Answer:
In this case, the horizontal velocity of the rocket starts from the acceleration, so if its velocity drops to zero,
Explanation:
When a satellite is in orbit the most important force is the docking of gravity with the Earth
F = m a
where the acceleration is centripetal and F is the force of universal attraction
centripetal acceleration is
a = v² / r
F = m v² / r
In this case, the horizontal velocity of the rocket starts from the acceleration, so if its velocity drops to zero, the force also drops to serious and the satellite steels to Earth.
The speed of the satellite is provides the speed, by local for smaller speeds in satellite, it descends in its orbits and when the speed is amate you have the energy to stop an orb to go to a higher orbit.
Answer:
A) hydrostatic force on top of cube = 882.9N
B) hydrostatic force on sides of cube = 0N
Explanation:
Detailed explanation and calculation is shown in the image below
Answer:
Nora's boat is moving towards Sam at 5 km/hr
Explanation:
The question says that Nora is few meters behind Nancy and is still with respect to her that means the relative velocity between Nora and Nancy is zero
Vrel = 5 - Vnora= 0 ⇒ Vnora = 5km/hr
Pictorially we can represent the given condition as:
Nora-------few meters------Nancy ----------------- Sam
5km/hr 5km/hr →
Hence, Nora's boat is moving towards Sam at 5km/hr.
Answer:
1838216 J
Explanation:
95 km/h = 26.39 m/s
40 km/h = 11.11 m/s
Initial kinetic energy
= .5 x 1600 x(26.39)²
= 557145.67 J
Final kinetic energy
= .5 x 1600 x ( 11.11)²
= 98745.68 J
Loss of kinetic energy
= 458400 J
Loss of potential energy
= mg x loss of height
= 1600 x 9.8 x 340 sin 15
= 1379816 J
Sum of Loss of potential energy and Loss of kinetic energy
= 1379816 + 458400
= 1838216 J
This is the work done by the friction . So this is heat generated.
To solve the problem it is necessary to use Newton's second law and statistical equilibrium equations.
According to Newton's second law we have to
where,
m= mass
g = gravitational acceleration
For the balance to break, there must be a mass M located at the right end.
We will define the mass m as the mass of the body, located in an equidistant center of the corners equal to 4m.
In this way, applying the static equilibrium equations, we have to sum up torques at point B,
Regarding the forces we have,
Re-arrange to find M,
Therefore the maximum additional mass you could place on the right hand end of the plank and have the plank still be at rest is 16.67Kg
