When the torques on the wheel and axle equal zero, the machine has <br> ____

Which of the following will increase the resistance of a wire? Check all that apply. Check all that apply. Increasing the cross-

Lostsunrise [7]

Correct answers:

- Increasing the resistivity of the material the wire is composed of will increase the resistance of the wire.

- Decreasing the cross-sectional area of the wire will increase the resistance of the wire.

- Increasing the length of the wire will increase the resistance of the wire.

Explanation:

The resistance of a wire is given by:

$R=\rho \frac{L}{A}$

where

$\rho$ is the resistitivity of the material

L is the length of the wire

A is the cross-sectional area of the wire

From the formula, we can notice that:

- The resistance (R) is directly proportional to both the resistivity ( $\rho$ ) and the length of the wire (L), so when one of these quantities increases, the resistance will increase as well

- The resistance (R) is inversely proportional to the cross-sectional area (A), therefore when the resistance increases when the cross-sectional area decreases.

Based on these observations, the correct statements are:

- Increasing the resistivity of the material the wire is composed of will increase the resistance of the wire.

- Decreasing the cross-sectional area of the wire will increase the resistance of the wire.

- Increasing the length of the wire will increase the resistance of the wire.

7 0

3 years ago

What is Newton's 1st Law?

Wittaler [7]

Answer:

An object in motion will stay in motion at the same speed and in the same direction (ie with constant velocity) unless a force makes it change.

Explanation:

3 0

2 years ago

A beam of light traveling through a liquid (of index of refraction n1 = 1.47) is incident on a surface at an angle of θ1 = 59° w

frosja888 [35]

Answer:

(a) $n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}$

(b) $n_{2} = 1.349$

(c) $v_{1} = 2.04\times 10^{8}\ m/s$

(d) $v_{2} = 2.22\times 10^{8}\ m/s$

Solution:

As per the question:

Refractive index of medium 1, $n_{1} = 1.47$

Angle of refraction for medium 1, $\theta_{1} = 59^{\circ}$

Angle of refraction for medium 2, $\theta_{1} = 69^{\circ}$

Now,

(a) The expression for the refractive index of medium 2 is given by using Snell's law:

$n_{1}sin\theta_{1} = n_{2}sin\theta_{2}$

where

$n_{2}$ = Refractive Index of medium 2

Now,

$n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}$

(b) The refractive index of medium 2 can be calculated by using the expression in part (a) as:

$n_{2} = \frac{1.47\times sin59^{\circ}}{sin69^{\circ}}$

$n_{2} = 1.349$

(c) To calculate the velocity of light in medium 1:

We know that:

$Refractive\ index,\ n = \frac{Speed\ of\ light\ in vacuum,\ c}{Speed\ of\ light\ in\ medium,\ v}$

Thus for medium 1

$n_{1} = \frac{c}{v_{1}$

$v_{1} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.47} = 2.04\times 10^{8}\ m/s$

(d) To calculate the velocity of light in medium 2:

For medium 2:

$n_{2} = \frac{c}{v_{2}$

$v_{2} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.349} = 2.22\times 10^{8}\ m/s$

5 0

3 years ago

Read 2 more answers

What did the asymptote say to the removable discontinuity worksheet answers?

ra1l [238]

“Don't hand that holier than thou line to me” is what the asymptote said to the removable discontinuity.

The distance between the curve and the line where it approaches zero as they tend to infinity is the line in the asymptote of a curve. This is unusual for modern authors but in some sources the requirement that the curve may not cross the line infinitely often is included.

The point that does not fit the rest of the graph or is undefined is called a removable discontinuity. By filling in a single point, the removable discontinuity can be made connected.

6 0

3 years ago

What is an isotope?

Troyanec [42]

Answer:

A. Atoms of the same element that have different numbers of nutrions.

8 0

3 years ago

When the torques on the wheel and axle equal zero, the machine has _____