Answer:
True
Explanation:
Pre-questioning may help a reader focus on information s/he hopes to find in the reading selection.
Answer:
you must throw 3 snowballs
Explanation:
We can solve this exercise using the concepts of conservation of the moment, let's define the system as formed by the refrigerator and all the snowballs. Let's write the moment
Initial. Before bumping that refrigerator
p₀ = n m v₀
Where n is the snowball number
Final. When the refrigerator moves
pf = (n m + M) v
The moment is preserved because the forces during the crash are internal
n m v₀ = (n m + M) v
n m (v₀ - v) = M v
n = M/m v/(vo-v)
Let's look for the initial velocity of the balls, suppose the person throws them with the maximum force if it slides in the snow (F = 100N), let's use the second law and Newton
F = m a
a = F / m
The distance the ball travels from zero speed to maximum speed is the extension of the arm (x = 1 m), let's look kinematically for the speed of the balls when leaving the arm
v₁² = v₀² + 2 a x
v₁² = 0+ 2 (100/1) 1
v₁ = 14.14 m / s
This is the initial speed for the crash
v₀ = v = 14.14 m / s
Let's calculate
n = M/m v/ (v₀-v)
n = 10/1 3 / (14.14 -3)
n = 2.7 balls
you must throw 3 snowballs
here in the given situation if monkey starts free fall at the same instant when veterinarian shoots towards it then we know that vertical component of motion of monkey and the dart will be same as under gravity
so here the dart will always hit the monkey because they both moves under same acceleration
so here for the angle we can use
now we have
H = 3 m
L = 87.5 m
now we will have
so angle will be 1.96 degree above the ground
Answer:
(a) f= 622.79 Hz
(b) f= 578.82 Hz
Explanation:
Given Data
Frequency= 600 Hz
Distance=1.0 m
n=120 rpm
Temperature =20 degree
Before solve this problem we need to find The sound generator moves on a circular with tangential velocity
So
Speed of sound is given by
c = √(γ·R·T/M)
............in an ideal gas
where γ heat capacity ratio
R universal gas constant
T absolute temperature
M molar mass
The speed of sound at 20°C is
c = √(1.40 ×8.314472J/molK ×293.15K / 0.0289645kg/mol)
c= 343.24m/s
The sound moves on a circular with tangential velocity
vt = ω·r.................where
ω=2·π·n
vt= 2·π·n·r
vt= 2·π · 120min⁻¹ · 1m
vt= 753.6 m/min
convert m/min to m/sec
vt= 12.56 m/s
Part A
For maximum frequency is observed
v = vt
f = f₀/(1 - vt/c )
f= 600Hz / (1 - (12.56m/s / 343.24m/s) )
f= 622.789 Hz
Part B
For minimum frequency is observed
v = -vt
f = f₀/(1 + vt/c )
f= 600Hz / (1 + (12.56m/s / 343.24m/s) )
f= 578.82 Hz
