Answer:
Solution:
we have given the equation of motion is x(t)=8sint [where t in seconds and x in centimeter]
Position, velocity and acceleration are all based on the equation of motion.
The equation represents the position. The first derivative gives the velocity and the 2nd derivative gives the acceleration.
x(t)=8sint
x'(t)=8cost
x"(t)=-8sint
now at time t=2pi/3,
position, x(t)=8sin(2pi/3)=4*squart(3)cm.
velocity, x'(t)=8cos(2pi/3)==4cm/s
acceleration, x"(t)==8sin(2pi/3)=-4cm/s^2
so at present the direction is in y-axis.
Decompose the forces acting on the block into components that are parallel and perpendicular to the ramp. (See attached free body diagram. Forces are not drawn to scale)
• The net force in the parallel direction is
∑ <em>F</em> (para) = -<em>mg</em> sin(21°) - <em>f</em> = <em>ma</em>
• The net force in the perpendicular direction is
∑ <em>F</em> (perp) = <em>n</em> - <em>mg</em> cos(21°) = 0
Solving the second equation for <em>n</em> gives
<em>n</em> = <em>mg</em> cos(21°)
<em>n</em> = (0.200 kg) (9.80 m/s²) cos(21°)
<em>n</em> ≈ 1.83 N
Then the magnitude of friction is
<em>f</em> = <em>µn</em>
<em>f</em> = 0.25 (1.83 N)
<em>f</em> ≈ 0.457 N
Solve for the acceleration <em>a</em> :
-<em>mg</em> sin(21°) - <em>f</em> = <em>ma</em>
<em>a</em> = (-0.457N - (0.200 kg) (9.80 m/s²) sin(21°))/(0.200 kg)
<em>a</em> ≈ -5.80 m/s²
so the block is decelerating with magnitude
<em>a</em> = 5.80 m/s²
down the ramp.
Answer:
The remaining air between the cup and the rough surface exert pressure on the inner wall of the cup, due to this it does not stick on the rough surface.
Explanation:
HOPE it helps:))))
Answer:
malesef herşey ingilizce yazıyor hiçbirşey anlayamadım
Explanation:
heeeeeem ben bunlardan anlamam giiiit başkasına sooooor bu arada en iyi seç benü bikere bari seçte bi en iyi cvp olsun