The initial velocity of the ball is 55.125 m/s.
<h3>Initial velocity of the ball</h3>
The initial velocity of the ball is calculated as follows;
During upward motion
h = vi - ¹/₂gt²
h = vi - 0.5(9.8)(3²)
h = vi - 44.1 ----------------- (1)
During downward motion
h = vi + ¹/₂gt²
h = 0 + 0.5(9.8)(1.5)²
h = 11.025 ----------- (2)
solve (1) and (2) together, to determine the initial velocity of the ball
11.025 = vi - 44.1
vi = 11.025 + 44.1
vi = 55.125 m/s
Thus, the initial velocity of the ball is 55.125 m/s.
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Via half-life equation we have:
Where the initial amount is 50 grams, half-life is 4 minutes, and time elapsed is 12 minutes. By plugging those values in we get:
There is 6.25 grams left of Ra-229 after 12 minutes.
From the calculations, the power expended is 43650 W.
<h3>What is the power expended?</h3>
Now we can find the acceleration from;
v = u + at
u = 0 m/s
v = 95 km/h or 26.4 m/s
t = 6.8 s
a = ?
Now
v = at
a = v/t
a = 26.4 m/s/ 6.8 s
a = 3.88 m/s^2
Force = ma = 850-kg * 3.88 m/s^2 = 3298 N
The distance covered is obtained from;
v^2 = u^2 + 2as
v^2 = 2as
s = v^2/2a
s = (26.4)^2/2 * 3.88
s = 696.96/7.76
s = 90 m
Now;
Work = Fs
Work = 3298 N * 90 m = 296820 J
Power = 296820 J/ 6.8 s
= 43650 W
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Answer:
x=4.06m
Explanation:
A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.
Vf=Vo+a.t (1)\\\\
{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\
X=Xo+ VoT+0.5at^{2} (3)\\
Where
Vf = final speed
Vo = Initial speed
T = time
A = acceleration
X = displacement
In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve
for this problem
Vf=7.6m/s
t=1.07
Vo=0
we can use the ecuation number one to find the acceleration
a=(Vf-Vo)/t
a=(7.6-0)/1.07=7.1m/s^2
then we can use the ecuation number 2 to find the distance
{Vf^{2}-Vo^2}/{2.a} =X
(7.6^2-0^2)/(2x7.1)=4.06m
Answer:
The appropriate solution is:
(a) 
(b) 
(c) 
Explanation:
According to the question, the value is:
Power of bulb,
= 60 W
Distance,
= 1.0 mm
Now,
(a)
⇒ 
On applying cross-multiplication, we get
⇒ 
⇒ 
⇒ 
(b)
As we know,
⇒ 
By putting the values, we get
⇒ 
(c)
⇒ 
⇒ 
⇒ 
