Answer:
a)
Y0 = 0 m
Vy0 = 15 m/s
ay = -9.81 m/s^2
b) 7.71 m
c) 3.06 s
Explanation:
The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards
Y(0) = 0 m
Vy(0) = 15 m/s
ay = -9.81 m/s^2 (negative because it points down)
Since acceleration is constant we can use the equation for uniformly accelerated movement:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
To find the highest point we do the first time derivative (this is the speed:
V(t) = Vy0 + a * t
We equate this to zero
0 = Vy0 + a * t
0 = 15 - 9.81 * t
15 = 9.81 * t
t = 0.654 s
At this time it will have a height of:
Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m
The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.
0 = Y0 + Vy0 * t + 1/2 * a * t^2
0 = 0 + 15 * t - 1/2 * 9.81 t^2
0 = 15 * t - 4.9 * t^2
0 = t * (15 - 4.9 * t)
t1 = 0 This is the moment it jumped into the air
0 = 15 - 4.9 * t2
15 = 4.9 * t2
t2 = 3.06 s This is the moment when it falls again.
3.06 - 0 = 3.06 s
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:
a =3.33 m/s²
Explanation:
given,
initial speed of Plane, u = 0 m/s
final speed of plane, v = 60 m/s
time of the acceleration, t = 18 s
average acceleration of the plane, a = ?
average acceleration is equal to change in velocity per unit time.



a =3.33 m/s²
Hence, average acceleration of the plane is equal to a =3.33 m/s²
Answer:
The total amount of energy that would have been released if the asteroid hit earth = The kinetic energy of the asteroid = 1.29 × 10¹⁵ J = 1.29 PetaJoules = 1.29 PJ
1 PJ = 10¹⁵ J
Explanation:
Kinetic energy = mv²/2
velocity of the asteroid is given as 7.8 km/s = 7800 m/s
To obtain the mass, we get it from the specific gravity and diameter information given.
Density = specific gravity × 1000 = 3 × 1000 = 3000 kg/m³
But density = mass/volume
So, mass = density × volume.
Taking the informed assumption that the asteroid is a sphere,
Volume = 4πr³/3
Diameter = 30 m, r = D/2 = 15 m
Volume = 4π(15)³/3 = 14137.2 m³
Mass of the asteroid = density × volume = 3000 × 14137.2 = 42411501 kg = 4.24 × 10⁷ kg
Kinetic energy of the asteroid = mv²/2 = (4.24 × 10⁷)(7800²)/2 = 1.29 × 10¹⁵ J