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3 years ago

Use a conversion factor to solve each situation:

Physics

1 answer:

klasskru [66]3 years ago

6 0

Enrique filled a bucket with 2,250 milliliters of water. How many liters of

water were in the bucket?

Answer:

2.25 liters

You might be interested in

Which type wire should be used to increase resistance

8_murik_8 [283]

Nichrome wire. That's the stuff that toasters are made from. The resistance is pretty high, considering the diameter. 1 meter is at about the same guage as that listed below for copper is about 96 ohms.

Most of the time you are trying to use wire with the least resistance.

A meter of copper has a listed resistance of 0.024 ohms / meter. The wire is a 19 guage wire which makes it pretty thin.

===============

I'm not sure what you are asking. If want the resistance of something in terms of what would increase the resistance of the same material for both calculations then

Rule 1: It you decrease the wire diameter, you increase the resistance

Rule 2: If you increase the length of the wire, you increase the resistance.

Both rules assume you are using something like copper.

7 0

3 years ago

Read 2 more answers

Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,

Marrrta [24]

Answer:

Solid Wire $I = 0.01237 \ A$

Stranded Wire $I_2 = 0.00978 \ A$

Solid Wire $R = 0.0149 \ \Omega$

Stranded Wire $R_1 = 0.0189 \ \Omega$

Explanation:

Considering the first question

From the question we are told that

The radius of the first wire is $r_1 = 1.53 mm = 0.0015 \ m$

The radius of each strand is $r_0 = 0.306 \ mm = 0.000306 \ m$

The current density in both wires is $J = 1750 \ A/m^2$

Considering the first wire

The cross-sectional area of the first wire is

$A = \pi r^2$

= > $A = 3.142 * (0.0015)^2$

= > $A = 7.0695 *10^{-6} \ m^2$

Generally the current in the first wire is

$I = J*A$

=> $I = 1750*7.0695 *10^{-6}$

=> $I = 0.01237 \ A$

Considering the second wire wire

The cross-sectional area of the second wire is

$A_1 = 19 * \pi r^2$

=> $A_1 = 19 *3.142 * (0.000306)^2$

=> $A_1 = 5.5899 *10^{-6} \ m^2$

Generally the current is

$I_2 = J * A_1$

=> $I_2 = 1750 * 5.5899 *10^{-6}$

=> $I_2 = 0.00978 \ A$

Considering question two

From the question we are told that

Resistivity is $\rho = 1.69* 10^{-8} \Omega \cdot m$

The length of each wire is $l = 6.25 \ m$

Generally the resistance of the first wire is mathematically represented as

$R = \frac{\rho * l }{A}$

=> $R = \frac{ 1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }$

=> $R = 0.0149 \ \Omega$

Generally the resistance of the first wire is mathematically represented as

$R_1 = \frac{\rho * l }{A_1}$

=> $R_1 = \frac{ 1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }$

=> $R_1 = 0.0189 \ \Omega$

3 0

3 years ago

Which of the following is an example of a properly written testable hypothesis? *

sveta [45]

Answer:

D. if it is dark, then an owl will find a mouse by the sound the mouse makes

3 0

4 years ago

A 30 kg student drops down from the monkey bars. The acceleration due to gravity is -9.8 m/s^2. Neglecting air drag, what is the

Sholpan [36]

The net force on the student is A) -294 N

Explanation:

Neglecting air resistance, there is only one force acting on the student: the force of gravity, which is given by

$F=mg$

where

m is the mass of the student

g is the acceleration of gravity

In this problem, we have:

m = 30 kg is the mass of the student

$g=-9.8 m/s^2$ is the acceleration of gravity, where the negative sign means the direction is downward

Substituting, we find the force of gravity on the student:

$F=(30)(-9.8)=-294 N$

And since this is the only force acting on the student, it is also the net force on him.

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

8 0

3 years ago

Tính chỉ số ampe kế thay ampe kế bằng vô kế có điện trở rất lớn thì vô kế chỉ bao nhiêu

ser-zykov [4K]

Answer:

I don't understand it.....

8 0

3 years ago