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3 years ago

A bird flies 2km to the south, then 0.2km to the north.What is the displacement of the bird ?

Physics

2 answers:

Lunna [17]3 years ago

6 0

Explanation:

It is given that,

A bird flies 2 km to the south, then 0.2 km to the north. The displacement of the bird is equal to the shortest path covered by it.

Initially, the bird is moving in south direction and then it move to north. Displacement can be calculated using difference between initial position and final position.

d = 2.2 km - 0.2 km = 1.8 km

Hence, the displacement of the bird is 1.8 km

vivado [14]3 years ago

3 0

It would be D because it is flying back to where it started 2-.2=1.8

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A thin, uniform stick of length 1.9 m and mass 3.1 kg is pinned through one end and is free to rotate. The stick is initially ha

nirvana33 [79]

Answer:

The acceleration is $\alpha = 7.10 \ rad/s^2$

Explanation:

From the question we are told that

The length of the stick is $d = 1.9 \ m$

The mass of the stick is $m = 3.1 \ kg$

The angular displacement is $\theta = 23.4^o$

Generally the torque of this uniform stick after this displacement is mathematically represented as

$\tau = \frac{1}{2} * d * [m*g]* sin (90 - \theta)$

$\tau = \frac{1}{2} * 1.9 * [3.1*9.8]* sin (90 - 23.4)$

$\tau = 26.49 \ kg\cdot m^2 \cdot s^{-2}$

Generally the moment of inertia of the uniform stick is mathematically represented as

$I = \frac{1}{3} * m * d^2$

=> $I = \frac{1}{3} * 3.1 * 1.9 ^2$

=> $I = 3.73 \ kg \cdot m^2$

Generally the angular acceleration is mathematically represented as

$\alpha = \frac{\tau}{I}$

=> $\alpha = \frac{26.49}{3.73}$

=> $\alpha = 7.10 \ rad/s^2$

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3 years ago

A student conducts an investigation on electricity and magnetism. Which relationship will the student discover between the curre

jekas [21]

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Elena-2011 [213]

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4 years ago

A 60-μC charge is held fixed at the origin and a −20-μC charge is held fixed on the x axis at a point x = 1.0 m. If a 10-μC char

Mila [183]

Answer:

Ek = 8,79 [J]

Explanation:

We are going to solve this problem, using the energy conservation principle

State 1 or initial state (charges at rest t=0)

E₁ = Ek + U₁

As charge are at rest Ek = 0

And U₁ has two components

U₁₂ = K * Q₁*Q₂ / 0,4 and U₃₂ = K*Q₃*Q₂ / 0,6

U₁₂ = 9*10⁹* 60*10⁻⁶*10*10⁻⁶/0,4 ⇒ U₁₂ = 9*60*10*10⁻³/0,4

U₃₂ = - 9*10⁹* 20*10⁻⁶*10*10⁻⁶/0,6 ⇒ U₃₂ = - 9*20*10*10⁻³/0,6

U₁₂ = 540*10⁻2/0,4 [J] ⇒13,5 [J]

U₃₂ = - 180*10⁻² /0,6 [J] ⇒ - 3 [J]

Then E₁ = E₁₂ + E₃₂

E₁ = 10,5 [J]

At the moment of Q₂ passing x = 40 cm or 0,4 m

E₂ = Ek + U₂

We can calculate the components of U₂ in this new configuration

U₂ = U₁₂ + U₃₂

U₁₂ = 9*10⁹* 60*10⁻⁶*10*10⁻⁶/0,7 ⇒ U₁₂ = 9*60*10*10⁻³/0,7

U₁₂ = 540*10⁻²/0,7 U₁₂ = 7,71 [J]

U₃₂ = - 9*10⁹* 20*10⁻⁶*10*10⁻⁶/0,3 ⇒ U₃₂ = - 9*20*10*10⁻³/0,3

U₃₂ = - 9*20*10⁻²/0,3

U₃₂ = - 6

U₂ = 7,71 -6

U₂ = 1,71 [J]

Then as

E₂ = Ek + U₂ and E₂ = E₁

Then

Ek + U₂ = E₁

Ek = 10,5 - U₂

Ek = 10,5 - 1,71

Ek = 8,79 [J]

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dsp73

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4 years ago

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