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3 years ago

What term best describes seawater?

2 answers:

4 0

Since you didn't tell us the choices, I can pick anything I like.

The one that always does it for me is " foaming brine " .

svlad2 [7]3 years ago

3 0

Answer:

Homogenous mixture

Explanation:

It is a mixture, and since we can not easily distinguish between the parts (salt and water) it is a homogenous mixture.

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A car travels at a constant velocity of 40 m/s for 5.0 s, determine its displacement during this 5.0 s.

zimovet [89]

Answer:200 m

Explanation:

V=40m/s

T=5 s

Displacement=v*t

40*5=200

3 0

4 years ago

Please help I need this fast

Ghella [55]

The solution has reacted.

5 0

3 years ago

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.

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

4 0

3 years ago

Bob is pushing a box across the floor at a constant speed of 1.7 m/s, applying a horizontal force whose magnitude is 70 N. Alice

OlgaM077 [116]

Answer:

a) 70 N, b) b. Each initially applied a force bigger than static friction to get the box moving and accelerating, then when the desired final speed was achieved they reduced the force to make the net force zero.

Explanation:

a) A constant speed means that magnitude of friction force is equal to the magnitude of the external force. The friction force is directly proportional to the normal force, which is equal to the weight of the box. Therefore, the magnitude of the force is 70 N.

b) Alice used initially a greater force to accelerate the box up to needed speed and later reduced the external force to keep speed constant. The right choice is option b.

3 0

3 years ago

One knife is longer from the cutting edge to the back of the blade. It's easier to push the longer knife through a pear and more

AlladinOne [14]

By the formula of torque balance we know that

$\tau = r\times F$

here we know that

r = distance of force

F = applied force

so here we know that as we will take large edge knife then the distance of force will increase

due to this the torque will also increase in it so it is easy to pear it

So here we can say

$F_{in}r_1 = F_{out}r_2$

$F_{out} = \frac{r_1}{r_2} F_{in}$

so output force will increase in this case

So here the correct answer would be

<em>b. the longer the knife, the stronger the output force</em>

6 0

3 years ago