How do you do question 2?

Physics

1 answer:

Inessa [10]3 years ago

7 0

Answer:

Look at explanation

Explanation:

a) Weight is another term for how much gravity is on an object and can be calculated by using the local gravity*mass so in order to find weight divide 1200 by g

m= 1200/9.8= 122.45kg

b) We know Fnet=ma and Fnet=Fapp+Fresistivity so Fapp+Fresistivity=ma

Plug in values

500+Fresistivity=122.45*2

solve for Fresistivty= 244.9-500=-255.1N (the reason it is negative is because it is in the opposite direction

c)Power= ΔE/Δt and we also know ΔE=ΔWork so Power= ΔWork/Δt. If a person pulls harder, they have a greater force and since mass is constant, acceleration is greater and since the amount of time needed to cover A to B is reduced since x-x0=v0t+1/2at²(v0=0, when you solve for t it will be lower because acceleration increases). If t decreases than Power increases by inverse proportionality. Work =Fd if the amount of Force increases by distance travelled remains constant than work also increases so power will also increase.

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A uniformly charged, one-dimensional rod of length L has total positive charge Q. Itsleft end is located at x = ????L and its ri

GREYUIT [131]

Answer:

$|\vec{F}| = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))$

Explanation:

The force on the point charge q exerted by the rod can be found by Coulomb's Law.

$\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r$

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.

In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.

We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.

Applying Coulomb's Law:

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qdq}{x + x_0}(\^x)$

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.

Now, we have to write 'dq' in term of the known quantities.

$\frac{Q}{L} = \frac{dq}{dx}\\dq = \frac{Qdx}{L}$

Now, substitute this into 'dF':

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qQdx}{L(x+x_0)}(\^x)$

Now we can integrate dF over the rod.

$\vec{F} = \int{d\vec{F}} = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}\int\limits^{L}_0 {\frac{1}{x+x_0}} \, dx = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))(\^x)$