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jekas [21]
3 years ago
11

What are five ways to decrease resistance?

Physics
1 answer:
s344n2d4d5 [400]3 years ago
8 0
Five ways to decrease resistance are:

1) Increase "Current"
2) Decrease "Potential difference"
3) Decrease "length" of conductor
4) Increase "Area of cross-section" of conductor
5) Switch to material of low "Resistivity"

Hope this helps!
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Please help asap 100 POINTS and BRAINLIEST to best answer.
andrew11 [14]

D. :)

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3 years ago
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APAKAH VOLUME 5 KG ES SAMA DENGAN VOLUME 5KG AIR APA YG MEMBEDAKAN KEDUA ZAT TERSEBUT
horrorfan [7]

Answer:c

Explanation:ganglandd

4 0
3 years ago
You and a partner sit on the floor and stretch out a coiled spring to a length of 7.2 meters. You shake the coil so you
vekshin1

Answer:

Approximately 5.9\; {\rm m\cdot s^{-1}} (assuming that the partner is holding the other end of the coil stationary.)

Explanation:

In a standing wave, an antinode is a point that moves with maximal amplitude, while a node is a point that does not move at all. There is an antinode between every two adjacent nodes. Likewise, there is a node between every two adjacent antinodes.

The side of the spring that is being shaken moving with maximal amplitude. Hence, that point on this spring would also be an antinode. In contrast, the side of the spring that is held still (does not move at all) would be a node.

There would be a node between:

  • the antinode at the end of the spring that is being shaken, and
  • the antinode between the two ends of this spring.

Overall, the nodes and antinodes on this spring would be:

  • node at the end that is being held still,
  • antinode (as mentioned in the question),
  • node (inferred, not mentioned in the question), and
  • antinode at the end that is being shaken.

The distance between two adjacent nodes is equal to one-half (that is, (1/2)) the wavelength of the wave. The distance between a node and an adjacent antinode is one-quarter (that is, (1/4)) of the wavelength of the wave.

Thus, if the wavelength of the wave in this question is \lambda, the length of this spring would be:

\displaystyle \frac{1}{2}\, \lambda + \frac{1}{4}\, \lambda = \frac{3}{4}\, \lambda.

The question states that the length of this coiled spring is 7.2\; {\rm m}. In other words, (3/4) \, \lambda = 7.2\; {\rm m}. The wavelength of this wave would be (7.2\; {\rm m}) / (3/4) = 9.6\; {\rm m}.

The frequency f of this wave is the number of cycles in unit time:

\begin{aligned} f &= \frac{10}{16.3\; {\rm s}} \approx 0.613\; {\rm s^{-1}}\end{aligned}.

Hence, the speed v of this wave would be:

\begin{aligned} v &= \lambda\, f \\ &=9.6\; {\rm m} \times 0.613\; {\rm s^{-1}} \\ &\approx 5.9\; {\rm m \cdot s^{-1}}\end{aligned}.

3 0
2 years ago
An 8-hour exposure to a sound intensity level of 90.0 dB may cause hearing damage. What energy in joules falls on a 0.800-cm-dia
11111nata11111 [884]

Answer:

1.4E-3J

Explanation:

Given that

Time = 8hrs = 28.8E3 seconds

Intensity= 90dB

D= 0.008m

Radius= 0.004m

So intensity is sound level Bis

10dBlog(I/Io)

I= 10 (B/10dB)Io

= 10( 90/10) x 10^-12

=0.001W/m²

But we know that

I = P/A

P= I πr²

= 5.02 x10^-8W

But energy is power x time

So E= 5.02E-8 x 28.8E3

= 1.4E-3J

5 0
3 years ago
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A white dwarf star has a density of about 1.0 x 10^9 kg/m3. If the earth were to suddenly become as dense as a white dwarf star,
GalinKa [24]

Answer:

R = 98304.75 m = 98.3 km

Explanation:

The density of an object is given as the ratio between the mass of that object and the volume occupied by that object.

Density = Mass/Volume

Now, it is given that the density of Earth has become:

Density = 1 x 10⁹ kg/m³

Mass = Mass of Earth (Constant) = 5.97 x 10²⁴ kg

Volume = 4/3πR³ (Volume of Sphere)

R = Radius of Earth = ?

Therefore,

1 x 10⁹ kg/m³ = (5.97 x 10²⁴ kg)/[4/3πR³]

4/3πR³ = (5.97 x 10²⁴ kg)/(1 x 10⁹ kg/m³)

R³ = (3/4)(5.97 x 10¹⁵ m³)/π

R = ∛[0.95 x 10¹⁵ m³]

<u>R = 98304.75 m = 98.3 km</u>

6 0
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