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Mars2501 [29]
3 years ago
15

Which organism is the primary consumer in this food chain?

Physics
1 answer:
VashaNatasha [74]3 years ago
3 0
Rabbit hope this u out hollow
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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
The age of the universe is thought to be about 14 billion years. Assuming two significant figures, write this in powers of ten i
ira [324]
A ) 14 billion years =  14,000,000,000
1.4 * 10^{10}
b ) 1 year = 365 · 24 · 3,600 = 31,536 · 10³ 
1.4 * 10 ^{10}*3.1536 * 10^{7}  =4.4 * 10 ^{17}

6 0
4 years ago
Read 2 more answers
What is a materials ability to be dissolved in a solvent
weeeeeb [17]
Answer: Solubility

Solubility is the ability to be dissolved. Saturation is when the concentration is too high(more than solubility) that when you add another material it won't dissolve. Solute is the material that dissolved. Solvent is the material that used for dissolving
5 0
4 years ago
A wheel 33 cm in diameter accelerates uniformly from 240 rpm to 360 rpm in 6.5 s. how far will a point on the edge of the wheel
melisa1 [442]

First we find the angular acceleration \alpha. The angular velocities are \frac{240}{60} =4 \;rps and \frac{360}{60} =6 \;rps The angular velocities and \alpha are related as

\omega=\omega_0+\alpha t\\ \alpha =\frac{\omega - \omega_0}{t} \\ \alpha =\frac{6-4}{(6.5)} \\ \alpha =0.3077 \;rps^2\\ \alpha =(0.3077 )2\pi =1.933 \;rad/s^2

Angle turned in 6.5 seconds is

2 \alpha\theta =   \omega^2-\omega_0^2\\ \theta =  \frac{ \omega^2-\omega_0^2}{2 \alpha}\\  \theta =  \frac{ 6^2-4^2}{2 (1.933)}\\  \theta = 5.173 \; rad

The distance traveled by a point on the edge of the wheel is r \theta = \frac{33}{2}(5.173)= 85.35 \;cm

7 0
3 years ago
An object moving at a constant velocity accelerates at a constant rate of
azamat

Answer:

70 m/s

Explanation:

v = u + at

100 = u + 5(120/5)

100 = u + 30

u = 70

3 0
3 years ago
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