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2 years ago

Physics!!! please help D:

Physics

1 answer:

Ugo [173]2 years ago

6 0

Answer: A

Explanation: isotopes of the same thing element have the same number of protons in the nucleus but differ in the number of neutrons.

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Three equal point charges, each with charge 1.45 μCμC , are placed at the vertices of an equilateral triangle whose sides are of

LUCKY_DIMON [66]

Answer:

U = 80.91 J

Explanation:

In order to calculate the electric potential energy between the three charges you use the following formula:

$U=k\frac{q_1q_2}{r_{1,2}}$ (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q1: q2 charge

r1,2: distance between charges 1 and 2.

For the three charges you have:

$U_T=k\frac{q_1q_2}{r_{1,2}}+k\frac{q_1q_3}{r_{1,3}}+k\frac{q_2q_3}{r_{2,3}}$ (2)

You use the fact that q1=q2=q3=q and that the distance between charges are equal. Then, in the equation (2) you have:

q = 1.45μC = 1.45*10^-6C

r = 0.700mm = 0.700*10^-3m

$U_T=3k\frac{q^2}{r}=3(8.98*10^9Nm^2/C^2)\frac{(1.45*10^{-6}C)}{0.700*10^{-3}m}\\\\U_T=80.91J$

The electric potential energy between the three charges is 80.91 J

7 0

3 years ago

A race car accelerates from 0 m/s ro 30 m/s with a displacement of 45 what is it acceleration

Minchanka [31]

Hdghehwhevevwgwhhwvsvsvsiosjwbwbwhshbshs

6 0

3 years ago

An object releasing energy is evidence of which type of chemical change?

Ray Of Light [21]

Exothermic is the answer to your question

8 0

2 years ago

A gymnast dismounts off the uneven bars in a tuck position with a radius of 0.3m (assume she is a solid sphere) and an angular v

kifflom [539]

Here we will say that there is no external torque on the system so we will have

$L_i = L_f$

here we know that

$L_i = I_1\omega_1$

where we know that

$I_1 = \frac{2}{5}mr^2$

Also we know that

$I_2 = \frac{1}{12}mL^2$

initial angular speed will be

$\omega_1 = 2\pi(2rev/s) = 4\pi rad/s$

now from above equation

$\frac{2}{5}mr^2 (4\pi) = \frac{1}{12}mL^2 \omega$

$0.4(0.3)^2(4\pi) = \frac{1}{12}(1.5)^2\omega$

$0.452 = 0.1875 \omega$

now we have

$\omega = 2.41 rad/s$

so final speed will be 2.41 rad/s

6 0

2 years ago

A toolbox of mass 3.2kg is lowered by a rope from the roof to the ground. Find the acceleration of the toolbox when the force of

Lisa [10]

Answer:

The answer to your question is:

a) 2.7 m/s²

b) -3.6 m/s²

Explanation:

Data

mass of the toolbox = 3.2 kg

a = ?

F = 40 N and F = 20 N

g = 9.81 m/s²

Formula

Second law of motion = F = ma

a + g = F / m

a = F/m - g

a) a = 40/3.2 - 9.81

a = 2.69 ≈ 2.7 m/s² positive up

b) a = 20/ 3.2 - 9.81

a = 6.25 - 9.81

= - 3.56 ≈ - 3.6 m/s² negative down

8 0

3 years ago