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IrinaVladis [17]

3 years ago

11

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

2 answers:

Ronch [10]3 years ago

5 0

Answer:

Electric current

Explanation:

The others do not pertain to the flow of electrons. Positive current is the closest but it refers to the positive flow of electrons not just the flow of electrons.

Lady_Fox [76]3 years ago

4 0

Answer:

Electric Current

Explanation:

Electric current: An electric current is the rate of flow of electric charge past a point or region.

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According to most car mechanics, plain water is the best coolant to use in an engine provided the engine is not being exposed to

Gelneren [198K]

Ethylene glycol reduces the cooling efficiency of water because it has a lower heat capacity as compared to water. A substance with low heat capacity would mean that the material only needs a small amount of energy to increase its temperature. So, for ethylene glycol, the heat transfer would be faster decreasing the cooling efficiency.

8 0

3 years ago

A brick lands 10.1 m from the base of a building. If it was given an initial velocity of 8.6 m/s [61º above the horizontal], how

Montano1993 [528]

<h2>Answer: 10.52m</h2><h2 />

First, we have to establish the <u>reference system</u>. Let's assume that the building is on the negative y-axis and that the brick was thrown at the origin (see figure attached).

According to this, the initial velocity $V_{o}$ has two components, because the brick was thrown at an angle $\alpha=61\º$ :

$V_{ox}=V_{o}cos\alpha$ (1)

$V_{ox}=8.6\frac{m}{s}cos(61\º)=4.169\frac{m}{s}$ (2)

$V_{oy}=V_{o}sin\alpha$ (3)

$V_{oy}=8.6\frac{m}{s}sin(61\º)=7.521\frac{m}{s}$ (4)

As this is a projectile motion, we have two principal equations related:

<h2>In the x-axis: </h2>

$X=V_{ox}.t$ (5)

Where:

$X=10.1m$ is the distance where the brick landed

$t$ is the time in seconds

If we already know $X$ and $V_{ox}$ , we have to find the time (we will need it for the following equation):

$t= \frac{X}{ V_{ox}}$ (6)

$t=2.42s$ (7)

<h2>In the y-axis: </h2>

$-y=V_{oy}.t+\frac{1}{2}g.t^{2}$ (8)

Where:

$y$ is the height of the building (<u>in this case it has a negative sign because of the reference system we chose)</u>

$g=-9.8\frac{m}{s^{2}}$ is the acceleration due gravity

Substituting the known values, including the time we found on equation (7) in equation (8), we will find the height of the building:

$-y=(7.521\frac{m}{s})(2.42s)+\frac{1}{2}(-9.8\frac{m}{s^{2}}).(2.42s)^{2}$ (9)

$-y=-10.52m$ (10)

Multiplying by -1 each side of the equation:

$y=10.52m$ >>>>This is the height of the building

3 0

3 years ago

Question 7 of 20

slamgirl [31]

Answer:

high number of coils in the wire

5 0

3 years ago

A plane fillies along a straight line path after taking off and it ends up 220 km farther east and 110km farther north relative

DENIUS [597]

330km hope this helped you.

3 0

3 years ago

A person drives a car around a circular cloverleaf with a radius of 63 m at a uniform speed of 10 m/s. (a) What is the accelerat

evablogger [386]

Answer:

acceleration is 1.59 m/s²

required % is 16.22 %

Explanation:

given data

radius = 63 m

speed = 10 m/s

to find out

acceleration of the car and acceleration due to gravity

solution

we know car is moving in circular path

so acceleration will be, a = $\frac{v^2}{r}$ .........1

put here value

a = $\frac{10^2}{63}$

a = 1.59 m/s²

so acceleration is 1.59 m/s²

so acceleration due to gravity as a percentage is 9.8 m/s²

required % = $\frac{1.59}{9.8}*100$

required % = 16.22 %

4 0

3 years ago