which of the following are vector quantities? check all that apply. a. time b. displacement c. force d. mass

Most of the resistance of the human body comes from the skin, as the interior of the body contains aqueous solutions that are go

ruslelena [56]

Answer:

a) Resistivity=R.A/L

Here, R=500000ohm, L=1.6m, A=2pi* (0.14/2)*0.002 m2=0.00088.........we have converted quantities into SI units

Resistivity=R.A/L

=500000*0.00088/1.6 = 440/1.6= 275 ohm.meter

b) pskin/pinterior =275/4.8=57.3

For second part, correct option is (A), because cross section area is smaller.

4 0

3 years ago

The rest mass of a proton is 1.673 × 10-27 kg. The relativistic mass of a proton when its velocity is 0.8c is approximately

sergiy2304 [10]

Do u know the formula
$m= \frac{ m_{0} }{ \sqrt{1- \frac{ v^{2} }{ c^{2} } } }$

7 0

3 years ago

The river narrows at a rapids from a width of 12 m to a width of only 5.8 m. The depth of the river before the rapids is 2.7 m;

Alisiya [41]

Answer:

7.89 m/s

Explanation:

Given that

Width of the river, b1 = 12 m

Width of the river, b2 = 5.8 m

Depth of the river, d1 = 2.7 m

Depth of the river, d2 = 0.85 m

Speed of the river, v1 = 1.2 m/s

Speed of the river, v2 = ?

Area of the river before the rapid, a1 = 12 * 2.7 = 32.4 m²

Area of the river after the rapid, a2 = 5.8 * 0.85 = 4.93 m²

To solve this question, we use a relation between the speed of the river and the volume of the river. We say,

Area1 * velocity1 = Area2 * velocity2, and when we substitute the values for each other we have

32.4 * 1.2 = 4.93 * v2

38.88 = 4.93v2

v2 = 38.88 / 4.93

v2 = 7.89 m/s

Therefore, the speed of the river after the rapid is 7.89 m/s

6 0

2 years ago

You are standing a distance of 17.0 meters from the center of a merry-go-round. The merry-go-round takes 9.50 seconds to go comp

GenaCL600 [577]

Answer:

(a)11.24 m/s

(b)7.44 m/s

(c)409 N

(d) $539.55\mu$

(e) 0

Explanation:

The period for 1 circle $2\pi$ of the merry go around is 9.5s. It means the angular speed is:

$\omega = \theta / t = 2\pi / 9.5 \approx 0.661 rad/s$

(a)The speed is

$v = \omega * R = 0.661 * 17 = 11.24 m/s$

(b) Centripetal acceleration:

$a = \frac{v^2}{R} = \frac{11.24^2}{17} = 7.44 m/s^2$

(c) Magnitude of the force that keeps you go around at this acceleration

$F = ma = 55 * 7.44 = 409 N$

(d) let the coefficient of friction by $\mu$ . The frictional force shall be this coefficient multiplied by normal force reverting gravity of the man

$F_f = mg\mu = 55*9.81\mu = 539.55\mu$

3 0

3 years ago

How does temperature affect chemical reaction

Elis [28]

When its temperature increases, the reactants have more kinetic energy so the frequency of effective collision increases, resulting in a faster rate of chemical reaction.

8 0

3 years ago