Use the scenario below for questions 4-7.

Later, the object slows down: this means that the magnitude of its velocity increases, therefore (since the velocity is negative) the curve must go upward, approaching and reaching the x-axis (which corresponds to zero velocity).

After that, the object's velocity keep increasing, but now it is positive: this means that the object is travelling in a direction opposite to the initial direction, so it has turned around.

Learn more about velocity:

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4 0

3 years ago

Positive Charge Q is distributed uniformly along the x-axis from x=0 to x=a. A positive point charge q is located on the positiv

deff fn [24]

Answer:

electric field E = - k Q (1 /r(r-a)), force F = - k Q qo / r (r-a) and force for r>>a F ≈ - k Q qo / r²

Explanation:

You are asked to find the electric field of a continuous charge distribution, so we must use the equation

E = k ∫dp /r²

Where k is the Coulomb constant that is worth 8.99 10⁹ N m² / C², r is the distance between the load distribution and the test charge, in this case everything is on the X axis.

We must find the charge differential (dq), let's use that uniformly distributed and create a linear charge density

λ = q / x

As it is constant, we can write it based on differentials

λ = dq / dx

dq = λ dx

We already have all the terms, let's integrate enter its limits, lower the distance from the left end of the distribution to the test charge (x = r) and the upper limit that is the distance from the left end of distribution to the test load ( x = r - a) where r> a

E = k ∫ λ dx / x²

E = k la (- 1 / x)

Let's get the negative sign from the parentheses

E = - k λ (1 / x)

E = - k λ (1 /(r-a) -1 /r) = - k λ [a / r (r-a)]

Let's change the charge density with the value of the total charge λ = Q / a

E = - k Q/a [a / r (r-a)]

E = - k Q (1 /r(r-a))

b) We calculate the force.

F = E qo

F = - k Q qo / r (r-a)

c) the force for charge porbe very far r >> a. In this case we can take r from the parentheses and neglect (a/r)

F = - k Qqo / r² (1 - a/r)

F ≈ - k Q qo / r²

6 0

3 years ago

6. Rock Ais thrown horizontally off a cliff with a velocity of 15 m/s. The rock lands 75m from

Harman [31]

Answer:

5 seconds

Explanation:

3 0

3 years ago