The particle's acceleration is 5.1 m/s²
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What is Acceleration ?</h3>
Acceleration can be defined as the rate at which velocity is changing. It is a vector quantity and it is measured in m/s²
Given that a particle is moving along a straight line with constant acceleration has a velocity of 2.35 m/s at t=3.42 s, and a velocity of -8.72 m/s at t=5.59s
The given parameters are;
Acceleration a = ΔV ÷ ΔT
a = (2.35 + 8.72) / (5.59 - 3.42)
a = 11.07 / 2.17
a = 5.1 m/s²
Therefore, the particle's acceleration is 5.1 m/s²
Learn more about Acceleration here: brainly.com/question/9069726
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Given:
Dy= 20 m
Vi = 5.0 m/s horizontally
A=9.81 m/s^2
Find:
Horizontal displacement
Solution:
D=ViT+(1/2)AT^2
Dy=(1/2)AT^2
T^2=Dy/(1/2)A
T=sqrt(Dy/(1/2)A)
T=sqrt(20/4.905)
T=2.0s
Dx=ViT
Dx=(5.0)(2.0)
Dx=10. meters
Gravitational Potential Energy GPE = mg∆h. 12. A 5.0 kg mass is initially sitting on the floor when it is lifted onto a table 1.15 meters high at.
First we need to find the speed of the dolphin sound wave in the water. We can use the following relationship between frequency and wavelength of a wave:

where
v is the wave speed

its wavelength
f its frequency
Using

and

, we get

We know that the dolphin sound wave takes t=0.42 s to travel to the tuna and back to the dolphin. If we call L the distance between the tuna and the dolphin, the sound wave covers a distance of S=2 L in a time t=0.42 s, so we can write the basic relationship between space, time and velocity for a uniform motion as:

and since we know both v and t, we can find the distance L between the dolphin and the tuna:
Answer: 10.3m/s
Explanation:
In theory and for a constant velocity the physics expression states that:
Eq(1): distance = velocity times time <=> d = v*t for v=constant.
If we solve Eq (1) for the velocity (v) we obtain:
Eq(2): velocity = distance divided by time <=> v = d/t
Substituting the known values for t=15s and d=155m we get:
v = 155 / 15 <=> v = 10.3