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3 years ago

Can someone pls help i’m not good at pysics

Physics

2 answers:

kykrilka [37]3 years ago

6 0

I’d say a stronger force because the moon doesn’t float away. But I’m
Not a scientist lol

Sonja [21]3 years ago

5 0

The earth pulls the moon with as much force as the moon pulls the earth.

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A wide sloping deposit of sediment formed where a stream leaves a mountain range is called

tatuchka [14]

An alluvial fan is a wide, sloping deposit of sediment formed where a stream leaves a mountain range. Make sure not to confuse it with a delta. A delta is a landform made of sediment that is deposited where a river flows into an ocean or lake. Hope this helped!

7 0

3 years ago

A charge q1 = +5.00 nC is placed at the origin of an xy-coordinate system, and a charge q2 = -2.00 nC is placed on the positive

Ivahew [28]

Answer:

The x- and y-components of the total force exerted is

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

The magnitude of the force is

$|F_{31 +32}| = 10.25 *10^{-5} N$

The direction of the force is

$\theta =327.43 ^o$ Clockwise from x-axis

Explanation:

From the question we are told that

The magnitude of the first charge is $q_1 = +5.00nC = 5.00*10^{-9}C$

The magnitude of the second charge is $q_2 = -2.00nC = -2.00*10^{-9}C$

The position of the second charge from the first one is $d_{12} = 4.00i \ cm = \frac{4.00i}{100} = 4.00i *10^{-2} m$

The magnitude of the third charge is $q_3 = +6.00nC = 6.00*10^{-9}C$

The position of the third charge from the first one is $\= d_{31} = (4i + 3j) cm = \frac{ (4i + 3j)}{100} = (4i + 3j) *10^{-2}m$

$|d_{31}| =(\sqrt{4 ^2 + 3^2}) *10^{-2} m$

$|d_{31}| =5 *10^{-2} m$

The position of the third charge from the second one is

$\= d_{32} = 3j cm = 3j *10^{-2}m$

$|d_{32}| =(\sqrt{ 3^2}) *10^{-2} m$

$|d_{32}| =3 *10^{-2} m$

The force acting on the third charge due to the first and second charge is mathematically represented as

$F_{31 +32} = \frac{kq_3 q_1}{|d_{31}| ^3} *\= d_{31} + \frac{kq_3 q_2}{|d_{32}| ^3} *\= d_{32}$

Substituting values

$F_{31 +32} = \frac{9 *10^9 * 6 *10^{-9} * 5*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2} \\ \ + \ \ \ \ \ \ \ \ \ \frac{9 *10^9 * 6 *10^{-9} * -2*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2}$

$F_{31 +32} = 2.16 *10^{-5} (4i + 3j) - 12*10^{-5} j$

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

The magnitude of $F_{31 +32}$ is mathematically evaluated as

$|F_{31 +32}| = \sqrt{(8.64^2 + 5.52 ^2) } *10^{-5}$

$|F_{31 +32}| = 10.25 *10^{-5} N$

The direction is obtained as

$tan \theta = \frac{-5.52 *10^{-5}}{8.64 *10^{-5}}$

$\theta = tan ^{-1} [-0.63889]$

$\theta = - 32.57 ^o$

$\theta = 360 - 32.57$

$\theta =327.43 ^o$

5 0

3 years ago

An electric current always produces a magnetic field that is ______ to the field. A. perpendicular B. parallel C. at a 37° angle

Aleks04 [339]

Answer:perpendicular

Explanation:

8 0

2 years ago

Which if, any, of these statements are true? (More than one may be true.) Assume the batteries are ideal. Check all that apply.

Maurinko [17]

Answer:

All are true except the last point that says that a battery is a current source and the current at the outlet is always the same.

Explanation:

A battery is an electro-chemical device which converts the chemical energy into usable electrical energy thus it provides electrical energy.
Since, the battery maintains a a constant potential difference between its terminals, once connected.
Since, the movement of electric current requires energy, which is supplied by the electric potential energy stored in the battery.
The current in the battery flows as per the Ohm's law and we can not say that the current leaving will always remain constant.
As the current is the flow of electric charge, and charges are not stored in batteries unlike capacitors, thus the current at the leaving end will depend on Ohm's law and will vary accordingly.

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2 years ago

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slamgirl [31]

D neutrons and protons

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3 years ago

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