 N) = 16.9 N

h) The magnitude of the resultant electric force acting on C, can be found just applying Pythagorean Theorem, as follows:

Fc = √(Fcx)²+(Fcy)² = (17.4)² + (16.9)²\sqrt{((17.4)^{2} +(16.9)^{2}} = 24.3 N

The angle from the horizontal can be found as follows:

Ф = arc tg (16.9 / 17.4) = 44.2º

4 0
3 years ago
A proton moves along the x-axis with vx=1.0×107m/s. As it passes the origin, what are the strength and direction of the magnetic
Sunny_sXe [5.5K]

Answer:

Magnetic field will be ZERO at the given position

Explanation:

As we know that the magnetic field due to moving charge is given as

B = \frac{\mu_0 qv sin\theta}{4\pi r^2}

so here we know that for the direction of magnetic field we will use

\hat B = \hat v \times \hat r

so we have

\hat B = \hat i \times (\hat i + 0\hat j + 0\hat k)

so magnetic field must be ZERO

So whenever charge is moving along the same direction where the position vector is given then magnetic field will be zero

3 0
3 years ago
Light incident on a surface at an angle of 45 degrees undergoes diffused reflection. At what angle will it reflect?
Alexus [3.1K]
 at any angle between -90 and 90 degrees. In the diffused reflection ,the surface is rough and the direction of light in not the same as the incident light. It can be reflected at any angle
4 0
3 years ago
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Normal conversation has a sound level of about 60 db. How many times more intense must a 100-hz sound be compared to a 1000-hz s
vampirchik [111]

Answer:

5.65 times

Explanation:

60 db sound is equal to 60 phons sound when frequency is kept at 1000Hz.

But when the frequency of sound  is changed to 100 Hz , according to equal loudness curves , the loudness level on phon scale will be 35 phons.

A decrease of 10 phon on phon- scale makes sound 2 times less loud

Therefore a decrease of 25 phons will make loudness less intense by a factor equal to 2²°⁵ or 5.65 less intense . Therefore intensity at 100 Hz

must be increased 5.65 times so that its intensity matches intensity of 60 dB sound at  1000 Hz  frequency.

6 0
4 years ago
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