Which would you expect to find in a home with passive solar heating?

A voltmeter was used to check the coolant and a reading of 0.2 volt with the engine off was measured. A reading of 0.8 volt was

Julli [10]

Answer:

C. Technician B

Explanation:

Excessive Galvanic activity:

To check for excessive galvanic activity, voltmeter is used to check the coolant. If the voltmeter is giving a reading greater than 0.5 V, there is excessive galvanic activity. Excessive galvanic activity is solved by flushing the coolant fluid from engine and refiling it.

Electrolysis problem:

When the system is not properly ground, the cooling system accepts stray current and the coolant becomes an electrolyte which might eat up the radiator. To test for excessive electrolysis, start the engine and turn on all electrical accessories, if the reading is more than 0.5 V, there is electrolysis problem. Ground wires and connections should be checked at this point to stop stray current.

In our case, the first reading is 0.2 V(engine turned off) which is normal and there is no excessive galvanic activity. This means that Technician A is not correct. The second reading is 0.8 V when the engine and all electrical accessories are turned on. This reading is greater than 0.5 V which means there is an electrolysis problem. This means that Technician B is correct and ground wires and connections should be inspected and repaired.

7 0

3 years ago

The roller coaster is designed according to safety regulations that prohibit the speed of the car from exceeding 20 m/s. find th

hram777 [196]

maximum allowed value of the speed in roller coaster is given as

$v = 20 m/s$

now from kinematics we can say

$v^2 - v_i^2 = 2 a s$

here initial speed will be

$v_i = 0$

acceleration is due to gravity

$a = 9.8 m/s^2$

now we can use this to find the height

$20^2 - 0^2 = 2 * 9.8* h$

$400 = 19.6 *h$

$h = 20.4 m$

so maximum allowed height will be 20.4 m

4 0

3 years ago

The atomic number, or ________ number, is the described as the number of _________ in the nucleus of an chemical element.

Leto [7]

The atomic number or proton number (symbol Z) of a chemical element is the number of protons found in the nucleus of an atom. It is identical to the charge number of the nucleus. The atomic number uniquely identifies a chemical element. In an uncharged atom, the atomic number is also equal to the number of electrons.

8 0

3 years ago

Two ladybugs sit on a rotating disk that is slowing down at a constant rate. The ladybugs are at rest with respect to the surfac

d1i1m1o1n [39]

The two ladybugs have same rotational (angular) speed

Explanation:

The rotational (angular) speed of an object in circular motion is defined as:

$\omega=\frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time interval considered

Here we have two ladybugs, which are located at two different distances from the axis. In particular, ladybug 1 is halfway between ladybug 2 and the axis of rotation. However, since they rotate together with the disk, and the disk is a rigid body, every point of the disk cover the same angle $\theta$ in the same time $t$ : this means that every point along the disk has the same angular speed, and therefore the two ladybugs also have the same angular speed.

On the other hand, the linear speed of the two ladybugs is different, because it follows the equation:

$v=\omega r$

where r is the distance from the axis: and since the two ladybugs are located at different $r$ , they have different linear speed.

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

3 0

3 years ago

What percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway? a plane accelerates from

ElenaW [278]

When is at the end of the runway the velocity of the plane is given by the equation $vf^{2}=0+2*a*s$ where s=1800 m is the runway length. Thus
$vf^{2}=2*5*1800=18000 (m/s)^{2}$
$vf =134.164 (m/s)$

At half runway the velocity of the plane is
$v^{2}=2*5* \frac{1800}{2}=9000 ( \frac{m}{s} )^{2}
$
$v= \sqrt{9000}=94.87 ( \frac{m}{s})$

Therefore at midpoint of runway the percentage of takeoff velocity is
‰ $P= \frac{v}{vf}= \frac{94.87}{134.164}=0.707$

6 0

3 years ago