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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

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Answer is power outlet in your house
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3 years ago
Joules constant, 4.1868 J/cal, is an equivalence relation for 4.1868 Joules of work for 1 calorie of heat delivered to a substan
Llana [10]

Answer:

The final answer is 6.1 BTU = 6.5621*103 KJ

Explanation:

1 BTU= 252 cal\\1 BTU =252 × 4.2 kJ\\6.1 BTU = 252 × 6.1×4.2 kJ                = 6562.08 kJ\\                = 6.5621 × 103 kJ\\

Hence, 6.1 BTU = 6.5621 × 103 kJ

7 0
3 years ago
On December 26, 2004, a great earthquake occurred off the coast of Sumatra and triggered immense waves (tsunami) that killed som
k0ka [10]

Answers:

a) 222.22 m/s

b) 800.00 km/h

Explanation:

The speed of a wave is given by the following equation:

v=f \lambda

Where:

v is the speed

f=\frac{1}{T} is the frequency, which has an inverse relation with the period T=1 h

\lambda=800 km is the wavelength

Solving with the given units:

v=\frac{1}{T}\lambda

v=\frac{1}{1 h}800 km

v=800.00 km/h This is the speed of the wave in km/h

Transforming this speed to m/s:

v=800.00 \frac{km}{h} \frac{1 h}{3600 s} \frac{1000 m}{1 km}

v=222.22 m/s This is the speed of the wave in m/s

5 0
3 years ago
When you take your 1900-kg car out for a spin, you go around a corner of radius 55 m with a speed of 15 m/s. The coefficient of
pentagon [3]

Answer:

7772.72N

Explanation:

When u draw your FBD, you realize you have 3 forces (ignore the force the car produces), gravity, normal force and static friction. You also realize that gravity and normal force are in our out of the page  (drawn with a frame of reference above the car). So that leaves you with static friction in the centripetal direction.

Now which direction is the static friction, assume that it is pointing inward so

Fc=Fs=mv²/r=1900*15²/55=427500/55=7772.72N

Since the car is not skidding we do not have kinetic friction so there can only be static friction. One reason we do not use μFn is because that is the formula for maximum static friction, and the problem does not state there is maximum static friction.

7 0
3 years ago
Two equally charged tiny spheres of mass 1.0 g are placed 2.0 cm apart. When released, they begin to accelerate away from each o
zhannawk [14.2K]

Answer:

The magnitude of the charge on each sphere is 0.135 μC

Explanation:

Given that,

Mass = 1.0

Distance = 2.0 cm

Acceleration = 414 m/s²

We need to calculate the magnitude of charge

Using newton's second law

F= ma

a=\dfrac{F}{m}

Put the value of F

a=\dfrac{kq^2}{mr^2}

Put the value into the formula

414=\dfrac{9\times10^{9}\times q^2}{1.0\times10^{-3}\times(2.0\times10^{-2})^2}

q^2=\dfrac{414\times1.0\times10^{-3}\times(2.0\times10^{-2})^2}{9\times10^{9}}

q^2=1.84\times10^{-14}

q=0.135\times10^{-6}\ C

q=0.135\ \mu C

Hence, The magnitude of the charge on each sphere is 0.135μC.

7 0
3 years ago
Read 2 more answers
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