Hi
The answer to this question is B. Reaction
Density-Dependent:
1<span><span><span><span>. </span>competition.</span><span>
<span>2. </span>overcrowding.</span><span>
3<span>. </span>predators.</span></span><span>
(These are a few from a test I took, hopefully they help you a bit >.<)</span></span>
Answer:
Explanation:
a ) Let let the frictional force needed be F
Work done by frictional force = kinetic energy of car
F x 107 = 1/2 x 1400 x 35²
F = 8014 N
b )
maximum possible static friction
= μ mg
where μ is coefficient of static friction
= .5 x 1400 x 9.8
= 6860 N
c )
work done by friction for μ = .4
= .4 x 1400 x 9.8 x 107
= 587216 J
Initial Kinetic energy
= .5 x 1400 x 35 x 35
= 857500 J
Kinetic energy at the at of collision
= 857500 - 587216
= 270284 J
So , if v be the velocity at the time of collision
1/2 mv² = 270284
v = 19.65 m /s
d ) centripetal force required
= mv₀² / d which will be provided by frictional force
= (1400 x 35 x 35) / 107
= 16028 N
Maximum frictional force possible
= μmg
= .5 x 1400 x 9.8
= 6860 N
So this is not possible.
Answer:
a= 23.65 ft/s²
Explanation:
given
r= 14.34m
ω=3.65rad/s
Ф=Ф₀ + ωt
t = Ф - Ф₀/ω
= (98-0)×
/3.65
98°= 1.71042 rad
1.7104/3.65
t= 0.47 s
r₁(not given)
assuming r₁ =20 in
r₁ = r₀ + ut(uniform motion)
u = r₁ - r₀/t
r₀ = 14.34 in= 1.195 ft
r₁ = 20 in = 1.67 ft
= (1.667 - 1.195)/0.47
0.472/0.47
u= 1.00ft/s
acceleration at collar p
a=rω²
= 1.67 × 3.65²
a = 22.25ft/s²
acceleration of collar p related to the rod = 0
coriolis acceleration = 2ωu
= 2× 3.65×1 = 7.3 ft/s²
acceleration of collar p
= 22.5j + 0 + 7.3i
√(22.5² + 7.3²)
the magnitude of the acceleration of the collar P just as it reaches B in ft/s²
a= 23.65 ft/s²
Answer:
<h2>Radiation</h2>
Explanation:
<h3>Greetings !</h3>
The Sun reaches us by propagating through the vacuum of space. Sunlight reaches the Earth in about 8 minutes and 20 seconds. When this energy reaches the Earth's atmosphere, both conduction and convection play key roles to scatter it throughout the planet.
