What are two things that determine our traits? If someone could answer asap, that would be awesome!
2 answers:
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Answer:
1.137278672 m/s
+5.9 cm or -5.9 cm
Explanation:
A = Amplitude = 6.25 cm
m = Mass of object = 225 g
k = Spring constant = 74.5 N/m
Maximum speed is given by
The maximum speed of the object is 1.137278672 m/s
Velocity is at any instant is given by
The locations are +5.9 cm or -5.9 cm
Answer:
Explanation:
The impulse equation is
Δp = FΔt, where Δp = final momentum - initial momentum, F is the Force exerted on an object, and Δt is the change in time. In this equation,the entire right side defines the impulse. In other words, FΔt is the impulse; thus the change in momentum an object experiences is due to its change in impulse and is directly proportional to it.
Therefore, once we find the change in momentum, that is the impulse the object experiences. Δp = final momentum - initial momentum, where
p = mv and p is momentum.
so
and
so
; therefore,
Δp = 25.0 - 17.5 = 7.5 which is the unit for momentum
A kangaroo will stay 0.699s in the air when he jumps.
<h3>What motion does the projectile make?</h3>A projectile is an object that is launched vertically into the air and moves while being pulled by gravity. A projectile is an object that is launched into the air horizontally and moves regardless of gravity.
<h3>How do you determine the rate of descent?</h3>G = -9.81 m/s2 is the acceleration of gravity close to the earth. Simply multiply the acceleration of gravity by the period since the object was released to determine the object's speed (or velocity). In other words, velocity is equal to -9.81 m/s2 * time, or V = gt.
<h3>According to the given information:</h3>From the conversion of energy =
t = √(2h/g)
= √((2 x 2.40) / 9.8)
=0.699 s
A kangaroo will stay 0.699s in the air when he jumps.
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