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2 years ago

A thin rod of length d on a frictionless surface is pivoted about one end

1 answer:

3 0

The magnitude of the angular momentum of the rod immediately after the collision is given as (pf + pi) * d. (Option A).

<h3>What is Angular Momentum?</h3>

Angular Momentum may be defined or described as the vector quantity of the rotation of a body, which is arrived at by multiplying its moment of inertia by its angular velocity.

The formula for Angular Momentum is given as:

L = mvr; Where

L = Angular Momentum

m = mass

v = velocity; and

r = radius.

Learn more about Angular Momentum at:
brainly.com/question/4126751

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Please answer this. Science 7th grade.

USPshnik [31]

I think the answer should be a I’m not sure tho

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3 years ago

A battery has a terminal voltage of 12.0 V when no current flows. Its internal resistance is 2.0 Ω. If a 4.6 Ω resistor is conne

rosijanka [135]

Answer:

Check attachment for solution

Explanation:

Given that 12V battery

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3 years ago

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25 POINTS

Alexandra [31]

pshyical change is a usually reversible change of a substance, as size or shape: Freezing a liquid is a physical change. Compare chemical change.

internal change is when the movement of the particles increases

specific latent heat is the amount of energy per kg (unit mass) required to change ice to water without change in temperature.

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3 years ago

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A spring with spring constant 11.5 N/m hangs from the ceiling. A 490 g ball is attached to the spring and allowed to come to res

Natalija [7]

Answer:

The time constant is $\tau = 17.5 \ s$

Explanation:

From the question we are told that

The spring constant is $k = 11.5 \ N/m$

The mass of the ball is $m_b = 490 \ g = 0.49 \ kg$

The amplitude of the oscillation t the beginning is $x = 6.70 cm = 0.067 \ m$

The amplitude after time t is $x_t = 2.20 cm = 0.022 \ m$

The number of oscillation is $N = 30$

Generally the time taken to attain the second amplitude is mathematically represented as

$t = N * T$ Here T is the period of oscillation

$t = N * 2\pi \sqrt{\frac{m}{k} }$

=> $t = 30 * 2 * 3.142 * \sqrt{\frac{ 0.490}{11.5} }$

=> $t = 38.88 \ s$

Generally the amplitude at time t is mathematically represented as

$x(t) = x e^{-\frac{at}{2m} }$

Here a is the damping constant so

at $t = 38.88 \ s$ , $x_t = 2.20 cm = 0.022 \ m$

$0.022 = 0.067 e^{-\frac{a * 38.88}{2 * 0.490} }$

=> $0.3284 = e^{-\frac{a * 38.88}{2 * 0.490} }$

taking natural log of both sides

=> $ln(0.3284 ) = -\frac{a * 38.88}{2 * 0.490} }$

=> $a = 0.028$

Generally the time constant is mathematically represented as

$\tau = \frac{m}{a}$

=> $\tau = \frac{0.490}{ 0.028}$

=> $\tau = 17.5 \ s$

4 0

3 years ago

A fuel pump sends gasoline from a car's fuel tank to the engine at a rate of 5.64 10-2 kg/s. the density of the gasoline is 735

Irina18 [472]

Given:
Gasoline pumping rate, R = 5.64 x 10⁻² kg/s
Density of gasoline, D = 735 kg/m³
Radius of fuel line, r = 3.43 x 10⁻³ m

Calculate the cross sectional area of the fuel line.
A = πr² = π(3.43 x 10⁻³ m)² = 3.6961 x 10⁻⁵ m²

Let v = speed of pumping the gasoline, m/s
Then the mass flow rate is
M = AvD = (3.6961 x 10⁻⁵ m²)*(v m/s)*(735 kg/m³) = 0.027166v kg/s

The gasoline pumping rate is given as 5.64 x 10⁻² kg/s, therefore
0.027166v = 0.0564
v = 2.076 m/s

Answer: 2.076 m/s
The gasoline moves through the fuel line at 2.076 m/s.

7 0

3 years ago

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