Answer:
The consecutive charge configuration has a more intense field than alternating
Explanation:
In each corner we place a different account there are only two different settings, see attached.
In the case of alternating charging (+ - + -) see diagram 1, the electric field in the center is canceled in pairs, resulting in a zero field
In the case of consecutive loads (+ + - -) in this case we have a result between the two charges, therefore the total field is
E = 2 k q / ra2 a cos 45
The consecutive charge configuration has a more intense field than alternating
Answer:
a. 0.000002 m
b. 0.00000182 m
Explanation:
36 cm = 0.36 m
15 cm = 0.15 m
a) We can start by calculating the air-water pressure of the bucket submerged 20m below the water surface:
Suppose air is ideal gas, then if the temperature stays the same, the product of its pressure and volume stays the same
Where P1 = 1.105 Pa is the atmospheric pressure, V_1 is the air volume in the bucket on the suface:
As the pressure increases, the air inside the bucket shrinks. But the crossection area stays constant, so only h, the height of air, decreases:
b) If the temperatures changes, we can still reuse the ideal gas equation above:
Answer:
47.76°
Explanation:
Magnitude of dipole moment = 0.0243J/T
Magnetic Field = 57.5mT
kinetic energy = 0.458mJ
∇U = -∇K
Uf - Ui = -0.458mJ
Ui - Uf = 0.458mJ
(-μBcosθi) - (-μBcosθf) = 0.458mJ
rearranging the equation,
(μBcosθf) - (μBcosθi) = 0.458mJ
μB * (cosθf - cosθi) = 0.458mJ
θf is at 0° because the dipole moment is aligned with the magnetic field.
μB * (cos 0 - cos θi) = 0.458mJ
but cos 0 = 1
(0.0243 * 0.0575) (1 - cos θi) = 0.458*10⁻³
1 - cos θi = 0.458*10⁻³ / 1.397*10⁻³
1 - cos θi = 0.3278
collect like terms
cosθi = 0.6722
θ = cos⁻ 0.6722
θ = 47.76°
Answer:
.c. −160°C
Explanation:
In the whole process one kg of water at 0°C loses heat to form one kg of ice and heat lost by them is taken up by ice at −160°C . Now see whether heat lost is equal to heat gained or not.
heat lost by 1 kg of water at 0°C
= mass x latent heat
= 1 x 80000 cals
= 80000 cals
heat gained by ice at −160°C to form ice at 0°C
= mass x specific heat of ice x rise in temperature
= 1 x .5 x 1000 x 160
= 80000 cals
so , heat lost = heat gained.
Answer:
a) W=2.425kJ
b) 
c) 
d) Q=-2.425kJ
Explanation:
a)
First of all, we need to do a drawing of what the system looks like, this will help us visualize the problem better and take the best possible approach. (see attached picture)
The problem states that this will be an ideal system. This is, there will be no friction loss and all the work done by the object is transferred to the water. Therefore, we need to calculate the work done by the object when falling those 10m. Work done is calculated by using the following formula:
Where:
W=work done [J]
F= force applied [N]
d= distance [m]
In this case since it will be a vertical movement, the force is calculated like this:
F=mg
and the distance will be the height
d=h
so the formula gets the following shape:
so now e can substitute:
which yields:
W=2.425kJ
b) Since all the work is tansferred to the water, then the increase in internal energy will be the same as the work done by the object, so:
c) In order to find the final temperature of the water after all the energy has been transferred we can make use of the following formula:
Where:
Q= heat transferred
m=mass
=specific heat
= Final temperature.
= initial temperature.
So we can solve the forula for the final temperature so we get:
So now we can substitute the data we know:
Which yields:
d)
For part d, we know that the amount of heat to be removed for the water to reach its original temperature is the same amount of energy you inputed with the difference that since the energy is being removed this means that it will be negative.
