Quantity of Charge , Q = ne
Where n = number of electrons
e = charge on one electron = -1.6 * 10 ^-19 C.
n = 50 * 10^31 electrons
Q = (50 * 10^31)*( -1.6 * 10 ^-19 ) = -8 * 10^13 C.
Note that the minus sign indicates that the charge is a negative charge.
Hello!!
Here we have a simple matter of conservation of energy. ME=PE+KE.
At point A we have PE=mgh and KE=1/2mv^2. At point A all we have is PE since the coaster isn’t rolling yet. But by conservation of energy, we know that it will have enough energy to roll down and get to and equal height on another hill. Providing we are neglecting friction and drag and resistance forces which we are in this case. So we can conclude that the KE will be greater at Point B since ME=PE+KE and for ME to remain the same and we know the PE is less on lower hill, so we can conclude that KE on lower hill is greater to keep ME the same and have conservation of energy.
Hope this helps you understand the concept!! Any questions please just ask!! Thank you so much!!
Answer:
15 cm
Explanation:
= Diameter of the coin = 15 mm
= Diameter of the image of coin = 5 mm
= distance of the coin from mirror = 15 cm
= distance of the image of coin from mirror = ?
Using the equation
= - 5 cm
= radius of curvature
Using the mirror equation
= - 15 cm
Answer:
Theta1 = 12° and theta2 = 168°
The solution procedure can be found in the attachment below.
Explanation:
The Range is the horizontal distance traveled by a projectile. This diatance is given mathematically by Vo cos(theta) t. Where t is the total time of flight of the projectile in air. It is the time taken for the projectile to go from starting point to finish point. This solution assumes the projectile finishes uts motion on the same horizontal level as the starting point and as a result the vertical displacement is zero (no change in height).
In the solution as can be found below, the expression to calculate the range for any launch angle theta was first derived and then the required angles calculated from the equation by substituting the values of the the given quantities.
The correct answer is B the total velocity is equal at both landing and launch because before your about launch you have 0 velocity then when you have landed you also have 0 velocity. Hope This Helps
