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3 years ago

Please I really need the help

Physics

1 answer:

Lunna [17]3 years ago

3 0

Answer:

7) there is a theory that these animals can see the magnetic field of the Earth due to a compass like mechanism inside their eyes in order to navigate.

8) Earth's magnetic field is a result of the currents found inside the outer core that consist of electricity.

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The jogger ran 3km east.

erma4kov [3.2K]

Answer: d

Explanation:

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4 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS AND I NEED ALL CORRECT ANSWERS ONLY!!!

Dominik [7]

the purpose of fuses and circuit breakers is (first answer)

7 0

3 years ago

Read 2 more answers

Answer the following questions for a mass that is hanging on a spring and oscillating up and down with simple harmonic motion. N

LiRa [457]

Answer:

1. equilibrium

2. bottom

3. bottom

4. nowhere

5. bottom

6. top & bottom

7. equilibrium

8. equilibrium

1. No

2. Yes

Explanation:

According to the following equation of motion for SHM:

$x(t) = A\cos(\omega t + \phi)$

where A is the amplitude, ω is the angular frequency, and ∅ is the phase angle.

Furthermore, the velocity and acceleration functions are as follows:

$y(t) = -\omega A\sin(\omega t + \phi)\\a(t) = -\omega^2 A\cos(\omega t + \phi)$

1. The acceleration is zero at the equilibrium. At the equilibrium, the net force on the object is zero. And according to Newton's Second Law, if the net force is zero, then the acceleration is zero as well.

2. The forces on the object in a vertical spring are the weight of the object and the spring force.

F = mg - kx

Since mg is constant along the motion, then the net force is maximum at the amplitude. For the special case in this question, the mass is always below the rest length of the spring. So the net force is maximum at the lower amplitude, because x is greater in magnitude at the lower amplitude. According to Newton's Second Law, acceleration is proportional to the net force, hence the acceleration is at a maximum at the bottom.

3. As explained above, the magnitude of the net force is at a maximum at the lower amplitude, that is bottom.

4. The spring force is defined by Hooke's Law: F = -kx. Since the oscillation is small enough so that the mass is always below the rest length of the spring, then x is always greater than zero, hence nowhere in the motion will the spring force becomes zero.

5. As explained above, the force of gravity is constant and the spring force is proportional to the displacement, x. Therefore, the spring force is at a maximum at the lower amplitude, that is bottom.

6. The speed is zero when the mass is instantaneously at rest, that is the amplitude.

7. The net force on the mass is zero at the equilibrium.

8. The speed is at a maximum at the equilibrium.

1. We will use the equation of motions given above. For simplicity, let's take ∅ = 0. At half its amplitude:

$\frac{A}{2} = A\cos(\omega t)\\\frac{1}{2} = \cos(\omega t)\\\omega t = \pi / 3$

Then the velocity at that point is

$v(t) = -\omega A\sin(\pi /3) = -\omega A (0.866)$

The maximum speed is where the acceleration is equal to zero:

$0 = -\omega^2 A\cos(\omega t)\\\omega t = \pi / 2\\v_{max} = -\omega A\sin(\pi /2) = -\omega A$

Comparing the maximum velocity to the velocity at A/2 yields that it is not half the maximum velocity:

$-\omega A(0.866) \neq -\omega A$

2. The maximum acceleration is at the amplitude.

$A = A\cos(\omega t)\\\omega t = 2\pi\\a_{max} = -\omega^2 A\cos(2\pi) = -\omega^2 A$

And the acceleration at A/2 is

$\frac{A}{2} = A\cos(\omega t)\\\omega t = \pi / 3\\a(t) = -\omega^2 A\cos(\pi / 3) = -\omega^2 A (0.5)$

Comparing these two results yields that the acceleration at half the amplitude is half the maximum acceleration.

5 0

3 years ago

What is the process of preparing a person to perform strenuous work and to recover from that work as quickly as possible?

vesna_86 [32]

The process of preparing a person to perform strenuous work and to recover from that work as quickly as possible is called training, I bet. Such w<span>arm up exercises as like stretching and other should lead person to the recovery. Hope it will help.</span>

3 0

4 years ago

A long line carrying a uniform linear charge density+50.0μC/m runs parallel to and 10.0cm from the surface of a large, flat plas

IRISSAK [1]

Answer:

The location of the points is at the distance above the line charge Option A.

Explanation:

Part A

The electric field due to a line of charge is expressed as :

Eₐ = λ /2πε₀L …… (1)

while the electric field due to a non-conducting elastic sheet is expressed as:

Еₙ = λ /2ε₀ …… (2)

In order for the particle to experience no force, the electric field at that point is zero.

Equating (1) and (2) together, we have:.

λ /2πε₀L = σ /2ε₀

Rearranging the above expression, we have:

L = λ /πσ

Substituting 50.0 μC/m for λ, 3.14 for π , and 100 μC/ m² for σ, we have:

L= 50.0 μC/m / (3.14)* 100 μC/ m²

L = 0.159m

From the L value obtained above, the location of all points where an particle would experience no force due to the arrangement of charged objects is L 0.159m from the line.

Part B

The electric field is zero at all the points above the line with the distance 0.159m from the line.

Thus, the location of the points is at the distance above the line charge Option A.

6 0

3 years ago