1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Oxana [17]
4 years ago
13

When a third object is brought in contact with the first object (after it gains the electrons), the resulting charge on the thir

d object is 0.9 C. What was its initial charge (in C)?
Physics
1 answer:
nirvana33 [79]4 years ago
8 0

Answer:

- Contact 1 with 3 ,  initial charge of 1.8 C.

- contact 1 with 2 and then 1 with 3 , first body should have 3.6 C

Explanation:

The excess charge on a body is distributed evenly throughout the body.

We can have two different configurations:

- Contact 1 with 3

When the third body was touched with the first, the initial charge was distributed between the two, so that when each one separated, it had half the charge, in this configuration the first body should have an initial charge of 1.8 C.

- contact 1 with 2 and then 1 with 3

Another possible configuration of the exercise is that the first body touches the second and the charge decrease to the half and then touches the third where it again decreases by half, so that the first body only gives it every ¼ of its initial load.

Therefore in this configuration if the third body has a load of 0.9C the first body should have 3.6 C

You might be interested in
A ski lift carries a 75.0-kg skier at 3.00 m/s for 1.50 min along a cable that is inclined at an angle of 40.0° above the horizo
Nookie1986 [14]

Answer

given,

mass of the ski = 75 Kg

speed of the skier, v = 3 m/s

time = 1.50 min = 90 s

angle of inclination, θ = 40°

distance = s x t

              = 3 x 90 = 270 m

a) W = F. d cos θ

   W = mg. d cos θ

   W = 75 x 9.8 x 270 x cos 40°

    W = 152021.52 J

work is done by the ski lift is equal to  152021.52 J

b) Power extended by the ski

 Power = \dfrac{Work\ done}{time}

Power = \dfrac{152021.52}{90}

      P = 1689.13 Watt.

power is expended by the ski lift is equal to 1689.13 W.

3 0
3 years ago
URGENT <br> Find the total<br> equivalent<br> resistance for the<br> circuit.
Nesterboy [21]

Answer:

20 + 10 = 30 \\ 30and50 in \: parallel \\  =  \frac{50 \times 30}{50 + 30}  \\  =  \frac{1500}{80}  \\ 18.75 \\ 30nd40parallel \\  \frac{30 \times 40}{70}  \\  \frac{1200}{70}  = 17.143 \\ total = 18.75 + 17.143 = 35.893ohm \\ thank \: u

3 0
3 years ago
AYOO I NEED HELP plz
oksano4ka [1.4K]

Answer:

1D

2C

3C

4C

5D

6D

7D

8D

9B

Explanation:

better give me points X﹏X

4 0
3 years ago
Read 2 more answers
A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turn
inn [45]

Answer:

The answer to the question is

The ladybug begins to slide

Explanation:

To solve the question we assume that the frictional force of the ladybug and the gentleman bug are the same

Where the  frictional force equals F_{Friction} = μ×N = m×g×μ

and the centripetal force is given by m·ω²·r

If we denote the properties of the ladybug as 1 and that of the gentleman bug as 2, we have

m₁×g×μ = m₁·ω²·r₁ ⇒ g×μ = ω²·r₁

and for the gentleman bug we have

m₂×g×μ = m₂·ω²·r₂ ⇒ g×μ = ω²·r₂

But r₁ = 2×r₂

Therefore substituting the values of r₁ =2×r₂ we have

g×μ = ω²·r₁ = g×μ = ω²·2·r₂

Therefore   ω²·r₂ = 0.5×g×μ for the ladybug. That is the ladybug has to overcome half the frictional force experienced by the gentleman bug before it start to slide

The ladybug begins to slide

6 0
3 years ago
The mass of the Earth is 6 × 1024 kg, the mass of the Moon is 7 × 1022 kg, and the center-to-center distance is 4 × 108 m. How f
Karolina [17]

Answer:

The center of mass of the Earth–Moon system is 4.613 × 10⁶ m from center of the Earth.

Explanation:

Let the reference point be the center of the Earth

X_{Cm = \frac{M_eX_e +M_mX_m}{M_e+M_m}

Where;

Xcm is the distance from center of the Earth =?

Me is the mass of the Earth = 6 × 10²⁴ kg

Xe is the center mass of the Earth = 0

Mm is the mass of the moon = 7 × 10²² kg

Xm is the center mass of the moon =  4 × 10⁸ m

X_{Cm = \frac{M_eX_e +M_mX_m}{M_e+M_m} =  \frac{M_e(0) +M_mX_m}{M_e+M_m} = \frac{ M_mX_m}{M_e+M_m}}\\\\X_C_m = \frac{7 X 10^{22}*4X10^8}{6X10^{24}+7X10^{22}} =\frac{28 X10^{30}}{607X10^{22}}\\\\  X_C_m = 4.613 X10^6 m

Therefore, the center of mass of the Earth–Moon system is 4.613 × 10⁶ m from center of the Earth.

8 0
4 years ago
Other questions:
  • Mike's car, which weighs 1,000 kg, is out of gas. Mike is trying to push the car to a gas station, and he makes the car go 0.05
    12·1 answer
  • Name at least three physical properties of the bowling ball.
    7·2 answers
  • A rocket with a mass of 2.0 Ã 106 kg is designed to take off from the surface of the earth by burning fuel and ejecting it with
    5·1 answer
  • What is the energy conversion in a hair dryer?
    5·1 answer
  • What can be said about the sign of the work done by the force f⃗ 1??
    9·1 answer
  • Why do you think the temperature does not change much during a phase change?
    6·2 answers
  • Can anyone help me with this question please <br> I’ll mark as brainliest <br> No links.
    7·2 answers
  • Cuál es el significado del diálogo intercultural​
    5·1 answer
  • 4. Kinetic or potential--a painting hanging on a wall in a frame?<br> kinetic<br> potential
    15·1 answer
  • Correctness the relation Where pressure=3accleration due to gravity/4pi G​
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!