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kvv77 [185]
3 years ago
13

The quantity of charge q (in coulombs) that has passed through a surface of area 2.00 cm2 varies with time

Physics
1 answer:
Makovka662 [10]3 years ago
7 0

Explanation:

We have,

Surface area, A=2\ cm^2=0.0002\ m^2

The current varies wrt time t as :

q(t) = 4t^3 + 5t + 6

(a) At t = 2 seconds, electrical charge is given by :

q(t) = 4t^3 + 5t + 6\\\\q(2) = 4(2)^3 + 5(2) + 6\\\\q=48\ C

(b) Current is given by :

I=\dfrac{dq}{dt}\\\\I=\dfrac{d(4t^3 + 5t + 6)}{dt}\\\\I=12t^2+5

Instantaneous current at t = 1 s is,

I=12(1)^2+5=17\ A

(c) Current is, I=12t^2+5

Current density is given by electric current per unit area.

J=\dfrac{I}{A}\\\\J=\dfrac{(12t^2+5)}{0.0002}\\\\J=5000(12t^2+5)\ A/m^2

Therefore, it is the required explanation.

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Continuous and aligned fiber-reinforced composite with cross-sectional area of 340 mm2 (0.53 in.2) is subjected to a longitudina
Alecsey [184]

(a) 23.4

The fiber-to-matrix load ratio is given by

\frac{F_f}{F_m}=\frac{E_f V_f}{E_m V_m}

where

E_f = 131 GPa is the fiber elasticity module

E_m = 2.4 GPa is the matrix elasticity module

V_f=0.3 is the fraction of volume of the fiber

V_m=0.7 is the fraction of volume of the matrix

Substituting,

\frac{F_f}{F_m}=\frac{(131 GPa)(0.3)}{(2.4 GPa)(0.7)}=23.4 (1)

(b) 44,594 N

The longitudinal load is

F = 46500 N

And it is sum of the loads carried by the fiber phase and the matrix phase:

F=F_f + F_m (2)

We can rewrite (1) as

F_m = \frac{F_f}{23.4}

And inserting this into (2):

F=F_f + \frac{F_f}{23.4}

Solving the equation, we find the actual load carried by the fiber phase:

F=F_f (1+\frac{1}{23.4})\\F_f = \frac{F}{1+\frac{1}{23.4}}=\frac{46500 N}{1+\frac{1}{23.4}}=44,594 N

(c) 1,906 N

Since we know that the longitudinal load is the sum of the loads carried by the fiber phase and the matrix phase:

F=F_f + F_m (2)

Using

F = 46500 N

F_f = 44594 N

We can immediately find the actual load carried by the matrix phase:

F_m = F-F_f = 46,500 N - 44,594 N=1,906 N

(d) 437 MPa

The cross-sectional area of the fiber phase is

A_f = A V_f

where

A=340 mm^2=340\cdot 10^{-6}m^2 is the total cross-sectional area

Substituting V_f=0.3, we have

A_f = (340\cdot 10^{-6} m^2)(0.3)=102\cdot 10^{-6} m^2

And the magnitude of the stress on the fiber phase is

\sigma_f = \frac{F_f}{A_f}=\frac{44594 N}{102\cdot 10^{-6} m^2}=4.37\cdot 10^8 Pa = 437 MPa

(e) 8.0 MPa

The cross-sectional area of the matrix phase is

A_m = A V_m

where

A=340 mm^2=340\cdot 10^{-6}m^2 is the total cross-sectional area

Substituting V_m=0.7, we have

A_m = (340\cdot 10^{-6} m^2)(0.7)=238\cdot 10^{-6} m^2

And the magnitude of the stress on the matrix phase is

\sigma_m = \frac{F_m}{A_m}=\frac{1906 N}{238\cdot 10^{-6} m^2}=8.0\cdot 10^6 Pa = 8.0 MPa

(f) 3.34\cdot 10^{-3}

The longitudinal modulus of elasticity is

E = E_f V_f + E_m V_m = (131 GPa)(0.3)+(2.4 GPa)(0.7)=41.0 Gpa

While the total stress experienced by the composite is

\sigma = \frac{F}{A}=\frac{46500 N}{340\cdot 10^{-6}m^2}=1.37\cdot 10^8 Pa = 0.137 GPa

So, the strain experienced by the composite is

\epsilon=\frac{\sigma}{E}=\frac{0.137 GPa}{41.0 GPa}=3.34\cdot 10^{-3}

3 0
3 years ago
A plastic rod 1.3 m long is rubbed all over with wool, and acquires a charge of -3e-08 coulombs. we choose the center of the rod
Anestetic [448]
(a) The plastic rod has a length of L=1.3m. If we divide by 8, we get the length of each piece:
L/8=1.3m/8=0.1625 m

(b) The center of the rod is located at x=0. This means we have 4 pieces of the rod on the negative side of x-axis, and 4 pieces on the positive side. So, starting from x=0 and going towards positive direction, we have: piece 5, piece 6, piece 7 and piece 8. Each piece is 0.1625 m long. Therefore, the center of piece 5 is at 0.1625m/2=0.0812 m. And the center of piece 6 will be shifted by 0.1625m with respect to this:
c_6 = 0.0812m+0.1625m=0.2437 m

(c) The total charge is Q=-3 \cdot 10^{-8}C. To get the charge on each piece, we should divide this value by 8, the number of pieces:
Q/8=-3\cdot 10^{-8}C/8=-3.75\cdot 10^{-9}C

(d) We have to calculate the electric field at x=0.7 generated by piece 6. The charge on piece 6 is the value calculated at point (c):
q= -3.75\cdot 10^{-9}C
If we approximate piece 6 as a single  charge, the electric field is given by
E=k_e  \frac{q}{d^2}
where k_e=8.99\cdot 10^9Nm^2C^{-2} and d is the distance between the charge (center of piece 6, located at 0.2437m) and point a (located at x=0.7m). Therefore we have
E= 8.99\cdot 10^9 Nm^2C^{-2} \frac{-3.75\cdot 10^9 C}{(0.2437m-0.7m)^2} =-161.9 V/m
poiting towards the center of piece 6, since the charge is negative.

(e) missing details on this question.
5 0
3 years ago
A hydrogen atom has its electron in the n = 6 level. the radius of the electron's orbit in the bohr model is 1.905 nm.
gayaneshka [121]
Are there any options or is it not multiple choice.
5 0
3 years ago
Kyle has the mass of 54 kg and is jogging at a velocity of m/s. What is Kyle’s kinetic energy
inysia [295]

m = mass = 54 kg

v = velocity = 3 m/s

KE =  (1/2)*m*v^2

KE =  (1/2) * 54 * (3)^2

KE = 243 J

Hope this helps!



3 0
3 years ago
Read 2 more answers
Give two environmental effects of using wood as an energy resource.
vladimir2022 [97]

Answer:

Using wood also helps keep carbon out of the atmosphere, helping to mitigate climate change.

Deforestation.  

Explanation:

Trees store carbon dioxide as they grow. After harvest, wood products continue to store much of this carbon. These benefits continue when wood is reclaimed to manufacture other products.

Extraction of wood within the supply areas clearly affects the forest and the environment. Some of these impacts can be seen in the study areas. The rainfall is increasingly irregular, biodiversity has been lost. Communities have experienced flooding and drought, which adversely affects the most important sector in the country's economy, that of agriculture. The local populations are concerned about these impacts, which affect not only the environment but the whole social economy.

6 0
2 years ago
Read 2 more answers
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