If the pH of a substance is 12.4, the substance is a ______

A pair of electrically charged objects attract each other with a force of 4 N when they are a distance of 3 m apart. If their ch

Rainbow [258]

Answer:

1 N

Explanation:

From coulomb's law,

The force of attraction between two charges is inversely proportional to the square of the distance between the charges.

From the question,

Assuming the charges are the same in both case,

F ∝ /r²....................... Equation 1

Fr² = k

F'r'² = Fr²........................... Equation 2

Where F' = First Force, r'² = First distance, F = second force, r² = second distance.

make F the subject of the equation,

F = F'r'²/r².................... Equation 3

Given: F' = 4 N, r' = 3 m, r = 6 m

Substitute into equation 3

F = 4(3²)/6²

F = 36/36

F = 1 N

7 0

3 years ago

What is the leading hypothesis for the origin of short gamma-ray bursts?

Arturiano [62]

Answer:

Gamma rays are the most energetic form of rays. A gamma ray burst can emit a large amount of energy in a very short interval of time. In fact, a short gamma-ray burst occurs in less than 2 seconds. A leading hypothesis states that when two neutron stars merge or when a neutron star merges with a black hole, short gamma-ray burst originates. Such merger are expected to produce kilonovoe.

6 0

3 years ago

Please fast answerrr thank you

Vedmedyk [2.9K]

Answer:

50J

Explanation:

At the top you have(A)

KE_a = O

PE_a = 100J

KE + PE = 100J

At the bottom you have (C)

KE_c= 100J

PE_c=0J

KE+PE = 100J

At point C:

You are at half the height.

We know that at H, PE =100J

PE_c = mgH

At C,

PE_c= mg (H/2) *at half the height

*m and g stay the same

Intuitively, the higher you are, the more potential energy you have.

If you decrease the height by a half, your PE will also decrease

At A:

PE_a / (mg) = H

At B:

PE_b / (mg) = H/2

to also get H on the right hand side, multiply by 2

2 (PE_b/ (mg))= H

2PE_b / (mg) = H

Ok, now that we have set up 2 equations (where H is isolated), find PE at B

AT A = AT B *This way you are saying that H = H (you compare both equations)

PE_a / (mg) = 2x PE_b / (mg)

*mg are the same for both cancel them (you can do that because of the = sign)

PE_a = 2PE_b

We know that PE_a = 100J

100J/2 = PE_b

PE at b = 50J

**FIND KE at b

We know that

KE_b + PE_b is always 100J

100J = 50J + KE_b

KE_b = 50J

4 0

2 years ago

Three +3.0-μC point charges are at the three corners of a square of side 0.50 m. The last corner is occupied by a −3.0-μC charge

kramer

Answer:

E = 440816.32 N/C

Explanation:

Given data:

Three point charge of charge equal to +3.0 micro coulomb

fourth point charge = - 3.0 micro coulomb

side of square = 0.50 m

$K =1/4 \pi \epsilon_0 = 8.99 \times 10^9$ N.m^2/c^2

Due to having equal charge on center of square, 2 charge produce equal electric field at center and other two also produce electric field at center of same value

So we have

$E_1 + E_3 = 0$