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2 years ago

Will give big prizes !!

Physics

1 answer:

In-s [12.5K]2 years ago

7 0

Answer:

it may be because of the movement of the atoms. i believe that the heat of the warm water makes it move faster, thus making the object float due to upthrust, or air pressure

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A uniformly charged ring of radius 10.0 cm has a total charge of 71.0 μC. Find the electric field on the axis of the ring at the

oee [108]

Answer:

General Expression: E = kql/(l² + r²)^(3/2)

(a) 6.3 MN/C

(b) 22.8 MN/C

(d) 0.63 MN/C

Explanation:

The general expression for electric field along axis of a uniformly charged ring is:

E = kqL/(L² + r²)^(3/2)

where,

E = Electric Field Strength = ?

k = Coulomb's Constant = 9 x 10⁹ N.m²/C²

q = Total Charge = 71 μC = 71 x 10⁻⁶ C

L = Distance from center on axis

r = radius of ring = 10 cm = 0.1 m

(a)

L = 1 cm = 0.01 m

Therefore,

E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(0.01 m)/[(0.01 m)² + (0.1 m)²]^(3/2)

E = (6390 N.m³/C)/(0.00101 m³)

E = 6.3 x 10⁶ N/C = 6.3 MN/C

(b)

L = 5 cm = 0.05 m

Therefore,

E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(0.05 m)/[(0.05 m)² + (0.1 m)²]^(3/2)

E = (31950 N.m³/C)/(0.00139 m³)

E = 22.8 x 10⁶ N/C = 27.4 MN/C

(c)

L = 30 cm = 0.3 m

Therefore,

E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(0.3 m)/[(0.3 m)² + (0.1 m)²]^(3/2)

E = (191700 N.m³/C)/(0.03162 m³)

E = 6.1 x 10⁶ N/C = 6.1 MN/C

(d)

L = 100 cm = 1 m

Therefore,

E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(1 m)/[(1 m)² + (0.1 m)²]^(3/2)

E = (639000 N.m³/C)/(1.015 m³)

E = 0.63 x 10⁶ N/C = 0.63 MN/C

8 0

2 years ago

which of the following best represent positive self-talk A. "I'm making progress". B. "its just not my day". "I've tried. I just

KatRina [158]

A. I'm making progress

5 0

3 years ago

Two loudspeakers emit sound waves of the same frequency along the x-axis. The amplitude of each wave is a. The sound intensity i

leonid [27]

Answer:

Explanation:

To find the amplitude of the sound, we must first determine the wavelength and the phase difference between the two speakers.

For the wavelength;

Recall that, the separation between two successive max. and min. intensity points are $\dfrac{\lambda}{2}$

Thus; for both speakers; the wavelength of the sound is:

$\dfrac{\lambda}{2} = (10+30) cm$

$\dfrac{\lambda}{2} = (40) cm$

λ = 80 cm

The relation between the path difference(Δx) and the phase difference(Δ∅) is:

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

where;

Δx = 10 cm

λ = 80 cm

Δ∅ = π rad

∴

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

$\pi \ rad = \dfrac{2 \pi}{80 \ cm}(10 \ cm) + \Delta \phi_o$

$\pi \ rad = \dfrac{2 \pi}{8}+ \Delta \phi_o$

$\pi \ rad = \dfrac{ \pi}{4}+ \Delta \phi_o$

$\Delta \phi_o = \pi -\dfrac{ \pi}{4}$

$\Delta \phi_o = \dfrac{ 4\pi - \pi}{4}$

$\Delta \phi_o = \dfrac{ 3\pi}{4} \ rad$

Suppose both speakers are placed side-by-side, then the path difference between the two speakers is: Δx = 0 cm

Thus, we have:

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

$\Delta \phi = \dfrac{2 \pi}{\lambda}(0 \ cm ) + \dfrac{3 \pi}{4} \ rad$

$\Delta \phi = \dfrac{3 \pi}{4} \ rad$

∴

The amplitude of the sound wave if the two speakers are placed side-by-side is:

$A = 2a \ cos \bigg (\dfrac{\Delta \phi }{2} \bigg)$

$A = 2a \ cos \bigg (\dfrac{\dfrac{3 \pi}{4} }{2} \bigg)$

$A = 2a \ cos \bigg ({\dfrac{3 \pi}{8} } \bigg)$

A = 0.765a

7 0

2 years ago

A student walks to the right 25-m along the 800 hall in 15-s. They turn around and walk 15-m to the left in 8.0-s. Calculate the

Ludmilka [50]

Answer:

Explanation:

Total distance covered = 25 + 15 = 40 m

Total time = 15 + 8 = 23 s

Average speed = total distance covered / total time

= 40 / 23

= 1.74 m / s

Total displacement = 25 - 15 = 10 m

Total time = 15 + 8 = 23 s

Average velocity = total displacement / total time

= 10 / 23

= .434 m / s to the right .

3 0

3 years ago

Place the following types of electromagnetic radiation in order from LOWEST energy

Alex

Placing the electromagnetic radiation in order from the lowest energy to the highest energy : ( 2 ) C,A,B

<h3>Electromagnetic spectrum </h3>

In the electromagnetic spectrum the electromagnetic radiation with the shorter wavelength possess a higher energy while the electromagnetic radiation with a longer wavelength possess the lower energy.

The electromagnetic radiation as listed in the question with the longest wavelength is the radio waves therefore it possess the lowest energy while the radiation with the shortest wavelength is the gamma rays therefore it possess the highest energy.

Hence we can conclude that Placing the electromagnetic radiation in order from the lowest energy to the highest energy : ( 2 ) C,A,B

Learn more about electromagnetic spectrum : brainly.com/question/25847009

3 0

1 year ago