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Fofino [41]
3 years ago
7

Scientists use the micron as a unit of length for very

Physics
1 answer:
r-ruslan [8.4K]3 years ago
6 0

Answer:

1 micron = 1.00E-6  m   is one way

1.00^-6 m    is another but is not usually considered scientific notation, but                      

                   often convenient to use.

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Bond [772]

Answer:

d: ice is less dense than liquid water.

Explanation:

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~mina

3 0
3 years ago
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An object moves in simple harmonic motion with amplitude 11m and period 4 minutes. At time t = 0 minutes, its displacement d fro
Sergio039 [100]

Answer:

Explanation:

Angular velocity ω = 2π / T = 2 x 3 . 14 / (4 x 60 ) = .026 rad / s.

d = D Sin ( ωt + θ )  where  D = amplitude = 11 m θ = initial phase.

-11 = 11 sin ( 0 + θ ) = sin θ = -1 , θ = - π /2

So, d = D Sin ( ωt - θ )

         d = 11  Sin ( .026 t -  π /2  ) is the required equation.

8 0
3 years ago
The difference between newtons first law of motion and newtons third law of motion
zmey [24]

The difference is two laws of motion.
6 0
3 years ago
A 15.0 kg sphere is at the origin and a 7.00 kg sphere is at x=20cm. At what position on the x-axis could you place a small mass
Zolol [24]

Answer:

The position of the small mass on the x- axis = 11.87 cm

Explanation:

From Newton's law of universal gravitation,

F₁ = Gm₁X/r₁² ................... Equation 1

Where F₁ = Force exerted on the on the small mass by the first sphere, m₁ = mass of the first sphere, X = mass of the small mass, r₁ = distance between the first sphere and the small mass.

Also,

F₂ = Gm₂X/r₂².................... Equation 2

Where F₂ = Force exerted by the second sphere on the small mass, m₂ = mass of the second sphere, X = mass of the small mass, r₂ = distance between the second sphere and the small mass.

<em>Note: F₁ = F₂ ( Net force on the small mass due to the sphere is zero)</em>

Therefore,

Gm₁X/r₁² = Gm₂X/r₂²

Equating the similar terms from both side of the equation,

m₁/r₁² = m₂/r₂².......................... Equation 3

<em>Given: m₁ = 15.0 kg, m₂ = 7.00 kg, let r₁ = y cm, then r₂ = (20-y) cm</em>

<em>Substituting these values into equation 3</em>

<em>15/y² = 7/(20-y)²</em>

15(20-y)² = 7y²

√{15(20-y)² = √7y²

√15×√(20-y)² = √7×√y² ( from the law of surd)

3.87(20-y) = 2.65y

77.4 - 3.87y = 2.65y

Collecting like terms and reordering the equation

2.65y+3.87y = 77.4

6.52y = 77.4

y = 77.4/6.52

y = 11.87 cm

Thus the position of the small mass on the x- axis = 11.87 cm

5 0
3 years ago
(1) A string of constant thickness and length I cm is stretched by a force of T Newton. A
kumpel [21]

Answer:

Explanation:

i) frequency is the square root of the quotient of Tension over linear density/ all divided by two times the length.

linear density and tension are constant, so the numerator remains constant

The length is doubled, so the frequency is halved.

f = 128 Hz

ii) Doubling the tension increases the frequency by √2

f = 256√2 = 362 Hz

2) If other factors remain constant, increasing tension increases frequency.

5 0
3 years ago
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