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kaheart [24]
3 years ago
10

Define these terms about speed and state their units speed​distance coveredtime taken

Physics
1 answer:
Neporo4naja [7]3 years ago
4 0

Explanation:

Speed is the rate of change of distance with time. It is a scalar quantity with magnitude both no direction.

  Speed  = \frac{distance}{time}

The unit is m/s or km/hr or mile/hr

Distance covered is simply the length of the path traveled.

 The unit is m or km or miles

Time taken is the duration of an event.

  The unit is seconds or minutes or hour.

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Albert's laboratory is filled with a constant uniform magnetic field pointing straight up. Albert throws some charges into this
guajiro [1.7K]

Answer:

\vec{F}=qB(v_y \hat{i} - v_x\hat{j})

Explanation:

The force excerted by a magnetic field on a charged particle is given by the Lorentz force:

\vec{F} = q \vec{v} \times \vec{B}

Lets consider the z-direction of our coordinate system the same direction of the magnetic field, that is:

\vec{B} = B \hat{k}

Let us consider that the velocity of a given particle is:

\vec{v} = v_x\hat{i} + v_y \hat{j} + v_z \hat{k}

Therefore, since k×k = 0

\vec{v} \times \vec{B} = (v_x\hat{i} + v_y \hat{j} + v_z \hat{k}) \times B\hat{k}\\\vec{v} \times \vec{B}  =  (Bv_x \hat{i} \times\hat{k}) + (Bv_y \hat{j}\times\hat{k})

And since  i, j , k are a rigth hand system:

i × j = k

j × k = i

k × i = j  --> i × k= -j

\vec{v} \times \vec{B}  = Bv_x (-\hat{j}) + Bv_y \hat{i} =  Bv_y \hat{i} - Bv_x\hat{j}

Threfore, if the particle has charge q and velocity v = (vx,vy,vz), the magnetic force it will feel will be

\vec{F}=qB(v_y \hat{i} - v_x\hat{j})

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If you repeat an experiment and the results are very different from the results you got the first time, the next step would be t
grigory [225]
Redo the previous experiments.

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A marble on a frictionless track, starting from point A in the drawing, is projected down the curved runway. (This means that th
EleoNora [17]

Answer:

v = 4.4 m / s

Explanation:

Unfortunately, the exercise scheme does not appear. Let's analyze the problem the marble leaves point A with an initial velocity, goes down and then rises to a given height where its velocity is zero, in the whole trajectory they tell us that the resistance is zero, so we can use the conservation relations of the enegy.

Starting point. Point A

          Em₀ = K + U = ½ m v2 + mg y_a

point B.

          Em_f = U = m g y

the energy is conserved

         Em₀ = Em_f

         ½ m v² + mg y_a = m g y

        ½ m v² = m g (y -y_a)

         v = \sqrt {2g ( y - y_a)}

         In the exercise the diagram is not seen, but the height of point A must be known, suppose that y_a = 4 m

       v = \sqrt{ 2 \  9.8 ( 5 -4)}

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