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Pani-rosa [81]
3 years ago
14

Suppose the electric field in problems 2 was caused by a point charge. The test charge is moved to a distance twice as far from

the charge. What is the magnitude of the force that the field exerts on the test charge now ?
Physics
1 answer:
mojhsa [17]3 years ago
8 0

Answer:

it is reduced four times.

Explanation:

By definition, the electric field is the force per unit charge created by a charge distribution.

If the charge creating the field is a point charge, the force exerted by it on a test charge, must obey Coulomb´s Law, so, it must be inversely proportional to the square of the distance between the charges.

So, if the distance increases twice, as the force is inversely proportional to the square of the distance, and the square of 2 is 4, this means that the magnitude of the force exerted on the test charge must be 4 times smaller.

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A weightlifter uses a force of 356 N to lift a set of weights 2.2 m off the ground. How much work did the weightlifter do? Answe
Valentin [98]

Answer:

783.2 N.m or Joules

Explanation:

W=F*d*cos(Ф)

Work done when a force F is applied to move an object by a displacement d meters

where cos(Ф) is the angle between applied force F and displacement d

Since the weightlifter is applying the force upward and the set of weight also move upward therefore, both are in same direction hence angle will be zero.

Now lets substitute the given values into the work equation:

W=356*2.2*cos(0)

since cos(0)=1

W=356*2.2*1

W=783.2 N.m

or

W=783.2 Joules

Since the unit of Force is Newton (N) and the unit of displacement is meters (m) therefore, unit of work done will be N.m

Note: 1 N.m is equivalent to 1 joule

7 0
3 years ago
The energy transferred between objects that are at different temperatures
sukhopar [10]

Answer:

We learned in the previous section that temperature is proportional to the average kinetic energy of atoms and molecules in a substance, and that the average internal kinetic energy of a substance is higher when the substance’s temperature is higher.

If two objects at different temperatures are brought in contact with each other, energy is transferred from the hotter object (that is, the object with the greater temperature) to the colder (lower temperature) object, until both objects are at the same temperature. There is no net heat transfer once the temperatures are equal because the amount of heat transferred from one object to the other is the same as the amount of heat returned. One of the major effects of heat transfer is temperature change: Heating increases the temperature while cooling decreases it. Experiments show that the heat transferred to or from a substance depends on three factors—the change in the substance’s temperature, the mass of the substance, and certain physical properties related to the phase of the substance.

The equation for heat transfer Q is

Q = mcΔT,

Explanation:

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2 years ago
If two charged balloons are 24cm apart and they feel a force of electrical repulsion of 20N, what would the force of electrical
OverLord2011 [107]

Answer:

Soory

Explanation:

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5 0
3 years ago
The sun's energy is stored in fossil fuels. true or false.
Scilla [17]
Yes,suns original energy are still stored at the bonds of fossils in which they taken long,long time ago and these fossils powered now industries,fuel for vehicles.In those chemical bonds holds much energy packed together for centuries.
8 0
3 years ago
A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413
natima [27]

Answer:

The value of  charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force F=14.413\ N

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

r=\sqrt{x_{2}^2+y_{2}^2}

Put the value into the formula

r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}

r=0.0876\ m

We need to calculate the magnitude of the charge q₃

Using formula of net force

F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})

Put the value into the formula

14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})

(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}

\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}

q_{3}=0.0210909\times(0.0438)^2

q_{3}=40.46\times10^{-6}\ C

q_{3}=40.46\ \mu C

Hence, The value of  charge q₃ is 40.46 μC.

5 0
3 years ago
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