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3 years ago

What does the following figure illustrate?

Physics

2 answers:

Verdich [7]3 years ago

5 0

Answer:

the answer would be c sry if it is wrong

Ilia_Sergeevich [38]3 years ago

4 0

The answer to this question is C

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Two charges (q1 = 3.8*10-6C, q2 = 3.2*10-6C) are separated by a distance of d = 3.25 m. Consider q1 to be located at the origin.

Sergio039 [100]

Answer:

The distance is 1.69 m.

Explanation:

Given that,

First charge $q_{1}= 3.8\times10^{-6}\ C$

Second charge $q_{2}=3.2\times10^{-6}\ C$

Distance = 3.25 m

We need to calculate the distance

Using formula of electric field

$E_{1}=E_{2}$

$\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(d-x)^2}$

$\dfrac{q_{1}}{q_{2}}=\dfrac{(x)^2}{(d-x)^2}$

$\sqrt{\dfrac{q_{1}}{q_{2}}}=\dfrac{x}{d-x}$

$x=(d-x)\times\sqrt{\dfrac{q_{1}}{q_{2}}}$

Put the value into the formula

$x=(3.25-x)\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x+x\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x=\dfrac{3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}}{(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})}$

$x=1.69\ m$

Hence, The distance is 1.69 m.

5 0

3 years ago

A railroad track and a road cross at right angles. An observer stands on the road and watches an eastbound train traveling at 60

mamaluj [8]

Answer:

After 4 s of passing through the intersection, the train travels with 57.6 m/s

Solution:

As per the question:

Suppose the distance to the south of the crossing watching the east bound train be x = 70 m

Also, the east bound travels as a function of time and can be given as:

y(t) = 60t

Now,

To calculate the speed, z(t) of the train as it passes through the intersection:

Since, the road cross at right angles, thus by Pythagoras theorem:

$z(t) = \sqrt{x^{2} + y(t)^{2}}$

$z(t) = \sqrt{70^{2} + 60t^{2}}$

Now, differentiate the above eqn w.r.t 't':

$\frac{dz(t)}{dt} = \frac{1}{2}.\frac{1}{sqrt{3600t^{2} + 4900}}\times 2t\times 3600$

$\frac{dz(t)}{dt} = \frac{1}{sqrt{3600t^{2} + 4900}}\times 3600t$

For t = 4 s:

$\frac{dz(4)}{dt} = \frac{1}{sqrt{3600\times 4^{2} + 4900}}\times 3600\times 4 = 57.6\ m/s$

4 0

3 years ago

high school physics, no need detail explain, just give the answer, but you have to make sure thank you

Goshia [24]

Answer:many questions add point

6 0

3 years ago

Need help girliess.

steposvetlana [31]

If I remember correctly, it is the 3rd answer choice.

8 0

3 years ago

I need to know what is a product and a reactant in this please

Alekssandra [29.7K]

Answer:

2Mg + O2 = Reactant | 2MgO = Product

Explanation:

It is pretty simple, look:

The elements that are being added (or in the majority of the cases, the one who are in the left side) are always the Reactant.

The other side is the product of the reactants.

It is like a mathematic formula

We some the numbers to get the result, the numbers are the reactants and the result is the product.

(I hope you didn´t get confused)

8 0

3 years ago