Using the Equation:
v² = vi² + 2 · a · s → Eq.1
where,
v = final velocity
vi = initial velocity
a = acceleration
s = distance
<span><span>We know that vi = 0 because the ball was at rest initially.
</span><span>
Therefore,
Solving Eq.1 for acceleration,
</span></span> v² = vi² + 2 · a · s
v² = 0 + 2 · a · s
v² = 2 · a · s
Rearranging for a,
a = v ²/2·<span>s
Substituting the values,
a = 46</span>²/2×1<span>
a = 1058 m/s</span>²
<span>Now applying Newton's 2nd law of motion,
</span>
<span>F = ma
= 0.145</span>×<span>1058
F = 153.4 N</span>
Unable to be noticed or not felt because of feeling slight
Answer:
We know there's two forces acting on a book while it sits on a table:the force of gravity pulling it down, and the normal force of the table acting upward on the book. The book isn't accelerating while it sits there. That's because the weight of the book is being counteracted by the normal force of the table.
Explanation:
There are two forces acting upon the book. One force - the Earth's gravitational pull - exerts a downward force. The other force - the push of the table on the book (sometimes referred to as a normal force) - pushes upward on the book.
Answer:
350 F to 100 F it take approx 87.33 min
Explanation:
given data
oven = 350◦F
cooling rack = 70◦F
time = 30 min
cake = 200◦F
solution
we apply here Newtons law of cooling
= -k(T-Ta)
= 
(T(t) -Ta)
= 
= -k(T-Ta)
-ky = -ky
T(t) -Ta = (To -Ta) 
T(t) = Ta+ (To -Ta)
put her value for time 30 min and T(t) = 200◦F and To =350◦F and Ta = 70◦F
so here
200 = 70 + ( 350 - 70 ) 
k = 0.025575
so here for T(t) = 100F
100 = 70 + ( 350 - 70 ) 
time = 87.33 min
so here 350 F to 100 F it take approx 87.33 min
