Answer:
The velocity of the ship relative to the earth V = 9.05 
Explanation:
The local ocean current is = 1.52 m/s
Direction
= 40°
Velocity component in X - direction
= 1.52
°
= 1.164 
Velocity component in Y - direction
= 8 + 1.52
°
= 8.97 
The velocity of the ship relative to the earth

Put the values of
and
we get,
⇒ 
⇒ V = 9.05 
This is the velocity of the ship relative to the earth.
Explanation:
Hey there!
Here,
Pascal is a unit of pressure.

Now, As per the formula the units are:
kg, m and s^2.
<em><u>Hope it helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>
This problem is a piece o' cake, IF you know the formulas for both kinetic energy and momentum. So here they are:
Kinetic energy = (1/2) · (mass) · (speed²)
Momentum = (mass) · (speed)
So, now ... We know that
==> mass = 15 kg, and
==> kinetic energy = 30 Joules
Take those pieces of info and pluggum into the formula for kinetic energy:
Kinetic energy = (1/2) · (mass) · (speed²)
30 Joules = (1/2) · (15 kg) · (speed²)
60 Joules = (15 kg) · (speed²)
4 m²/s² = speed²
Speed = 2 m/s
THAT's all you need ! Now you can find momentum:
Momentum = (mass) · (speed)
Momentum = (15 kg) · (2 m/s)
<em>Momentum = 30 kg·m/s</em>
<em>(Notice that in this problem, although their units are different, the magnitude of the KE is equal to the magnitude of the momentum. When I saw this, I wondered whether that's always true. So I did a little more work, and I found out that it isn't ... it's a coincidence that's true for this problem and some others, but it's usually not true.)</em>
Applying Newtons version of Kepler's third law or the orbital velocity law to the star orbiting 40000 light years from the center of the Milky Way Galaxy allows us to determine the mass of the Milky Way Galaxy that lies within 40000 light years in the galactic center.
<h3>
</h3><h3>What is orbital velocity law?</h3>
The orbital velocity law states that, the orbital velocity is directly proportional to the mass of the body for which it is being calculated and inversely proportional to the radius of the body. Earths orbital velocity near its surface is around 8km/sec if the air resistance is disregarded.
In space exploration, orbital velocity is a crucial topic. Space authorities heavily rely on it to comprehend how to launch satellites. It aids scientists in figuring out the velocities at which satellites must orbit a planet or other celestial body to prevent collapsing into it. The speed at which one body orbits the other body is known as the orbital velocity. The term "orbit" refers to an object's consistent circular motion around the Earth. The distance between the object and the earth's centre determines the orbit's velocity.
To know more about orbital velocity law, refer brainly.com/question/11353717
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Answer:
a) 
b) 
c) 
Explanation:
From the exercise we know the initial velocity of the projectile and its initial height

To find what time does it take to reach maximum height we need to find how high will it go
b) We can calculate its initial height using the following formula
Knowing that its velocity is zero at its maximum height



So, the projectile goes 1024 ft high
a) From the equation of height we calculate how long does it take to reach maximum point



Solving the quadratic equation



So, the projectile reach maximum point at t=2s
c) We can calculate the final velocity by using the following formula:


Since the projectile is going down the velocity at the instant it reaches the ground is:
