Answer:
,Assume that the average volume of an adult human body is one-tenth
cubic meter (0.10 m) and that there are two billion (2.0 x 109)
adults in the world.
a. What would be the total volume of all the adults in the world?
b. Compute the length of one edge of a cubic container that has a
volume equal to the volume of all the adults in the world.
Answer:
(a) 2.85 m
(b) 16.5 m
(c) 21.7 m
(d) 22.7 m
Explanation:
Given:
v₀ₓ = 19 cos 71° m/s
v₀ᵧ = 19 sin 71° m/s
aₓ = 0 m/s²
aᵧ = -9.8 m/s²
(a) Find Δy when t = 3.5 s.
Δy = v₀ᵧ t + ½ aᵧ t²
Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²
Δy = 2.85 m
(b) Find Δy when vᵧ = 0 m/s.
vᵧ² = v₀ᵧ² + 2 aᵧ Δy
(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy
Δy = 16.5 m
(c) Find Δx when t = 3.5 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²
Δx = 21.7 m
(d) Find Δx when Δy = 0 m.
First, find t when Δy = 0 m.
Δy = v₀ᵧ t + ½ aᵧ t²
(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²
0 = t (18.0 − 4.9 t)
t = 3.67
Next, find Δx when t = 3.67 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²
Δx = 22.7 m
Answer:
option (a) 0.61 s
Explanation:
Given;
Time taken by the ball to reach the ground = 0.50 s
Let us first calculate the distance through which the ball falls on the ground
from the Newton's equation of motion, we have

where,
s is the distance
a is the acceleration
t is the time
here it is the case of free fall
thus, a = g = acceleration due to gravity
u = initial speed of the ball = 0
on substituting the values, we get

or
s = 1.225 m
Now,
when the elevator is moving up with speed of 1.0 m/s
the initial speed of the ball = -1.0 m/s (as the elevator is moving in upward direction)
thus , we have

or

or
4.9t^2 - t - 1.225 = 0
or
t = 0.612 s
hence, the correct answer is option (a) 0.61 s
Answer:
TEMPERATURE
Explanation:
When a wave is absorbed by a material medium, different phenomena occur, but the collisions with the other particles causes the energy to be transformed into internal energy in the atoms and molecules of the material, with TEMPERATURE measurements the increase in the internal energy of the material.