How does water in a hot spring get hot?

REY [17]

Answer:

Atoms found in nature are either stable or unstable. ... An atom is unstable (radioactive) if these forces are unbalanced; if the nucleus has an excess of internal energy. Instability of an atom's nucleus may result from an excess of either neutrons or protons

7 0

2 years ago

Consider an insulating sphere of radius 6 cm surrounded by a conducting sphere of inner radius 18 cm and outer radius 26 cm. Fur

dimulka [17.4K]

Answer:

$-1.7908787542\times 10^{-9}\ C$

$3.4260289211\times 10^{-9}\ C$

$1.7908787542\times 10^{-9}\ C$

$1.6351501669\times 10^{-9}\ C$

Explanation:

r = Radius

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Electric field is given by

$E=-\dfrac{kq}{r^2}\\\Rightarrow q=-\dfrac{Er^2}{k}\\\Rightarrow q=-\dfrac{1610\times 0.1^2}{8.99\times 10^9}\\\Rightarrow q=-1.7908787542\times 10^{-9}\ C$

The charge is $-1.7908787542\times 10^{-9}\ C$

$Q+q=\dfrac{Er^2}{k}\\\Rightarrow Q=\dfrac{Er^2}{k}-q\\\Rightarrow Q=\dfrac{120\times 0.35^2}{8.99\times 10^9}-(-1.7908787542\times 10^{-9})\\\Rightarrow Q=3.4260289211\times 10^{-9}\ C$

The charge is $3.4260289211\times 10^{-9}\ C$

The charge inside will have the polarity changed

$q=+1.7908787542\times 10^{-9}\ C$

Outside the charge will be

$3.4260289211\times 10^{-9}-1.7908787542\times 10^{-9}\\ =1.6351501669\times 10^{-9}\ C$

3 0

3 years ago

A long, hollow, cylindrical conductor (inner radius 3.4 mm, outer radius 7.3 mm) carries a current of 36 A distributed uniformly

Elden [556K]

Answer:

a. $B= 9.45 \times10^{-3} T$

b. $B= 0.820 T$

c. $B= 0.0584 T$

Explanation:

First, look at the picture to understand the problem before to solve it.

a. d1 = 1.1 mm

Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:

To solve the equations we have to convert all units to those of the international system. (mm→m)

$B=\frac{u_{0}I_{w}}{2\pi d_{1}} =\frac{52 \times4\pi \times10^{-7} }{2\pi 1.1 \times 10^{-3}} =9.45 \times10^{-3} T\\$

μ0 is the constant of proportionality

μ0=4πX10^-7 N*s2/c^2

b. d2=3.6 mm

Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:

J: current density

c: outer radius

b: inner radius

The cilinder's current is negative, as it goes on opposite direction than the wire's current.

$J= \frac {-I_{c}}{\pi(c^{2}-b^{2} ) }}$

$J=\frac{-36}{\pi(5.33\times10^{-5}-1.16\times10^{-5}) } =-274.80\times10^{3} A/m^{2}$

$B=\frac{u_{0}(I_{w}-JA_{s})}{2\pi d_{2} } \\A_{s}=\pi (d_{2}^{2}-b^2)=4.40\times10^{-6} m^2\\$

$B=\frac{6.68\times10^{-5}}{8.14\times10^{-5}} =0.820 T$

c. d3=7.4 mm

Here, the point is located out of the cilinder. Therefore, we have to consider both, the conductor's current and the wire's current as follows:

$B=\frac{u_{0}(I_w-I_c)}{2\pi d_3 } =\frac{2.011\times10^-5}{3.441\times10^{-4}} =0.0584 T$

As we see, the magnitud of the magnetic field is greater inside the conductor, because of the density of current and the material's nature.

3 0

3 years ago

Technician A says a short to voltage may or may not blow a fuse and may or may not affect more than one circuit. Technician B sa

Tomtit [17]

Answer:

C. Both A and B

Explanation:

Fuses are rated by the amperage they can carry before heat melts the element. The fuse is ideal for protection against short circuits. Short circuits produce enough amperage to vaporize a fuse element and break connection in one cycle of a 60-cycle system.

Specifically, the voltage rating determines the ability of the fuse to suppress the internal arcing that occurs after a fuse link melts and an arc is produced.

6 0

3 years ago

The apparent weight of a student in alift is 564N . if the mass of the student is 60.3kg, what is the acceleration of the lift ?

Yuki888 [10]

Answer:

-.457 m/s^2

Explanation:

Actual weight = 60 .3 (9.81) = 591.54 N

Accel of lift changes this to 60.3 ( 9.81 - L) where L - accel of lift

60.3 ( 9.81 - L ) = 564

solve for L = .457 m/s^2 DOWNWARD

so L = - .457 m/s^2

4 0

2 years ago