Which are true of gamma radiation?

Terrorists carried out their attacks on September 11, 2001, by

dedylja [7]

Im pretty sure its b hijacking planes

8 0

4 years ago

What would be the final temperature of the system if all the heat lost by the copper block were absorbed by the water in the cal

ivanzaharov [21]

<span>Answer: q Cu = q H2O so, (121g)(0.385 J/g-K)(100.4C-x) = (150g)(4.18 J/g-K)(x-25.1C) x= 30.3C</span>

5 0

4 years ago

Which of the following expressions gives the ratio of the energy density of the magnetic field to that of the electric field jus

miss Akunina [59]

Answer:

$(d) \ \ \frac{\mu_o}{\epsilon_o} (\frac{L}{2\pi r*R} )^2$

Explanation:

Energy density in magnetic field is given as;

$U_B = \frac{1}{2 \mu_o} B^2$

where;

B is the magnetic field strength

Energy density of electric field

$U_E = \frac{1}{2}\epsilon E^2$

where;

E is electric field strength

Take the ratio of the two fields energy density

$\frac{U_B}{U_E} = \frac{1}{2\mu_o} B^2 / \frac{1}{2}\epsilon E^2\\\\\frac{U_B}{U_E} = \frac{B^2}{2\mu_o} *\frac{2}{\epsilon E^2} \\\\\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B^2}{E^2})$

$\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B}{E})^2$

But, Electric field potential, V = E x L = IR (I is current and R is resistance)

$\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B*L}{E*L})^2$

Now replace E x L with IR

$\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B*L}{IR})^2$

Also, B = μ₀I / 2πr, substitute this value in the above equation

$\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{\mu_oI*L}{2\pi r* IR})^2$

cancel out the current "I" and factor out μ₀

$\frac{U_B}{U_E} = \frac{\mu_o^2}{\mu_o \epsilon} (\frac{L}{2\pi r* R})^2$

Finally, the equation becomes;

$\frac{U_B}{U_E} = \frac{\mu_o}{\epsilon} (\frac{L}{2\pi r*R })^2$

Therefore, the correct option is (d) μ₀/ϵ₀ (L /R 2πr)²

3 0

3 years ago

Wire 1 has a resistance R1. Wire 2 is made of the same material, with length 1/2 as long and diameter 1/4. What is the resistanc

hodyreva [135]

-- reduce the length of a wire to 1/2 . . . cut the resistance in half

-- reduce the diameter to 1/4 . . . reduce the cross-section area by (1/4²) . . . increase the resistance by 16x .

-- R2 = (R1) · (1/2) · (16) = 8 · R1

5 0

3 years ago

A student wearing frictionless in-line skates on a horizontal surface is pushed by a friend with a constant force of 45 N. How f

kiruha [24]

Work is obtained by multiplying the force and the object's displacement. The force and displacement and force should be in the same direction in order to have work.
W = F x d
d = W / F
Substituting the known values,
d = 352 J / 45 N = 7.82 m
Thus, the displacement of the student is 7.82 m.

8 0

4 years ago