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3 years ago

6

URGANT!!!!!!! NEED ANSWERS TODAY!!!!

1 answer:

natima [27]3 years ago

5 0

2H2 + O2 = 2H2O + Energy

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True or false forces in a force pair never cancel each other out

pishuonlain [190]

The answer is false

3 0

3 years ago

A car, starting from the origin, travels

sweet [91]

Answer:

The net displacement of the car is 3 km West

Explanation:

Please see the attached drawing to understand the car's trajectory: First in the East direction for 4 km (indicated by the green arrow that starts at the origin (zero), and stops at position 4 on the right (East).

Then from that position, it moves back towards the West going over its initial path, it goes through the origin and continues for 3 more km completing a moving to the West a total of 7 km. This is indicated in the drawing with an orange trace that end in position 3 to the left (West) of zero.

So, its NET displacement considered from the point of departure (origin at zero) to the final point where the trip ended, is 3 km to the west.

8 0

3 years ago

Read 2 more answers

What is the mass of a wooden block with a density 2.0 and a volume of 7.0?

Karolina [17]

Answer:

mass = 14

Explanation:

equation ; p = m / v

2 = m / 7

to get the m, your variable, by itself you would multiply 7 onto the other side of the equation, getting this;

2 * 7 = m

and 2 * 7 = 14, therefore m = 14, so your mass would be 14!

Hope this helps!

3 0

3 years ago

Two point charges are on the y-axis. A 3.0 µC charge is located at y = 1.15 cm, and a -2.28 µC charge is located at y = -2.00 cm

Ghella [55]

Answer:

Total electric potential, $V=1.32\times 10^6\ volts$

Explanation:

It is given that,

First charge, $q_1=3\ \mu C=3\times 10^{-6}\ C$

Second charge, $q_2=-2.28\ \mu C=-2.28\times 10^{-6}\ C$

Distance of first charge from origin, $r_1=1.15\ cm=0.0115\ m$

Distance of second charge from origin, $r_2=2\ cm=0.02\ m$

We need to find the total electric potential at the origin. The electric potential at the origin is given by :

$V=\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}$

$V=k(\dfrac{q_1}{r_1}+\dfrac{q_2}{r_2})$

$V=9\times 10^9(\dfrac{3\times 10^{-6}}{0.0115}+\dfrac{-2.28\times 10^{-6}}{0.02})$

V = 1321826.08 V

or

$V=1.32\times 10^6\ volts$

So, the total electric potential at the origin is $1.32\times 10^6\ volts$ . Hence, this is the required solution.

3 0

3 years ago

The fragment of an asteroid or any interplanetary material is known as a

Elodia [21]

The fragment of an asteroid or any interplanetary material is known as A. METEROID

8 0

4 years ago