Answer:
(a). The initial velocity is 28.58m/s
(b). The speed when touching the ground is 33.3m/s.
Explanation:
The equations governing the position of the projectile are


where
is the initial velocity.
(a).
When the projectile hits the 50m mark,
; therefore,

solving for
we get:

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

which gives

(b).
The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

the vertical component of the velocity is

which gives a speed
of


Explanation:
It is given that,
Mass of the ball, m = 1 lb
Length of the string, l = r = 2 ft
Speed of motion, v = 10 ft/s
(a) The net tension in the string when the ball is at the top of the circle is given by :



F = 18 N
(b) The net tension in the string when the ball is at the bottom of the circle is given by :



F = 82 N
(c) Let h is the height where the ball at certain time from the top. So,


Since, 

Hence, this is the required solution.
D. Atoms.
Explanation:
All the matter is made of elementary particles called "atoms".
Further, an atom is made of electrons, protons and neutrons. The electrons & protons are again made of the fundamental sub-particles, electrons (leptons) and the protons(quarks).
The classification of particles is shown in the figure attached
Answer:
a) the values of the angle α is 45.5°
b) the required magnitude of the vertical force, F is 41 lb
Explanation:
Applying the free equilibrium equation along x-direction
from the diagram
we say
∑Fₓ = 0
Pcosα - 425cos30° = 0
525cosα - 368.06 = 0
cosα = 368.06/525
cosα = 0.701
α = cos⁻¹ (0.701)
α = 45.5°
Also Applying the force equation of motion along y-direction
∑Fₓ = ma
Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)
525sin45.5° + F + 212.5 - 600 = 27.95
374.46 + F + 212.5 - 600 = 27.95
F - 13.04 = 27.95
F = 27.95 + 13.04
F = 40.99 ≈ 41 lb
Answer:
The energy of the system is 15 J.
Explanation:
Given that,
Energy E = 2.5 J
Amplitude = 10 cm
We need to calculate the spring constant
Using formula of mechanical energy of the system

Put the value into the formula



If the block is replaced by a block with twice the mass of the original block
Amplitude = 6 cm
We need to calculate the energy
Using formula of mechanical energy

Put the value into the formula


Hence, The energy of the system is 15 J.