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My name is Ann [436]
1 year ago
13

According to Coulomb's law, the electrical force between two charged objects:A.is zero if they are opposite charges.B.increases

with increasing charge.C.does not depend on the amount of charge.D.increases with increasing distance.
Physics
1 answer:
kondor19780726 [428]1 year ago
7 0

Answer and explanation:

A correct option is an option (B).

The electrical force between two charges is given as,

F=\frac{1}{4\pi\epsilon_0}\frac{q_1q_1}{r^2}

The electrical force is directly proportional to the product of two charges. Thus Force will depend on two charges irrespective of their signs.

Option (A) is incorrect because if charges are opposite, the value of force will not be zero. It will be -ve.

Option (C) is incorrect because the force is directly proportional to the product of charges, it depends on the amount of charge.

Option (D) is also incorrect because the force in inversly proportional to the distance between two charges. Thus, if the distance between two charges is increased, the force between two charges will decrease.

Concllusion:

The correct option is option (B).

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When approaching an intersection, bridge, or railroad crossing, you should never drive (pass on the left half of the roadway whe
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3 years ago
A tank initially holds 100 gallons of salt solution in which 50 lbs of salt has been dissolved. A pipe fills the tank with brine
olga_2 [115]

Answer:

A. 171.24 Ibs

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To find the amount of salt in the tank,

Let Q = Amount of salt in the mixture

And let 100 + (3-2)t = 100 + t be the volume of mixture at anytime t.

Rate of gain - Rate of loss = dQ / dt

Concentration of salt = Q / (100+t)

For the linear differential equation,

dQ / dt = 3(2) - 2 [Q/ (100 + t)]

dQ /dt + Q [2 / (100 + t)] = 6

The general solution of the linear differential equation is:

Q (i.f) = ∫ A(t) (i.f) dt + C

Therefore,

i.f = e ^ ∫ P(t) dt

And P(t) = 2 / (100 + t)

i.f = e ^ ∫ 2 / (100 + t)

  = e ^ 2㏑ (100 + t)

     = e ^ ㏑ (100 + t) ^2 = (100 + t) ^2

Q(100 + t) ^ 2 = ∫6 (100 + t) ^2 dt + C

 Q(100 + t) ^2 = 2(100 + t) ^ 3 + C

  When t = 0, Q = 50

Therefore,

50( 100) ^2 = 2(100) ^3 + C

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therefore, when t = 30,

Q (100 + 30) ^2 = 2(100 + 30) ^3 - 1.5 * 10 ^6

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7 0
3 years ago
At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?
sergey [27]

1) Initial upward acceleration: 6.0 m/s^2

2) Mass of burned fuel: 0.10\cdot 10^4 kg

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude F_g = mg, in the downward direction, where

m=1.9\cdot 10^4 kg is the rocket's mass

g=9.8 m/s^2 is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

T=3.0\cdot 10^5 N

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

T-mg=ma (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

a'=6.9 m/s^2

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

T-m'g=m'a'

where

m' is the new mass of the rocket

Re-arranging the equation and solving for m', we find

m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg

And since the initial mass of the rocket was

m=1.9 \cdot 10^4 kg

This means that the mass of fuel burned is

\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg

3 0
3 years ago
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