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vlabodo [156]
2 years ago
9

Hope you can answer this: A Student Visits A Farm And Makes These Notes In Her Journal.

Physics
1 answer:
Strike441 [17]2 years ago
8 0

Answer:

Common Sense

Explanation:

The chick has probably seen other chicks get caught by the Hawk and knows not to go near, or saw a giant bird flying straight towards it and used common sense to identify it as danger and run away. Although if this is for a test or a grade or something, please do not use the answer, it is most likely incorrect. This is honestly my best answer.

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Driving a motor vehicle requires many coordinated functions which are impacted by alcohol and other drugs
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I am assuming this is a true or false question, to which the answer would be True.

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A charge Q is distributed uniformly along the x axis from x1 to x2. What would be the magnitude of the electric field at x0 on t
Lena [83]

Answer:

  E =  k Q    1 / (x₀-x₂) (x₀-x₁)

Explanation:

The electric field is given by

              dE = k dq / r²

In this case as we have a continuous load distribution we can use the concept of linear density

              λ= Q / x = dq / dx

              dq = λ dx

We substitute in the equation

           ∫ dE = k ∫ λ dx / x²

We integrate

           E = k λ (-1 / x)

We evaluate between the lower limits x = x₀- x₂ and higher x = x₀-x₁

           E = k λ (-1 / x₀-x₁ + 1 / x₀-x₂)

           E = k λ  (x₂ -x₁) / (x₀-x₂) (x₀-x₁)

We replace the density

             E = k (Q / (x₂-x₁)) [(x₂-x₁) / (x₀-x₂) (x₀-x₁)]

             E =  k Q    1 / (x₀-x₂) (x₀-x₁)

8 0
2 years ago
A balloon is at a height of 81m and is ascending upwards with a velocity of 12m/s. A body of 2kg weight is dropped from it. If g
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Irrigation channels that require regular flow monitoring are often equipped with electromagnetic flowmeters in which the magneti
san4es73 [151]

A) 1.36\cdot 10^{-4}T

The magnetic field at the center of a coil of N turns is given by

B=\frac{\mu_0 N I}{2R}

where

I is the current in the coil

N is the number of turns

R is the radius of the coil

Here we have

I = 6.5 A is the current in the coil

N = 100 is the number of turns

R=\frac{6.0 m}{2}=3.0 m is the radius of the coil

Substituting,

B=\frac{(4\pi \cdot 10^{-7} H/m)(100)(6.5 A)}{2(3.0 m)}=1.36\cdot 10^{-4}T

B) The force points north

The direction of the force on a positive ion in water can be found by using the right-hand rule. In fact, we have:

- Index finger: direction of motion of the ion --> towards east

- Middle finger: direction of magnetic field --> downward

- Thumb: direction of the force --> towards north

So, the force points north.

C) 3.26\cdot 10^{-23}N

The magnitude of the magnetic force on a charged particle moving perpendicularly to the field is

F=qvB

where

q is the charge of the particle

v is the velocity

B is the magnitude of the magnetic field

In this case, we have

q=+e=1.6\cdot 10^{-19} C is the charge

v=1.5 m/s is the velocity

B=1.36\cdot 10^{-4}T is the magnetic field strength

Substituting,

F=(1.6\cdot 10^{-19} C)(1.5 m/s)(1.36\cdot 10^{-4}T)=3.26\cdot 10^{-23}N

8 0
2 years ago
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