Answer:
1.5024
Explanation:
Draw a diagram. Put the two cells in series. Now draw 3 resistors. Two of them equal 0.26 ohms each. The third one is the lightbulb which is 12 ohms.
R = 0.26 + 0.26 + 12 = 12.52
The bulb has a voltage of 2.88 volts across it. You can get the current from that.
i = E / R
i = 2.88 / 12 =
i = 0.24 amps.
Now you can get the voltage drop across the two cells.
E = ?
R = 0.26
i = 0.24 amps
E = 0.26 * 0.24
E = 0. 0624
Finally divide the 2.88 by 2 to get 1.44
Each cell has an emf of 1.44 + 0.0624 = 1.5024
The electric force between two charge objects is calculated through the Coulomb's law.
F = kq₁q₂/d²
The value of k is 9.0 x 10^9 Nm²/C² and the charge of proton is 1.602 x10^-19 C. Substituting the known values from the given,
2.30x10^-26 = (9.0 x 10^9 Nm²/C²)(1.602 x10^-19C)²/d²
The value of d is equal to 0.10 m.
Force on the particle is defined as the application of the force field of one particle on another particle. The magnitude and direction of the electrical force will be 4.05×10⁴N towards the north.
<h3>What is electrical force?</h3>
Force on the particle is defined as the application of the force field of one particle on another particle. It is a type of virtual force.
The given data in the problem is
q₁ is the negative charge = 6 µC=6×10⁻⁶ C
q₂ is the positive charge = 3 µC=3×10⁻⁶ C
r is the distance between the charges=0.002 m
is the electric force =?
The value of electric force will be;
Hence the magnitude and direction of the electrical force will be 4.05×10⁴N towards the north.
To learn more about the electrical force refer to the link;
brainly.com/question/1076352
Doesn't seem like we know much here, but we can answer it. Let's talk about what we know. We know it takes 3.24 s for the ball to go up and drop back down again. We know that gravity is the only force acting after the ball leaves the hand, so a = 9.8 m/s^2 (we'll say it's negative in our equations because down being negative is intuitive). We also know that it stops moving for a brief moment at the top of the arc, where v = 0 m/s. Because gravity is the only force, and it slows it down on the way up at the same rate it speeds it up on the way down and the distance covered in upward and downward motion is the same, we can confidently say that it will reach the top of its arc (where v = 0 and it turns around) in half the total time it is in the air, so it takes 1.62 s to reach the peak. Now we can use a kinematics equation, let's use vf = vi + a*t, where vf is final velocity and is 0, vi is initial velocity and is some unknown v we need to solve for, a is acceleration and is -9.8 m/s^2 and t is time and since this is just to the top of the arc, we'll use half the time so 1.62 s. We can solve for vi and plug stuff in like so: v = -a*t = -(-9.8m/s^2)*(1.62s) = 15.876 m/s.
