Question 12 of 15 are pictures of relationships. Answer here

A magnetic field is uniform over a flat, horizontal circular region with a radius of 1.50 mm, and the field varies with time. In

Daniel [21]

Answer: $81.57\mu V$

Explanation:

Given

radius of circular region r=1.50 mm

$A=\pi r^2=7.069\times 10^{-6}\ m^2$

Magnetic Field $B=1.50\ T$

time t=130 ms

Flux is given by

$\phi =B\cdot A$

change in Flux $d\phi =(B_f-B_i)A$

Emf induced is $e=\frac{\mathrm{d} \phi}{\mathrm{d} t}$

$e=\frac{(1.5)\cdot 7.069\times 10^{-6}}{130\times 10^{-3}}$

$e=81.57 \mu V$

3 0

2 years ago

Help, It has multiple answers for this question.

evablogger [386]

The answer is ultra violet radiation. From the air

4 0

2 years ago

When a body is moving with a uniform velocity, the acceleration is ___?

denis23 [38]

According to the formula :
a=ΔV / ΔT
When a body is moving with a uniform velocity, the acceleration is zero. That's it. You should remember, that velocity is not constant whereas speed is constant.

4 0

2 years ago

Potassium is a crucial element for the healthy operation of the human body. Potassium occurs naturally in our environment and th

Mariulka [41]

Answer:

a) 0.0288 grams

b) $2.6*10^{-10} J/kg$

Explanation:

Given that:

A typical human body contains about 3.0 grams of Potassium per kilogram of body mass

The abundance for the three isotopes are:

Potassium-39, Potassium-40, and Potassium-41 with abundances are 93.26%, 0.012% and 6.728% respectively.

a)

Thus; a person with a mass of 80 kg will posses = 80 × 3 = 240 grams of potassium.

However, the amount of potassium that is present in such person is :

0.012% × 240 grams

= 0.012/100 × 240 grams

= 0.0288 grams

b)

the effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body is calculate as follows:

First the Dose in (Gy) = $\frac{energy \ absorbed }{mass \ of \ the \ body}$

= $\frac{1.10*10^6*1.6*10^{-14}}{80}$

= $2.2*10^{-10} \ J/kg$

Effective dose (Sv) = RBE × Dose in Gy

Effective dose (Sv) = $1.2 *2.2*10^{-10} \ J/kg$

Effective dose (Sv) = $2.6*10^{-10} J/kg$

5 0

2 years ago

Identical point charges (+50 x 10 power -6C) are placed at the corners of a square with sides of 2.0-m length. How much external

Gnesinka [82]

Answer:

636.4 J

Explanation:

The potential energy between one of the charges at the corner of the square and the fifth identical charge is U = kq²/r where q = charge = +50 × 10⁻⁶ C and r = distance from center of square. = √2 m (since the midpoint of the sides = 1 m, so the distance from the charge at the corner to the center is thus √(1² + 1²) = √2)

Since we have four charges, the additional potential energy to move the charge to the centre of the square is U' = 4U = 4kq²/r

U' = 4kq²/r

= 4 × 9 × 10⁹ Nm²/C² (+50 × 10⁻⁶ C)²/√2 m

= 900 Nm²/√2 m

= 636.4 J

8 0

3 years ago