Answer:
The answer to your question is:
a) t1 = 2.99 s ≈ 3 s
b) vf = 39.43 m/s
Explanation:
Data
vo = 10 m/s
h = 74 m
g = 9.81 m/s
t = ? time to reach the ground
vf = ? final speed
a) h = vot + (1/2)gt²
74 = 10t + (1/2)9.81t²
4.9t² + 10t -74 = 0 solve by using quadratic formula
t = (-b ± √ (b² -4ac) / 2a
t = (-10 ± √ (10² -4(4.9(-74) / 2(4.9)
t = (-10 ± √ 1550.4 ) / 9.81
t1 = (-10 + √ 1550.4 ) / 9.81 t2 = (-10 - √ 1550.4 ) / 9.81
t1 = (-10 ± 39.38 ) / 9.81 t2 = (-10 - 39.38) / 9.81
t1 = 2.99 s ≈ 3 s t2 = is negative then is wrong there are
no negative times.
b) Formula vf = vo + gt
vf = 10 + (9.81)(3)
vf = 10 + 29.43
vf = 39.43 m/s
The answer is 0.000824653J
You need to use the formula Mass * Velocity^2 over 2
<span> Beryllium has an exclusive </span>+2<span> oxidation state in all of its compounds</span>
The speed of the object increases
Explanation:
We can answer this question by applying the work-energy theorem, which states that the work done on an object is equal to the change in kinetic energy of the object. Mathematically:
where
W is the work done on the object
are the final and initial kinetic energy of the object, respectively
m is the mass of the object
v is its final speed
u is its initial speed
In this case, the force does a positive amount of work on the object, so
This also implies that
And so
And therefore
which means that the speed of the object increases.
Learn more about work:
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Answer:
82.7 m
Explanation:
u= 22m/s
a= 2.4 m/s^2.
t= 3.2 secs
Therefore the distance travelled can be calculated as follows
S= ut + 1/2at^2
= 22 × 3.2 + 1/2 × 2.4 × 3.2^2
= 70.4 + 1/2×24.58
= 70.4 + 12.29
= 82.7 m
Hence the distance travelled by the truck is 82.7 m
