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lana66690 [7]
3 years ago
13

Each of the boxes starts at rest and is then pulled for 2.0 m across a level, frictionless floor by a rope with the noted force.

Which box has the highest final speed
Physics
1 answer:
Yuki888 [10]3 years ago
3 0

Answer:

Explanation:

d

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Two long parallel wires carry currents of 3.35 A and 6.99 A . The magnitude of the force per unit length acting on each wire is
Nady [450]

Answer:

244mm

Explanation:

I₁ = 3.35A

I₂ = 6.99A

μ₀ = 4π*10^-7

force per unit length (F/L) = 6.03*10⁻⁵N/m

B = (μ₀ I₁ I₂ )/ 2πr .........equation i

B = F / L ..........equation ii

equating equation i & ii,

F / L = (μ₀ I₁ I₂ )/ 2πr

Note F/L = B = F

F = (μ₀ I₁ I₂ ) / 2πr

2πr*F = (μ₀ I₁ I₂ )

r = (μ₀ I₁ I₂ ) / 2πF

r = (4π*10⁻⁷ * 3.35 * 6.99) / 2π * 6.03*10⁻⁵

r = 1.4713*10⁻⁵ / 6.03*10⁻⁵

r = 0.244m = 244mm

The distance between the wires is 244m

7 0
3 years ago
Read 2 more answers
What is true of transverse waves?
topjm [15]
<h2>Answer:D</h2>

Explanation:

Option A:

Surface waves are neither transverse nor longitudinal.They traverse perpendicularly or parallel to the wave's motion along the interface between different media.

Option B:

Transverse waves vibrate perpendicularly to the direction of the propagation of the wave.

Option C:

Sound is a longitudinal wave.Not a transverse wave.

Option D:

Transverse waves don't require a medium for propagation.But they propagate in medium too.

7 0
3 years ago
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I need help with life a child is adopted and I’m a duck where is my left shoe
Pavel [41]

just swim In water and find your shoe

8 0
3 years ago
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Consider the hydrogen atom. How does the energy difference between adjacent orbit radii change as the principal quantum number i
Kisachek [45]

Answer:

the energy difference between adjacent levels decreases as the quantum number increases

Explanation:

The energy levels of the hydrogen atom are given by the following formula:

E=-E_0 \frac{1}{n^2}

where

E_0 = 13.6 eV is a constant

n is the level number

We can write therefore the energy difference between adjacent levels as

\Delta E=-13.6 eV (\frac{1}{n^2}-\frac{1}{(n+1)^2})

We see that this difference decreases as the level number (n) increases. For example, the difference between the levels n=1 and n=2 is

\Delta E=-13.6 eV(\frac{1}{1^2}-\frac{1}{2^2})=-13.6 eV(1-\frac{1}{4})=-13.6 eV(\frac{3}{4})=-10.2 eV

While the difference between the levels n=2 and n=3 is

\Delta E=-13.6 eV(\frac{1}{2^2}-\frac{1}{3^2})=-13.6 eV(\frac{1}{4}-\frac{1}{9})=-13.6 eV(\frac{5}{36})=-1.9 eV

And so on.

So, the energy difference between adjacent levels decreases as the quantum number increases.

5 0
3 years ago
Consider two waves X and Y traveling in the same medium. The two carry the same amount of energy per unit time, but X has one-se
RideAnS [48]

Answer:

7 / 1

Explanation:

The ratio of their amplitude = one-seventh and the ratio of their amplitude = the ratio of their wavelength

Ax / Ay = λx / λy  = 1 / 7

λy / λx = 7 / 1

7 0
3 years ago
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