Answer:
0.500 T
Explanation:
Since the change in time and the number of coils are both 1, I set the problem up to be 1.3=(1.5(x)-13(x)). I then plugged in numbers for x until I got the answer to be 1.3 V.
Answer:
The current will decrease.
Explanation:
When another bulb is added, the resistance is going to increase. Keep in mind that the current is inversely proportional to the resistance (<em>Ohm's law: R= </em><em>V</em><em>/</em><em>I</em><em> </em><em>).</em> Therefore when the resistance increase, the current running in the circuit will decrease.
If you are pushing the coin across the table at a constant rate, the friction of the table and the horizontal force of your hand pushing are equal, and the coin itself moves at a constant rate. If you push a coin and let it go, there is no horizontal force keeping the coin going. Friction slows the coin to a stop. In both cases, the gravitational downward pull of Earth is equally but oppositely resisted by the upward push of table on the coin.
Observe that the object below moves in the negative direction with a changing velocity. An object which moves in the negative direction has a negative velocity. If the object is slowing down then its acceleration vector is directed in the opposite direction as its motion (in this case, a positive acceleration). The dot diagram shows that each consecutive dot is not the same distance apart (i.e., a changing velocity). The position-time graph shows that the slope is changing (meaning a changing velocity) and negative (meaning a negative velocity). The velocity-time graph shows a line with a positive (upward) slope (meaning that there is a positive acceleration); the line is located in the negative region of the graph (corresponding to a negative velocity). The acceleration-time graph shows a horizontal line in the positive region of the graph (meaning a positive acceleration).
I don't know how I can show you the figure
Answer: q = -52.5 μC
Explanation:
The complete question is given thus;
A point charge Q moves on the x-axis in the positive direction with a speed of 280 m/s. A point P is on the y-axis at y=+70mm. The magnetic field produced at the point P, as the charge moves through the origin, is equal to -0.30uTk. What is the charge Q? (uo=4pi x 10^-7 T m/A).
SOLVING:
from the given parameters we can solve this problem.
Given that the
Speed = 280 m/s
y = 70mm
B = -30 * 10⁻⁶T
Using the equation for magnetic field we have;
Β = μqv*r / 4πr²
making q (charge) the subject of formula we have that;
q = B * 4 *πr² / μqv*r
substituting the values gives us
q = (-0.3*10⁻⁶Tk * 4π * 0.07²) / (4π*10⁻⁷ * 280 ) = - [14.7 * 10⁻¹⁰k / 2.8 * 10⁻⁵ k ]
q = -52.5 μC
cheers i hope this helped !!!
