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slega [8]
3 years ago
11

A team of scientists is researching the best way to use tides to produce energy. One team includes environmental

Physics
2 answers:
Xelga [282]3 years ago
6 0
Geology because It is the study of processes that shape Earth.
Zolol [24]3 years ago
4 0

Answer:

agree with above answer

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Dorji weight 1500n.if the total surface area of soles of his feet is 0.5 m2 what is the pressure exerted by his body on the grou
Fantom [35]

Answer:

P = 3000 Pa

Explanation:

Weight of Dorji, W = F = mg = 1500 N

The total surface area of soles of his feet is 0.5 m²

We need to find the pressure exerted by his body on the ground​. The pressure is equal to the force acting per unit area. So,

P=\dfrac{1500\ N}{0.5\ m^2}\\\\=3000\ Pa

So, the pressure exerted by his body on the ground​ is 3000 Pa.

3 0
3 years ago
A uniform line charge of density λ lies on the x axis between x = 0 and x = L. Its total charge is 7 nC. The electric field at x
DedPeter [7]

Answer:

The electric field at x = 3L is 166.67 N/C

Solution:

As per the question:

The uniform line charge density on the x-axis for x, 0< x< L is \lambda

Total charge, Q = 7 nC = 7\times 10^{- 9} C

At x = 2L,

Electric field, \vec{E_{2L}} = 500N/C

Coulomb constant, K = 8.99\times 10^{9} N.m^{2}/C^{2}

Now, we know that:

\vec{E} = K\frac{Q}{x^{2}}

Also the line charge density:

\lambda = \frac{Q}{L}

Thus

Q = \lambda L

Now, for small element:

d\vec{E} = K\frac{dq}{x^{2}}

d\vec{E} = K\frac{\lambda }{x^{2}}dx

Integrating both the sides from x = L to x = 2L

\int_{0}^{E}d\vec{E_{2L}} = K\lambda \int_{L}^{2L}\frac{1}{x^{2}}dx

\vec{E_{2L}} = K\lambda[\frac{- 1}{x}]_{L}^{2L}] = K\frac{Q}{L}[frac{1}{2L}]

\vec{E_{2L}} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{2L}] = \frac{63}{L^{2}}

Similarly,

For the field in between the range 2L< x < 3L:

\int_{0}^{E}d\vec{E} = K\lambda \int_{2L}^{3L}\frac{1}{x^{2}}dx

\vec{E} = K\lambda[\frac{- 1}{x}]_{2L}^{3L}] = K\frac{Q}{L}[frac{1}{6L}]

\vec{E} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{6L}] = \frac{63}{6L^{2}}

Now,

If at x = 2L,

\vec{E_{2L}} = 500 N/C

Then at x = 3L:

\frac{\vec{E_{2L}}}{3} = \frac{500}{3} = 166.67 N/C

4 0
4 years ago
A vertical wall (8.7 m x 3.2 m) in a house faces due east. A uniform electric field has a magnitude of 210 N/C. This field is pa
Dominik [7]

Answer:

\phi=4344.72Nm^2/c

Explanation:

From the question we are told that:

Dimension of Wall:

 (L*B)=(8.7 m * 3.2 m)

Electric field B=210 N/C

Angle \theta =42 \textdegree North

Generally the equation for electric Flux is mathematically given by

 \phi=EAcos\theta

 \phi=210*(8.7*3.2)*cos 42

 \phi=4344.72Nm^2/c

3 0
3 years ago
A gas expands at a constant pressure of 3 atm from a volume of 0.02 cubic meters to 0.10 cubic meters. In the process it experie
Nonamiya [84]

a) 2.4\cdot 10^4 J

For a gas transformation occuring at a constant pressure, the work done by the gas is given by

W=p(V_f -V_i)

where

p is the gas pressure

V_f is the final volume of the gas

V_i is the initial volume

For the gas in the problem,

p=3 atm = 3\cdot 1.013\cdot 10^5 Pa = 3.039\cdot 10^5 Pa is the pressure

V_i = 0.02 m^2 is the initial volume

V_f = 0.10 m^3 is the final volume

Substituting,

W=(3.039\cdot 10^5 Pa)(0.10 m^3-0.02m^2)=24312 J = 2.4\cdot 10^4 J

b) 3.24\cdot 10^5 J

The heat absorbed by the gas can be found by using the 1st law of thermodynamics:

\Delta U = Q-W

where

\Delta U is the change in internal energy of the gas

Q is the heat absorbed

W is the work done

Here we have

\Delta U = 3.0\cdot 10^5 J

W=2.4\cdot 10^4 J

So we can solve the equation to find Q:

Q=\Delta U + W = 3.0\cdot 10^5 J +2.4\cdot 10^4 J = 3.24\cdot 10^5 J

And this process is an isobaric process (=at constant pressure).

8 0
3 years ago
For a charged solid metal sphere with total charge Q and radius R centered on the origin: Select "True" or "False" for each stat
mario62 [17]

Answer:

the answer is true

Explanation:

7 0
3 years ago
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