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lubasha [3.4K]
2 years ago
14

I need the solution to this

Physics
1 answer:
posledela2 years ago
6 0

Answer:

He could jump 2.6 meters high.

Explanation:

Jumping a height of 1.3m requires a certain initial velocity v_0. It turns out that this scenario can be turned into an equivalent: if a person is dropped from a height of 1.3m in free fall, his velocity right before landing on the ground will be v_0. To answer this equivalent question, we use the kinematic equation:

v_0 = \sqrt{2gh}=\sqrt{2\cdot 9.8\frac{m}{s^2}\cdot 1.3m}=5.0\frac{m}{s}

With this result, we turn back to the original question on Earth: the person needs an initial velocity of 5 m/s to jump 1.3m high, on the Earth.

Now let's go to the other planet. It's smaller, half the radius, and its meadows are distinctly greener. Since its density is the same as one of the Earth, only its radius is half, we can argue that the gravitational acceleration g will be <em>half</em> of that of the Earth (you can verify this is true by writing down the Newton's formula for gravity, use volume of the sphere times density instead of the mass of the Earth, then see what happens to g when halving the radius). So, the question now becomes: from which height should the person be dropped in free fall so that his landing speed is 5 m/s ? Again, the kinematic equation comes in handy:

v_0^2 = 2g_{1/2}h\implies \\h = \frac{v_0^2}{2g_{1/2}}=\frac{25\frac{m^2}{s^2}}{2\cdot 4.9\frac{m}{s^2}}=2.6m

This results tells you, that on the planet X, which just half the radius of the Earth, a person will jump up to the height of 2.6 meters with same effort as on the Earth. This is exactly twice the height he jumps on Earth. It now all makes sense.

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Answer:

a. b- x= y

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Explanation:

a. x components:

dE = \frac{kdq}{(a+r-x^{2}) } \\

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Integrating and solving gives:

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b. the force is given by the equation derived from (a.):

F = \frac{-kQqi}{r (a+r)}

c. Given that r>>a, the expression becomes:

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Explanation:

When the size of the charge distribution is less than the distance to the deviation point of the charge then the charge distribution would produce the same effect such as a linear charge.

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3 years ago
The Heaviside function H is defined by H(t)={0 if t&lt;0, 1 if t≥0 It is used in the study of electric circuits to represent the
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Answer:

V(t)= 240V* H(t-5)

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The heaviside function is defined as:

H(t) =1 \quad t\geq 0\\H(t) =0 \quad t

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Thus we define a function f:

f(t) = H(t-5)

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Now we just need to add a scale up factor of 240 V, because thats the voltage applied after the heaviside function switches on. (H(t-5) =1 when t\geq 5, so it becomes just a 1, which we can safely ignore.)

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V(t)= 240V* H(t-5)

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- the vertical motion is an uniformly accelerated motion, with constant acceleration g=9.81 m/s^2, initial position h (the height of the building) and initial vertical velocity v_0 \sin \alpha (with a negative sign, since it points downwards)

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c) We want the ball to reach a point 10.0 meters below the level of launching, so we want to find the time t such that 
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Answer:

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