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Ede4ka [16]
4 years ago
13

A small, positively charged ball is moved close to a large, positively charged ball. which describes how the small ball likely r

esponds when it is released?it will move toward the large ball because like charges repel.it will move toward the large ball because like charges attract.it will move away from the large ball because like charges repel.it will move away from the large ball because like charges attract.
Physics
2 answers:
NNADVOKAT [17]4 years ago
6 0

Answer;

-it will move away from the large ball because like charges repel.

Explanation;

-Electric force is the force that pushes apart two like charges, or that pulls together two unlike charges. The basic law of electrostatics Like charges of electricity repel each other, whereas unlike charges attract each other.

When small, positively charged ball is moved close to a large, positively charged ball it would be pushed away from the large positively charged ball since they are both positively charged. One has to put in energy to try to move the small ball closer to the large ball. The closer one try to move it to the large ball, the more energy one has to put in, so the more electrical potential energy the small ball would have.

MariettaO [177]4 years ago
4 0

Answer:

C. It will move away from the large ball because like charges repel.

Explanation:

I just did the test and got it right

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Guys answer with a clear explanation and plzz don't spam.
timama [110]

Answer:

20.7N

Explanation:

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3 0
3 years ago
The​ half-life of a certain radioactive substance is 12 hours. There are 19 grams present initially. a. Express the amount of su
neonofarm [45]

Answer:

(a) N=19\times e^{-\lambda t}

(b) 15 hours

Explanation:

half life, T = 12 hours

No = 19 g

(a) Let N be the amount remaining after time t.

Let λ be the decay constant.

\lambda =\frac {0.6931}{T}

The equation of radioactivity used here is given by

N=N_{o}e^{-\lambda t}

N=19\times e^{-\lambda t}

(b) N = 8 gram

Substitute the values in above equation

\lambda =\frac {0.6931}{12}

λ = 0.0577 per hour

So, 8=19\times e^{-0.577t}

e^{-0.0577t}=0.421

Take natural log on both the sides

- 0.0577 t = - 0.865

t = 15 hours

4 0
3 years ago
A cubic metal box with sides of 17 cm contains air at a pressure of 1 atm and a temperature of 278 K. The box is sealed so that
const2013 [10]

Answer:

F = 3.98 kN

Explanation:

GIVEN DATA:

sides of box = 17 cm

pressure = 1 atm = 101325 N/m2

T2 = 378K

T1 = 278 K

final pressure can be calculate by using below relation

\frac{P_{1}}{P_{2}}=\frac{T_{1}}{T_{2}}

we know that

force = pressure * area

therefore force is

F =(\frac{T_{1}}{T_{2}}*P_{1})A

F =(\frac{378}{278}*101325)(17*10^{-2})^{2}

F = 3.98 kN

5 0
4 years ago
A thin hoop is hung on a wall, supported by a horizontal nail. The hoop's mass is M=2.0 kg and its radius is R=0.6 m. What is th
boyakko [2]

Answer:

Explanation:

Given that,

Mass of the thin hoop

M = 2kg

Radius of the hoop

R = 0.6m

Moment of inertial of a hoop is

I = MR²

I = 2 × 0.6²

I = 0.72 kgm²

Period of a physical pendulum of small amplitude is given by

T = 2π √(I / Mgd)

Where,

T is the period in seconds

I is the moment of inertia in kgm²

I = 0.72 kgm²

M is the mass of the hoop

M = 2kg

g is the acceleration due to gravity

g = 9.8m/s²

d is the distance from rotational axis to center of of gravity

Therefore, d = r = 0.6m

Then, applying the formula

T = 2π √ (I / MgR)

T = 2π √ (0.72 / (2 × 9.8× 0.6)

T = 2π √ ( 0.72 / 11.76)

T = 2π √0.06122

T = 2π × 0.2474

T = 1.5547 seconds

T ≈ 1.55 seconds to 2d•p

Then, the period of oscillation is 1.55seconds

6 0
3 years ago
Visible light waves are ______ waves<br> A.mechanical <br> B.electromagnetic
motikmotik
B.Electromagnetic
explanation:
4 0
3 years ago
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