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BlackZzzverrR [31]
3 years ago
6

Two massive particles with identical charge are launched into the uniform field between two plates from the same launch point wi

th the same velocity. They both impact the positively charged plate, but the second one does so four times as far as the first. What sign is the charge? What physical difference would give them different impact points (quantify as a percent)? How does this compare to the gravitational projectile motion case?
Physics
1 answer:
Svetach [21]3 years ago
6 0
<h2><em><u>A</u></em><em><u>n</u></em><em><u>s</u></em><em><u>w</u></em><em><u>e</u></em><em><u>r</u></em><em><u>n</u></em><em><u>s</u></em><em><u>w</u></em><em><u>e</u></em><em>:</em></h2>

<h2>The interaction between charged object is a non-contact force that acts over some distance of separation.Charge,charge and distance.Every <em>e</em><em>l</em><em>e</em><em>c</em><em>t</em><em>r</em><em>i</em><em>c</em><em>a</em><em>l</em><em> </em><em>i</em><em>n</em><em>t</em><em>e</em><em>r</em><em>a</em><em>c</em><em>t</em><em>i</em><em>o</em><em>n</em><em> </em><em>i</em><em>n</em><em>v</em><em>o</em><em>l</em><em>v</em><em>e</em><em>s</em><em> </em><em>a</em><em> </em><em>f</em><em>o</em><em>r</em><em>c</em><em>e</em><em> </em><em>t</em><em>h</em><em>a</em><em>t</em><em> </em><em>h</em><em>i</em><em>g</em><em>h</em><em>l</em><em>i</em><em>g</em><em>h</em><em>t</em><em>s</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>i</em><em>m</em><em>p</em><em>o</em><em>r</em><em>t</em><em>a</em><em>n</em><em>c</em><em>e</em><em> </em><em>o</em><em>f</em><em> </em><em>t</em><em>h</em><em>e</em><em>s</em><em>e</em><em> </em><em>t</em><em>h</em><em>r</em><em>e</em><em>e</em><em> </em><em>v</em><em>a</em><em>r</em><em>i</em><em>a</em><em>b</em><em>l</em><em>e</em><em>s</em></h2>

er

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Carol is farsighted ( presbyopia) and cannot see objects clearly that are closer to her eyes than about meter. She sees objects
egoroff_w [7]

Answer:

1.0 dioptres

Explanation:

Farsightedness is an eye defect in which a person can see far objects clearly but not near objects. That implies that the patients' near point is farther than 25cm which is the normal least distance of distinct vision.

Farsightedness results from the eyeball being too long or the crystalline lens not being sufficiently converging.

Carol is farsighted with a near point of about a meter (100cm). We desire to make a lens to enable her near point be reduced to about 50cm. The focal length and power of this lens is calculated in the image attached.

The power of a lens is the inverse of its focal length in meters hence the 100 in the formula for power of the lens.

7 0
3 years ago
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Autorhythmic cells can generate action potentials spontaneously because they have Autorhythmic cells can generate action potenti
OlgaM077 [116]

Answer:

They can generate potentials spontaneously because they have Unstable Membrane Potentials.

Explanation:

Autorythmic cells or Pacemaker cells are cells that provide Action potentials (electrical impulses) that starts off the cardiac cycle.

N:B This action potential is created spontaneously.

To explain further, the heart originate in specialized cardiac muscle cells, called autorhythmic cells, that can excite themselves and therefore are able to generate an action potential without external stimulation by nerve cells. And this sets the cardiac cycle i

(Pumping of the heart) into motion. (The pace maker potential)

The Autorhythmic cells create an action potential spontaneously

And this is possible because they have an UNSTABLE RESTING POTENTIAL that is continuously depolarizing, while it drifts slowly toward threshold. As Na+ ions enter the cell, the inner surface of the plasma membrane becomes less negative gradually, thus generating the pacemaker potential.

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3 years ago
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Coulomb’s law and static point charge ensembles (15 points). A test charge of 2C is located at point (3, 3, 5) in Cartesian coor
fenix001 [56]

Answer:

a) F_{r}= -583.72MN i + 183.47MN j + 6.05GN k

b) E=3.04 \frac{GN}{C}

Step-by-step explanation.

In order to solve this problem, we mus start by plotting the given points and charges. That will help us visualize the problem better and determine the direction of the forces (see attached picture).

