Which of these is NOT a common building material?

A 3 mm inside diameter tube is placed in a fluid with a surface tension of 600 mN/m and density of 3.7 g/cm3. The contact angle

Aleks04 [339]

Answer: The height of the fluid rise is 0.01m

Explanation:

Using the equation

h = (2TcosѲ )/rpg

h= height of the fluid rise

diameter of the tube =3mm

radius of the tube= 3/2 =1.5mm=0.0015

T= surface tension = 600mN/m=0.6N/m

Ѳ = contact angle = $60^o$ C

p= density =3.7g/cm3= 3700kg/m3

g= acceleration due to gravity =9.8m/s2

h = ( 2*0.6*0.5)/(0.0015*3700*9.8)

h = 0.6/54.39

h= 0.01m

Therefore,the height of the fluid rise is 0.01m

8 0

3 years ago

Refer to a long, straight wire carrying constant current I. What can be concluded about the magnitude of the magnetic field at d

sergij07 [2.7K]

Answer:

<em>"the magnitude of the magnetic field at a point of distance a around a wire, carrying a constant current I, is inversely proportional to the distance a of the wire from that point"</em>

Explanation:

The magnitude of the magnetic field from a long straight wire (A approximately a finite length of wire at least for close points around the wire.) decreases with distance from the wire. It does not follow the inverse square rule as is the electric field from a point charge. We can then say that<em> "the magnitude of the magnetic field at a point of distance a around a wire, carrying a constant current I, is inversely proportional to the distance a of the wire from that point"</em>

From the Biot-Savart rule,

B = μI/2πR

where B is the magnitude of the magnetic field

I is the current through the wire

μ is the permeability of free space or vacuum

R is the distance between the point and the wire, in this case is = a

5 0

3 years ago

If you were to drive an average velocity of 65 mi/hr to Traverse City, which is

Amanda [17]

Find the number of hours by dividing the distance by mph. The number of hours will be to the left of the decimal point:

250 miles / 65 mph
= 3.846153846
= 3 hours

2) Find the number of minutes by multiplying what is remaining from step 1 by 60 minutes. The minutes will be to the left of the decimal point:

0.846153846 x 60
= 50.76923076
= 50 minutes

3) Find the number of seconds by multiplying what is remaining from step 2 by 60 seconds. The seconds will be to the left of the decimal point:

0.76923076 x 60
= 46.1538456
= 46 seconds

So 3 hours 50 mins and 46 seconds

3 0

3 years ago

5/137 Under the action of its stern and starboard bow thrusters, the cruise ship has the velocity vB = 1 m/s of its mass center

quester [9]

The image is missing, so i have attached it;

Answer:

A) V_rel = [-(2.711)i - (0.2588)j] m/s

B) a_rel = (0.8637i + 0.0642j) m/s²

Explanation:

We are given;

the cruise ship velocity; V_b = 1 m/s

Angular velocity; ω = 1 deg/s = 1° × π/180 rad = 0.01745 rad/s

Angular acceleration;α = -0.5 deg/s² = 0.5 x π/180 rad = -0.008727 rad/s²

Now, let's write Velocity (V_a) at A in terms of the velocity at B(V_b) with r_ba being the position vector from B to A and relative velocity (V_rel)

Thus,

V_a = V_b + (ω•r_ba) + V_rel

Now, V_a = 0. Thus;

0 = V_b + (ω•r_ba) + V_rel

V_rel = -V_b - (ω•r_ba)

From the image and plugging in relevant values, we have;

V_rel = -1[(cos15)i + (sin15)j] - (0.01745k * -100j)

V_rel = - (cos15)i - (sin15)j - 1.745i

Note that; k x j = - i

V_rel = [-(2.711)i - (0.2588)j] m/s

B) Let's write the acceleration at A with respect to B in terms of a_b.

Thus,

a_a = a_b + (α*r_ba) + (ω(ω•r_ba)) + (2ω*v_rel) + a_rel

a_a and a_b = 0.

Thus;

0 = (α*r_ba) + (ω(ω•r_ba)) + (2ω*v_rel) + a_rel

a_rel = - (α*r_ba) - (ω(ω•r_ba)) - (2ω*v_rel)

Plugging in the relevant values with their respective position vectors, we have;

a_rel = - (-0.008727k * -100j) - (0.01745k(0.01745k * -100j)) - (2*0.01745k * [-(2.711)i - (0.2588)j])

a_rel = 0.8727i - (0.01745² x 100)j + 0.0946j - 0.009i

Note that; k x j = - i and k x i = j

Thus,simplifying further ;

a_rel = 0.8637i + 0.0642j m/s²

4 0

3 years ago

Calculate the minimum average power output necessary for a person to run up a 12.0 m long hillside, which is inclined at 25.0° a

Viktor [21]

Answer:

Power, P = 924.15 watts

Explanation:

Given that,

Length of the ramp, l = 12 m

Mass of the person, m = 55.8 kg

Angle between the inclined plane and the horizontal, $\theta=25^{\circ}$

Time, t = 3 s

Let h is the height of the hill from the horizontal,

$h=l\ sin\theta$

$h=12\times \ sin(25)$

h = 5.07 m

Let P is the power output necessary for a person to run up long hill side as :

$P=\dfrac{E}{t}$

$P=\dfrac{mgh}{t}$

$P=\dfrac{55.8\times 9.8\times 5.07}{3}$

P = 924.15 watts

So, the minimum average power output necessary for a person to run up is 924.15 watts. Hence, this is the required solution.

3 0

3 years ago