∑F = ma
a = ∑F/m
a = 35 N / 5.4 kg
a = 6.5 m/s²
Answer:
Case A
Explanation:
given,
size of bacteria = 1 mm x 1 mm
velocity = 20 mm/s
size of the swimmer = 1.5 m x 1.5 m
velocity of swimmer = 3 m/s
Viscous force
for the bacteria
for the swimmer
from the above force calculation
In case B inertial force that represent mass is more than the inertial force in case of bacteria.
Viscous force is dominant in case of bacteria.
So, In Case A viscous force will be dominant.
Answer: 3.75 m
Explanation:
5 squirts in 1 second
So, 1 squirt in 1/5 second which is 0.2 second.
The difference in timing of two consecutive squirt is 0.2 second, so
time (t) = 0.2 s.
speed (s) = 15 m/s
Distance of separation (d) = ?
Now, formula for distance is
d = s × t
d = 15 × 0.2
d = 3.75 m
The second one if it’s on edge
First
let us imagine the projectile launched at initial velocity V and at angle
θ relative to the horizontal. (ignore wind resistance)
Vertical component y:
The
initial vertical velocity is given as Vsinθ
The moment the projectile reaches the maximum
height of h, the vertical velocity
will be 0, therefore the time t taken to attain this maximum height is:
h = Vsinθ - gt
0 = Vsinθ - gt
t = (Vsinθ)/g
where
g is acceleration due to gravity
Horizontal component x:
The initial horizontal velocity is given as Vcosθ. However unlike
the vertical component, this horizontal velocity remains constant because this is unaffected by gravity. The time to travel the
horizontal distance D is twice the value of t times the horizontal velocity.
D = Vcosθ*[(2Vsinθ)/g]
D = (2V²sinθ cosθ)/g
D = (V²sin2θ)/g
In order for D (horizontal distance) to be
maximum, dD/dθ = 0
That is,
2V^2 cos2θ / g = 0
And since 2V^2/g must not be equal to zero, therefore cos(2θ) = 0
This is true when 2θ = π/2 or θ = π/4
Therefore it is now<span> shown that the maximum horizontal travelled is attained when
the launch angle is π/4 radians, or 45°.</span>
