Answer:

Explanation:
v = Velocity of Ferdinand = 5 m/s
= Angle of jump
T = Time taken = 0.6 s
R = Distance between lily pads = 2.4 m
Horizontal range is given by

The angle at which Ferdinand make each of his jumps is 
Answer:
Neon (Ne), chemical element, inert gas of Group 18 (noble gases) of the periodic table, used in electric signs and fluorescent lamps. Colourless, odourless, tasteless, and lighter than air, neon gas occurs in minute quantities in Earth's atmosphere and trapped within the rocks of Earth's crust.
Explanation:
I am smart
Answer:
Explanation:
Basically the star slowly burns its hydrogen into Helium. Depending on the mass, the star will have a turbulent core where the Helium will be fully mixed or a radiative core where the helium will settle at the centre (remember it's heavier than Hydrogen). The second case is what happens in the Sun.
Answer:
2.7
Explanation:
The following data were obtained from the question:
Mass (m) of box = 100 Kg
Length (L) of ramp = 4 m
Height (H) of ramp = 1.5 m
Mechanical advantage (MA) of ramp =?
Mechanical advantage of a ramp is simply defined as the ratio of the length of the ramp to the height of the ramp. Mathematically, it is given by:
Mechanical Advantage = Lenght / height
MA= L/H
With the above formula, we can obtain the mechanical advantage of the ramp as follow:
Length (L) of ramp = 4 m
Height (H) of ramp = 1.5 m
Mechanical advantage (MA) of ramp =?
MA = 4/1.5
MA = 2.7
Therefore, the mechanical advantage of the ramp is 2.7
Answer:
Part a)

Part b)

Explanation:
Part a)
Electric field due to large sheet is given as


now the electric field is given as


Now acceleration of an electron due to this electric field is given as



Now work done on the electron due to this electric field



So work done is given as



Part b)
Now we know that work done by all forces = change in kinetic energy of the electron
so we will have


