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Ludmilka [50]
3 years ago
8

Solve each of these problems, remembering to include formula, calculations, and units!

Physics
1 answer:
KATRIN_1 [288]3 years ago
5 0

Answer:

I nEeD heLp

Explanation:

HelP

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OlgaM077 [116]
A.is an example of decomposition reaction.
6 0
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A positive charge +q1 is located to the left of a negative charge -q2. On a line passing through the two charges, there are two
AfilCa [17]

Answer:

please the answer below

Explanation:

(a) If we assume that our origin of coordinates is at the position of charge q1, we have that the potential in both points is

V_1=k\frac{q_1}{r-1.0}-k\frac{q_2}{1.0}=0\\\\V_2=k\frac{q_1}{r+5.2}-k\frac{q_2}{5.2}=0\\\\

k=8.89*10^9

For both cases we have

k\frac{q_1}{r-1.0}=k\frac{q_2}{1.0}\\\\q_1(1.0)=q_2(r-1.0)\\\\r=\frac{q_1+q_2}{q_2}\\\\k\frac{q_1}{r+5.2}=k\frac{q_2}{5.2}\\\\q_1(5.2)=q_2(r+5.2)\\\\r=\frac{5.2q_1-5.2q_2}{q_2}

(b) by replacing this values of r in the expression for V we obtain

k\frac{q_1}{\frac{5.2(q_1-q_2)}{q_2}+5.2}=k\frac{q_2}{5.2}\\\\\frac{q_1}{q_2}=\frac{(q_1-q_2)}{q_2}-1.0=\frac{q_1-q_2-q_2}{q_2}=\frac{q_1-2q_2}{q_2}

hope this helps!!

3 0
3 years ago
Read 2 more answers
A satellite’s velocity is 30000m/s. After 60 secs, it’s velocity shows to 15000m/a. What is the satellite’s acceleration?
Orlov [11]

Answer:

Acceleration = 9 × 10^5 m/s^2 ( deceleration )

Explanation:

From the first equation of motion:

V = u + at

15000 = 30000 + 60a

a = ( 15000-30000)/60

a = 9 × 10^5 m/s^2

4 0
3 years ago
A EXAMPLE OF WHEN BOTH PHYSICAL CHANGE AND CHEMICAL CHANGE OCCUR 3 EXAMPLES
kvasek [131]

Answer:

I know 1, that is in the case of a burning of a candle.

Explanation:

5 0
3 years ago
A 6.75 nC charge is located 1.99 m from a 4.46 nC point charge.
sladkih [1.3K]

Explanation:

Given that,

Charge 1, q_1=6.75\ nC=6.75 \times 10^{-9}\ C

Charge 2, q_2=4.46\ nC=4.46\times 10^{-9}\ C

The distance between charges, r = 1.99 m

To find,

The electrostatic force and its nature

Solution,

(a) The electric force between two charges is given by :

F=\dfrac{kq_1q_2}{r^2}

F=\dfrac{9\times 10^9\times 6.75\times 10^{-9}\times 4.46\times 10^{-9}}{(1.99)^2}

F=6.84\times 10^{-8}\ N

(b) As the magnitude of both charges is positive, then the force between charges will be repulsive.

Therefore, this is the required solution.

5 0
3 years ago
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