In the reaction 2H, + O, → H,0, what coeficient should be placed in front of H,0 to balance the reaction?

Two charged objects separated by some distance attract each other. If the charges on both objects are doubled with no change in

Serggg [28]

Answer:

(a) The force between them quadruples

Explanation:

According to coulomb's law, initial force between the two charged objects is given as;

$F_1=\frac{Kq_1q_2}{r^2}$

where;

k is coulomb's constant

q₁ is the charge on the first object

q₂ is the charge on the second object

r is the distance between the two objects

When the charges on both objects are doubled, then;

q₁ = 2q₁

q₂ = 2q₂

Force between the two charged objects will become

$F_2 = \frac{K2q_12q_2}{r^2} = \frac{4Kq_1q_2}{r^2} = 4(\frac{Kq_1q_2}{r^2}) = 4F_1$

Therefore, the force between them quadruples

4 0

3 years ago

PLEASE HELP!! DUE SOON!! THANKS BRAINLIEST OFFERED!!

pav-90 [236]

Answer:

real, inverted, and smaller than the object

Explanation:

When the object is placed beyond the center of curvature, the image will formed between the focus and the center of curvature. The size of the image is diminished and its nature is real and inverted.

The whole description is shown in the attached figure. It is clear that the size of the image is smaller than the object.

7 0

3 years ago

What is the change in internal energy if 60J of heat are released from a system and 20J of work is done on the system? Use U=Q-W

Ganezh [65]

D because 60-20= 40J

5 0

2 years ago

How would a small bar magnet be oriented when placed at position X

Alex Ar [27]

The answer is right here

4 0

2 years ago

Read 2 more answers

On the surface of the earth the weight of an object is 200 lb. Determine the height of the

siniylev [52]

Answer:

The height of the object is 5007.4 miles.

Explanation:

Given that,

Weight of object = 200 lb

We need to calculate the value of $Gmm_{e}$

Using formula of gravitational force

$F=\dfrac{Gmm_{e}}{r^2}$

Put the value into the formula

$200=\dfrac{Gmm_{e}}{(3958.756)^2}$

$200\times(3958.756)^2=Gmm_{e}$

$Gmm_{e}=3.134\times10^{9}$

We need to calculate the height of the object

Using formula of gravitational force

$F=\dfrac{Gmm_{e}}{r^2}$

Put the value into the formula

$125=\dfrac{200\times(3958.756)^2}{r^2}$

$r^2=\dfrac{200\times(3958.756)^2}{125}$

$r^2=25074798.5$

$r=\sqrt{25074798.5}$

$r=5007.4\ miles$

Hence. The height of the object is 5007.4 miles.

7 0

3 years ago