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3 years ago

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What is the amount of electric field passing through a surface called? A. Electric flux.B. Gauss’s law.C. Electricity.D. Charge

surface density.E. None of the above.

1 answer:

rjkz [21]3 years ago

3 0

Answer:

A. Electric flux

Explanation:

Electric flux is the rate of flow of the electric field through a given area (see ). Electric flux is proportional to the number of electric field lines going through a virtual surface.

Electric flux has SI units of volt metres (V m), or, equivalently, newton metres squared per coulomb (N m2 C−1). Thus, the SI base units of electric flux are kg·m3·s−3·A−1.

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Consider four point charges arranged in a square with sides of length L. Three of the point charges have charge q and one of the

nydimaria [60]

Answer: $F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]$

Explanation:

Given

Three charges of magnitude q is placed at three corners and fourth charge is placed at last corner with -q charge

Force due to the charge placed at diagonally opposite end on -q charge

$F_1=\frac{kq(-q)}{(L\sqrt{2})^2}$

where $L\sqrt{2}=$ Distance between the two charges

$F_1=-\frac{kq^2}{2L^2}$

negative sign indicates that it is an attraction force

Now remaining two charges will apply the same amount of force as they are equally spaced from -q charge

$F_2=\frac{kq(-q)}{(L)^2}$

The magnitude of force by both the charge is same but at an angle of $90^{\circ}$

thus combination of two forces at 2 and 3 will be

$F'=\sqrt{2}\frac{kq^2}{2L^2}$

Now it will add with force due to 1 charge

Thus net force will be

$F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]$

6 0

2 years ago

A concrete block (B-36 x10 °C-') of volume 100 mat 40°C is cooled to

ruslelena [56]

T1=40°C=313K
T_2=-10°C=263K

Applying Charles law

$\\ \sf\Rrightarrow \dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

$\\ \sf\Rrightarrow \dfrac{100}{313}=\dfrac{V_2}{263}$

$\\ \sf\Rrightarrow V_2=\dfrac{26300}{313}$

$\\ \sf\Rrightarrow V_2=84.02ml$

6 0

1 year ago

A 0.25-kg ball sits on the roof of a building that is 10 meters tall. Find the GPE. (Gravity = 9.8 on Earth)

Tasya [4]

Gravitational potential energy = mgh or mass times acceleration due to gravity times the height

Here the mass is 0.25kg, the height is 10m, and gravity is 9.8m/s^2 so...

GPE = (0.25)(10)(9.8)

GPE = 24.5 J

7 0

3 years ago

The Great Sandini is a 60 kg circus performer who is shotfrom a cannon (actually a spring gun). You don't find many men ofhis ca

d1i1m1o1n [39]

Answer:

V=15.3 m/s

Explanation:

To solve this problem, we have to use the energy conservation theorem:

$U_e+K_i+U_{gi}+W_{friction}=K_f+U_{gf}$

the elastic potencial energy is given by:

$U_e=\frac{1}{2}*k*x^2\\U_e=\frac{1}{2}*1100N/m*(4m)^2\\U_e=8800J$

The work is defined as:

$W_{friction}=F_f*d*cos(\theta)\\W_{friction}=40N*2.5m*cos(180)\\W_{friction}=-100J$

this work is negative because is opposite to the movement.

The gravitational potencial energy at 2.5 m aboves is given by:

$U_{gf}=m*g*h\\U_{gf}=60kg*9.8*2.5\\U_{gf}=1470J$

the gravitational potential energy at the ground and the kinetic energy at the begining are 0.

$8800J+0+0+-100J=\frac{1}{2}*62kg*v^2+1470J\\v=\sqrt{\frac{2(8800J-100J-1470J)}{62kg}}\\v=15.3m/s$

3 0

2 years ago

(a) What is the fluid speed in a fire hose with a 9.00-cm diameter carrying 80.0 L of water per second? (b) What is the flow rat

son4ous [18]

Answer:

12.5752053801 m/s

$80\times 10^{-3}\ m^3/s$

No.

Explanation:

Q = Volume flow rate = $80\ L/s=80\times 10^{-3}\ m^3/s$

d = Diameter of pipe = 9 cm

A = Area = $\dfrac{\pi}{4}d^2$

Volume flow rate is given by

$Q=Av\\\Rightarrow v=\dfrac{Q}{A}\\\Rightarrow v=\dfrac{80\times 10^{-3}}{\dfrac{\pi}{4} (9\times 10^{-2})^2}\\\Rightarrow v=12.5752053801\ m/s$

Velocity of fluid is 12.5752053801 m/s

The volume flow rate in m³/s is $80\times 10^{-3}\ m^3/s$

The flow of fluid does not depend on the type of water used. Hence the answers would be same. If Q is constant v will be the same irrespective of the type of water used.

8 0

2 years ago