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Grace [21]
3 years ago
12

Which molecules from food and air ends up with the cell in the body

Physics
1 answer:
ozzi3 years ago
4 0

Answer:

Hmm look it up

Explanation:

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What would be most useful to help make a simple compass a nonmetal bar,a round metal can,a small iron nail,or a Quartz needle
pav-90 [236]
Iron nail. the rest of those are not iron or some form of magnetic material.
5 0
3 years ago
Read 2 more answers
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
3 years ago
A ride-sharing car moving along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed
fgiga [73]

Answer:

Explanation:

Time taken to accelerate to 28 m /s

= 28 / 2 = 14 s

a ) Total length of time in motion

= 14 + 41 + 5

= 60 s .

b )

Distance covered while accelerating

s = ut + 1/2 at²

= 0 + .5 x 2 x 14²

= 196 m .

Distance covered while moving in uniform motion

= 28 x 41

= 1148 m

distance covered while decelerating

v = u - at

0 = 28 - a x 5

a = 5.6 m / s²

v² = u² - 2 a s

0 = 28² - 2 x 5.6 x s

s = 28² / 2 x 5.6

= 70 m .

Total distance covered

= 196 + 1148 + 70

= 1414 m

total time taken = 60 s

average velocity

= 1414 / 60

= 23.56 m /s .

8 0
3 years ago
The mass of the uniform cantilever is 1100 kg. Determine the force on the beam at A. Determine the force on the beam at B. Use C
Mashcka [7]

Answer:

Force A=-−2,697.75 N

Force B=13, 488.75 N

Explanation:

Taking moments at point A, the sum of clockwise and anticlockwise moments equal to zero.

25 mg-20Fb=0

25*1100g=20Fb

Fb=25*1100g/20=1375g

Taking g as 9.81 then Fb=1375*9.81=13,488.75 N

The sum of upward and downward forces are same hence Fa=1100g-1375g=-275g

-275*9.81=−2,697.75. Therefore, force A pulls downwards

Note that the centre of gravity is taken to be half the whole length hence half of 50 is 25 m because center of gravity is always at the middle

8 0
3 years ago
Two pipes of equal length are each open at one end. Each has a fundamental frequency of 470 Hz at 310 K. In one pipe the air tem
Tresset [83]

Answer:

Given:

Fundamental frequency: 470Hz

T1:310k,T2:315k

Calculating velocity

Recall v=(331m/s)✓[T1/273k)

V=331✓(310/273)

V1=331*(1.0656)=352.72m/s

V2=331✓(315/273)=355.5m/s

Fundamental frequency=4L

F2=F1(V2/V1)

F2=470(355.5/352.72)=474.4Hz

Beat=[F2-F1]=474.4-470=4.4Hz

Explanation:

7 0
3 years ago
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