Explanation:
"Static friction is a force that keeps an object at rest. It must be overcome to start moving the object."
(556 x 0.68) = static friction of 378.08N. before movement occurs.
The forces (a) and (b) will not move it.
Each will incur a frictional force preventing movement equal to itself, = 222N. and 334N. respectively.
Forces (c) and (d) will move it, and accelerate it.
Forces (c) and (d) will both encounter friction of (556 x 0.56) = 311.36N. when the cabinet is moving.
Answer:
Explanation:
GIVEN
diameter = 15 fm =
m
we use here energy conservation
there will be some initial kinetic energy but after collision kinetic energy will zero
on solving these equations we get kinetic energy initial
J ..............(i)
That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2 e
and gains kinetic energy K =e∆V ..........(ii)
by accelerating through a potential difference ∆V
Thus the alpha particle will
just reach the 
nucleus after being accelerated through a potential difference ∆V
equating (i) and second equation we get
e∆V = 35.33 Me V
Answer:
289.1kg
Explanation:
because W=M*G, and M=W/G, so 2891N/10m/s²=289.1kg
Answer:
The back end of the vessel will pass the pier at 4.83 m/s
Explanation:
This is purely a kinetics question (assuming we're ignoring drag and other forces) so the weight of the boat doesn't matter. We're given:
Δd = 315.5 m
vi = 2.10 m/s
a = 0.03 m/s^2
vf = ?
The kinetics equation that incorporates all these variables is:
vf^2 = vi^2 + 2aΔd
vf = √(2.1^2 + 2(0.03)(315.5))
vf = 4.83 m/s
