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3 years ago

"How do changes in voltage or resistance affect current in an electric circuit?"

1 answer:

5 0

We can answer the question by looking at the Ohm's law, which gives us the relationship between voltage (V), current (I) and resistance (R) of a circuit:
$V=IR$
equivalently, we can rewrite it as
$I= \frac{V}{R}$

by looking at the equation, we can make the following observations:
1) The current is proportional to the voltage: therefore, if the voltage increases, the current increases as well; if the voltage decreases, the current decreases too.
2) The current is inversely proportional to the resistance: if the resistance increases, the current decreases, and if the resistance decreases, the current increases.

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In Hooke's law, what does k represent?

kicyunya [14]

For Plato, K represents the stiffness of the spring. Hope this helps!

3 0

3 years ago

Read 2 more answers

food does not pass through the liver or the pancreas why are these organs dtill important parts of the digestive system?

kirill115 [55]

Your pancreas makes a digestive juice that has enzymes that break down carbohydrates, fats, and proteins. The pancreas delivers the digestive juice to the small intestine through small tubes called ducts. Liver. Your liver makes a digestive juice called bile that helps digest fats and some vitamins.

5 0

2 years ago

An aluminum wire having a cross-sectional area equal to 3.90 10-6 m2 carries a current of 6.00 A. The density of aluminum is 2.7

stich3 [128]

Answer:

Vd = 1.597 ×10⁻⁴ m/s

Explanation:

Given: A = 3.90×10⁻⁶ m², I = 6.00 A, ρ = 2.70 g/cm³

To find:

Drift Velocity Vd=?

Solution:

the formula is Vd = I/nqA (n is the number of charge per unit volume)

n = No. of electron in a mole ( Avogadro's No.) / Volume

Volume = Molar mass / density ( molar mass of Al =27 g)

V = 27 g / 2.70 g/cm³ = 10 cm³ = 1 × 10 ⁻⁵ m³

n= (6.02 × 10 ²³) / (1 × 10 ⁻⁵ m³)

n= 6.02 × 10 ²⁸

Now

Vd = (6A) / ( 6.02 × 10 ²⁸ × 1.6 × 10⁻¹⁹ C × 3.9×10⁻⁶ m²)

Vd = 1.597 ×10⁻⁴ m/s

6 0

3 years ago

A 75kg hockey player is skating across the ice at a speed of 6.0m/s. What is the magnitude of the average force required to stop

liq [111]

Answer:

692.31 N

Explanation:

Applying,

F = ma............... Equation 1

Where F = Average force required to stop the player, m = mass of the player, a = acceleration of the player

But,

a = (v-u)/t............ Equation 2

Where v = final velocity, u = initial velocity, t = time.

Substitute equation 2 into equation 1

F = m(v-u)/t............ Equation 3

From the question,

Given: m = 75 kg, u = 6.0 m/s, v = 0 m/s (to stop), t = 0.65 s

Substitute these values into equation 3

F = 75(0-6)/0.65

F = -692.31 N

Hence the average force required to stop the player is 692.31 N

6 0

3 years ago

A spring is 6.0cm long when it is not stretched, and 10cm long when a 7.0N force is applied. What force is needed to make it 20c

Artist 52 [7]

Answer:

Approximately $25\; {\rm N}$ (assuming that this spring is ideal.)

Explanation:

The displacement of a spring is the new length of the spring relative to the original length.

For example:

When the $6.0\; {\rm cm}$ -spring in this question is stretched to $10\; {\rm cm}$ , the displacement is $x = (10\; {\rm cm} - 6.0\; {\rm cm})$ .
Likewise, if this spring is stretched to $20\; {\rm cm}$ , the displacement would be $(20\; {\rm cm} - 6\; {\rm cm})$ .

If this spring is ideal, the force on the spring would be proportional to the displacement of the spring. In other words, if a force of $F_{\text{a}}$ displaces this spring by $x_{\text{a}}$ , while a force of $F_{\text{b}}$ displaces this spring by $x_{\text{b}}$ , then:

$\displaystyle \frac{F_{\text{a}}}{x_{\text{a}}} = \frac{F_{\text{b}}}{x_{\text{b}}}$ .

In this question, it is given that a force of $F_{\text{a}} = 7.0 \; {\rm N}$ would stretch this spring by $x_{\text{a}} = (10\; {\rm cm} - 6.0\; {\rm cm})$ . Thus, the force $F_{\text{b}}$ required to stretch this spring by $x_{\text{a}} = (20\; {\rm cm} - 6.0\; {\rm cm})$ would satisfy:

$\displaystyle \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}}= \frac{F_{\text{b}}}{20\; {\rm cm} - 6.0\; {\rm cm}}$ .

Rearrange and solve for $F_{\text{b}}$ :

$\begin{aligned} F_{\text{b}} &= \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}} \, (20\; {\rm cm} - 6.0\; {\rm cm}) \\ &\approx 25\; {\rm N}\end{aligned}$ .

7 0

2 years ago