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2 years ago

Hydrogen and oxygen react to form water. At an atomic level, how do we know that a chemical reaction has occurred?

Physics

1 answer:

pshichka [43]2 years ago

6 0

The hydrogen and oxygen atoms regroup into a new molecule.

Hope this helps!!

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You have just moved into a new apartment and are trying to arrange your bedroom. You would like to move your dresser of weight 3

LenaWriter [7]

Answer:

0 Joules

Explanation:

The work done is given by

$W=F\times s\times cos\theta$

where,

F = Force applied

s = Displacement of the object = 0 m

$\theta$ = Angle between the force applied and the horizontal = 0

$W=F\times 0\times cos0\\\Rightarrow W=0\ J$

Work is only observed when there is a displacement.

The work done by me is 0 Joules as I was unable to move it.

6 0

2 years ago

The figure (Figure 1) shows the velocity of a solar-powered motorhome (RV) as a function of time. The driver accelerates from a

SpyIntel [72]

Explanation :

It is given that, the driver accelerates from a stop sign, cruises for 20 s at a constant speed of 60 km/h, and then brakes to come to a stop 40 s after leaving the stop sign.

We know that acceleration is defined as the rate of change of velocity.

$a=\dfrac{dv}{dt}$

Where

dv is the change in velocity, dv = 0 - 60 m/s = -60 m/s

dt is the change in time, dt = 40 s - 30 s = 10 s

So, $a=\dfrac{-60\ m/s}{10\ s}$

$a = -6\ m/s^2$

From the graph it is clear that, from 30 s to 40 s the car is decelerating. So, at every second within this time the value of acceleration will be same i.e. $-6\ m/s^2$ .

6 0

2 years ago

if you have an isotope of the element carbon (C) that has the mass number of 14, what other information would help you determine

n200080 [17]

The number of protons
number of neutrons=the mass number- number of protons
14-6=8

4 0

2 years ago

Saturn moves in an orbit around the Sun with radius 10 AU. How many degrees does it move on the Celestial in one year? (Hint: Ca

Lana71 [14]

Answer:

B. About 12 degrees

Explanation:

The orbital period is calculated using the following expression:

T = 2π*( $\sqrt{\frac{r^3}{Gm}}$ )

Where r is the distance of the planet to the sun, G is the gravitational constant and m is the mass of the sun.

Now, we don't actually need to solve the values of the constants, since we now that the distance from the sun to Saturn is 10 times the distance from the sun to the earth. We now this because 1 AU is the distance from the earth to the sun.

Now, we divide the expression used to calculate the orbital period of Saturn by the expression used to calculate the orbital period of the earth. Notice that the constants will cancel and we will get the rate of orbital periods in terms of the distances to the sun:

$\frac{Tsaturn}{Tearth}$ = $\sqrt{\frac{rSaturn^3}{rEarth^3} } = \sqrt{10^3}}$