Once we drew the points, we can start calculating the forces:

r_{AP}^{2}=(3-0)^{2}+(3-0)^{2}+(5+0)^{2}

which yields:

r_{AP}^{2}= 43 m^{2}

(I will assume the positions are in meters)

Next, we can make use of the force formula:

F=k_{e}\frac{q_{1}q_{2}}{r^{2}}

so we substitute the values:

F_{AP}=(8.99x10^{9})\frac{(1C)(2C)}{43m^{2}}

which yields:

F_{AP}=418.14 MN

Now we can find its components:

F_{APx}=418.14 MN*\frac{3}{\sqrt{43}}i

F_{APx}=191.30 MNi

F_{APy}=418.14 MN*\frac{3}{\sqrt{43}}j

F_{APy}=191.30MN j

F_{APz}=418.14 MN*\frac{5}{\sqrt{43}}k

F_{APz}=318.83 MN k

And we can now write them together for the first force, so we get:

F_{AP}=(191.30i+191.30j+318.83k)MN

We continue with the next force. The procedure is the same so we get:

r_{BP}^{2}=(3-1)^{2}+(3-1)^{2}+(5+0)^{2}

which yields:

r_{BP}^{2}= 33 m^{2}

Next, we can make use of the force formula:

F_{BP}=(8.99x10^{9})\frac{(4C)(2C)}{33m^{2}}

which yields:

F_{BP}=2.18 GN

Now we can find its components:

F_{BPx}=2.18 GN*\frac{2}{\sqrt{33}}i

F_{BPx}=758.98 MNi

F_{BPy}=2.18 GN*\frac{2}{\sqrt{33}}j

F_{BPy}=758.98MN j

F_{BPz}=2.18 GN*\frac{5}{\sqrt{33}}k

F_{BPz}=1.897 GN k

And we can now write them together for the second, so we get:

F_{BP}=(758.98i + 758.98j + 1897k)MN

We continue with the next force. The procedure is the same so we get:

r_{CP}^{2}=(3-5)^{2}+(3-4)^{2}+(5-0)^{2}

which yields:

r_{CP}^{2}= 30 m^{2}

Next, we can make use of the force formula:

F_{CP}=(8.99x10^{9})\frac{(7C)(2C)}{30m^{2}}

which yields:

F_{CP}=4.20 GN

Now we can find its components:

F_{CPx}=4.20 GN*\frac{-2}{\sqrt{30}}i

F_{CPx}=-1.534 GNi

F_{CPy}=4.20 GN*\frac{2}{\sqrt{30}}j

F_{CPy}=-766.81 MN j

F_{CPz}=4.20 GN*\frac{5}{\sqrt{30}}k

F_{CPz}=3.83 GN k

And we can now write them together for the third force, so we get:

F_{CP}=(-1.534i - 0.76681j +3.83k)GN

So in order to find the resultant force, we need to add the forces together:

F_{r}=F_{AP}+F_{BP}+F_{CP}

so we get:

F_{r}=(191.30i+191.30j+318.83k)MN + (758.98i + 758.98j + 1897k)MN + (-1.534i - 0.76681j +3.83k)GN

So when adding the problem together we get that:

F_{r}=(-0.583.72i + 0.18347j +6.05k)GN

which is the answer to part a), now let's take a look at part b).

b)

Basically, we need to find the magnitude of the force and divide it into the test charge, so we get:

F_{r}=\sqrt{(-0.583.72)^{2} + (0.18347)^{2} +(6.05)^{2}}

which yields:

F_{r}=6.08 GN

and now we take the formula for the electric field which is:

E=\frac{F_{r}}{q}

so we go ahead and substitute:

E=\frac{6.08GN}{2C}

E=3.04\frac{GN}{C}

7 0
4 years ago
Question 3 of 10
Eva8 [605]

Answer:

D. Newton's second law

Explanation:

Newton's second law of motion states that force of an object is a product of its mass and its acceleration.

Mathematically, F= ma where  m is mass and a is acceleration

So from the statement above : The acceleration of an object is proportional to the force applied to it and  inversely proportional to its mass , it can be seen from the formula variation as;

F= ma -----making a the subject of the formula

a= F/ m

a= 1/m * F --------- a  is inversely related to m  as you can see from 1/m but directly related to F  hence;

Increase in mass with the same force applied causes the body to accelerate slower where as when force increases, the body accelerates faster.

5 0
3 years ago
suppose you want to determine the surface area of this sugar cube. it has edges that are each 2 cm long. if you cut the cube in
ivolga24 [154]

Answer:

Half: 6 cm^2    Whole: 12 cm^2

Explanation:

First, we know that the edges of the cube are 2 cm long. So there are 6 faces on a cube. We do 2x6=12 cm^2 as our total surface area. Then it asks for each half. So you would divide it by 2 and get 6 cm^2 as your half.

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