The coin's acceleration is <u>0.37 m/s²</u>
Acceleration is the rate of change of the velocity of an item with appreciation to time. Accelerations are vector portions. The orientation of an item's acceleration is given by the orientation of the net pressure appearing on that object.
<u>Calculation:-</u>
<u />
V² = U -2aS
a = U/2S
= 2/2×2.7
= <u>0.37 m/s²</u>
Acceleration is the charge at which velocity modifications with time, in terms of each speed and route. A factor or an object moving in a straight line is accelerated if it quickens or slows down. movement on a circle is extended despite the fact that the rate is consistent because the course is continually changing.
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Answer:
Explanation:
a)Magnitude = 
84=
x= +50.67 or -50.67 units
b) We are given that the resultant is entirely in the -ve x direction which means that the y-component of the resultant is 0; It means that the y-component of the next vector = -ve of the y component of the initial vector i.e 67.
To make the magnitude 80 units in the negative x direction where the y component is 0, the x component must be -130.67(-50.67 - 80) as the x component is + 50.67units.
Magnitude = 
= 146.85 units
c) The direction vector = 67/146.85 i - 130.67/146.85 j where i corresponds to the vector in y direction and j corresponds to the vector in x direction. Or this vector is at an angle of 180 - 
i.e 152.85 degrees from the +ve x-axis.
Answer:
(a) 3.44 x 10^-3 m^3/s
(b) 8.4 m/s
Explanation:
area of water line, A = 5.29 x 10^-3 m
number of holes, N = 15
Speed of water in line, V = 0.651 m/s
(a) Volume flow rate is given by
V = area of water line x speed of water in water line
V = 5.29 x 10^-3 x 0.651 = 3.44 x 10^-3 m^3/s
(b) area of one hole, a = 4.13 x 10^-4 m
Let v be the velocity of water in each hole
According to the equation of continuity
A x V = a x v
5.29 x 10^-3 x 0.651 = 4.1 x 10^-4 x v
v = 8.4 m/s
Current is defined as the rate of charge flowing a point every second. Having a current of 1 Ampere signifies 1 Coulomb is flowing in a circuit every second. It is measured by the use of an ammeter which is positioned in series to the component to be measured. The current in the problem is calculated as follows:
I = 2.0 x 10^-4 C / 5.0 x 10^-5 s
<span>I = 4 A or 4.0 x 10^0 A</span>
