Answer:
C) 50 m/s
Explanation:
With the given information we can calculate the acceleration using the force and mass of the box.
Newton's 2nd Law: F = ma
- 5 N = 1 kg * a
- a = 5 m/s²
List out known variables:
- v₀ = 0 m/s
- a = 5 m/s²
- v = ?
- Δx = 250 m
Looking at the constant acceleration kinematic equations, we see that this one contains all four variables:
Substitute known values into the equation and solve for v.
- v² = (0)² + 2(5)(250)
- v² = 2500
- v = 50 m/s
The final velocity of the box is C) 50 m/s.
Answer:
The options are not shown, so let's derive the relationship.
For an object that is at a height H above the ground, and is not moving, the potential energy will be:
U = m*g*H
where m is the mass of the object, and g is the gravitational acceleration.
Now, the kinetic energy of an object can be written as:
K = (1/2)*m*v^2
where v is the velocity.
Now, when we drop the object, the potential energy begins to transform into kinetic energy, and by the conservation of the energy, by the moment that H is equal to zero (So the potential energy is zero) all the initial potential energy must now be converted into kinetic energy.
Uinitial = Kfinal.
m*g*H = (1/2)*m*v^2
v^2 = 2*g*H
v = √(2*g*H)
So we expressed the final velocity (the velocity at which the object impacts the ground) in terms of the height, H.
Answer:
3rd order polynomial
Explanation:
Given that the increase in the order of the polynomial the error between the curve fit and measured data will decreases hence :
The polynomial order that is best to use is the 3rd order polynomial, this is because using a 3rd order polynomial will produce a less variance and a low Bias
Answer:
<em>The depth will be equal to</em> <em>6141.96 m</em>
<em></em>
Explanation:
pressure on the submarine 
= 62 MPa = 62 x 10^6 Pa
we also know that = ρgh
where
ρ is the density of sea water = 1029 kg/m^3
g is acceleration due to gravity = 9.81 m/s^2
h is the depth below the water that this pressure acts
substituting values, we have
= 1029 x 9.81 x h = 10094.49h
The gauge pressure within the submarine 
= 101 kPa = 101000 Pa
this gauge pressure is balanced by the atmospheric pressure (proportional to 101325 Pa) that acts on the surface of the sea, so it cancels out.
Equating the pressure , we have
62 x 10^6 = 10094.49h
depth h = <em>6141.96 m</em>
Answer:
https://physicsabout.com/acceleration-and-velcoity/
