When a car's starter is in use, it draws a large current. The car's lights draw much less current. As a certain car is starting, the current through the battery is 54.0 A and the potential difference across the battery terminals is 9.63 V. When only the car's lights are used, the current through the battery is 2.06 A and the terminal potential difference is 12.8 V. Find the battery's emf.

ε = 12.9257 V

Explanation:

To do this we need to see what is happening in both cases, when the lights are not being used and when the lights are used.

We will call case 1, the part when the lights are not being used, and case 2 when they're used.

Case 1:

Here, we should write the expression that will help us to determine the emf of the battery, which is:

ε - I₁R = V₁   (1)

Where:

ε: Emf of the battery

I: current of the car

R: Resistance

V: Voltage

Applying this expression with the given data, we have the following:

ε - 54R = 9.63    (2)

For the moment, we can't do anything because we have two incognites. Let's see what happen when we write the expression for case 2.

Case 2:

Applying the same as case 1 but using the other data we have:

ε - I₂R = V₂

Replacing we have:

ε - 2.06R = 12.8    (3)

Now, we can use (2) and (3) and solve for ε calculating the resistance:

ε = 12.8 + 2.06R   (4)

ε = 9.63 + 54R     (5)

Equalling both:

12.8 + 2.06R = 9.63 + 54R

12.8 - 9.63 = 54R - 2.06R

3.17 = 51.94R

R = 3.17 / 51.94 = 0.061 Ω

With this value, we can calculate the emf, using equation (4) or (5):

ε = 12.8 + 2.06(0.061)

Hope this helps

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases . this process does not change the mass number.

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as -decay. in this the mass number remains same.

from the above information, we see that only alpha-decay is producing a change in the mass number of the nucleus.