Initial speed of the coin (u)= 0 (As the coin is released from rest)
Acceleration due to gravity (a) = g = 9.81 m/s²
Time of fall (t) = 1.5 s
From equation of motion we have:
By substituting values in the equation, we get:
v = 0 + 9.81 × 1.5
v = 14.715 m/s
Speed of the coin as it hits the ground/Final speed of the coin = 14.715 m/s
Answer:
Explanation:
Given
velocity=100 ft/s
height of landing zone=10 ft
Equation of 
Maximum height=35 feet
at maximum height
At
It can be calculated using Boyle's Law. A.
The resultant motion is given by pithagoras, since the two components (north and east) are perpendicular to each other.
They are asking you about the direction so you have to use trigonometry, finding that the direction is Ф=arctan(3.8/12)=17.57° north of east.
<u>Answer:</u>
<em>Newtons II law: </em>
<em> </em>It is defined as<em> "the net force acting on the object is a product of mass and acceleration of the body"</em> . Also it defines that the <em>"acceleration of an object is dependent on net force and mass of the body".</em>
Let us assume that,a string is attached to the cart, which passes over a pulley along the track. At another end of the string a weight is attached which hangs over the pulley. The hanging weight provides tension in the spring, and it helps in accelerating the cart. We assume that the string is massless and no friction between pulley and the string.
Whenever the hanging weight moves downwards, the cart will accelerate to right side.
<em>For the hanging weight/mass</em>
When hanging weight of mass is m₁ and accelerate due to gravitational force g.
Therefore we can write F = m₁ .g
and the tension acts in upward direction T (negetive)
Now, Fnet = m₁ .g - T
= m₁.a
So From Newtons II law<em> F = m.a</em>
