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3 years ago

Why does a lead of a pencil becomes smaller after use brainly people plplplsss help

1 answer:

6 0

Pencil lead is made up of graphite. These graphite particles stick to the fiber in the paper when you write. So what happens is it transfers from the pencil to the paper. When you write you're literally rubbing the graphite off the pencil.

I hope this explanation will help you out.

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A series circuit contains a 9-volt battery, a 3-ohm resistor and a 2-ohm resistor. What is the voltage drop across the 2-ohm res

BARSIC [14]

Since everything in the circuit is in series .. .

-- The total resistance is (3 + 2) = 5 ohms.

-- The voltage across the 3-ohm resistor is 3/5 of the total voltage.

-- The voltage across the 2-ohm resistor is 2/5 of the total voltage.

(2/5) of (9 volts) = 18/5 = 3.6 volts .

7 0

4 years ago

Read 2 more answers

What is the velocity of a 2000 kg truck with a momentum of 48,000 kg•m/s?

Goshia [24]

Answer:

Explanation:

momentum= mass* velocity

velocity= momentum/ mass

5 0

3 years ago

A carbon fiber car bumper is hit by another car with the stress of 106,483 Pa. If carbon fiber has a Young's modulus of 228 x 10

hoa [83]

Answer:

Stress = 4.67 * 10^-7 N/m²

Explanation:

Young's modulus of the material = Stress/Strain

Given

Young's modulus = 228 x 10^9 Pa

Stress = 106,483 Pa

Required

Strain

From the formula;

Strain = Stress/Young modulus

Strain = 106,483 /228 x 10^9

Stress = 4.67 * 10^-7 N/m²

3 0

3 years ago

What part of the gun position equipment supports all of the elevating parts of the gun?

BARSIC [14]

The gunstock it’s also called stock or shoulder stock

5 0

4 years ago

A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical ener

defon

Complete question:

A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical energy of the spring–load system is 2.00 J. Find

(a) the force constant of the spring and (b) the amplitude of the motion.

Answer:

(a) the force constant of the spring = 47 N/m

(b) the amplitude of the motion = 0.292 m

Explanation:

Given;

mass of the spring, m = 200g = 0.2 kg

period of oscillation, T = 0.410 s

total mechanical energy of the spring, E = 2 J

The angular speed is calculated as follows;

$\omega = \frac{2\pi}{T} \\\\\omega = \frac{2\pi}{0.41} \\\\\omega = 15.33 \ rad/s$

(a) the force constant of the spring

$\omega = \sqrt{\frac{k}{m} } \\\\\omega^2 = \frac{k}{m} \\\\k = m \omega^2\\\\k = (0.2)(15.33)^2\\\\k = 47 \ N/m$

(b) the amplitude of the motion

E = ¹/₂kA²

2E = kA²

A² = 2E/k

$A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2\times 2}{47} }\\\\A = 0.292 \ m$

7 0

3 years ago