Answer:
a) ω = 9.86 rad/s
b) ac = 194. 4 m/s²
c) minimum coefficient of static friction, µs = 19.8
Explanation:
a) angular speed, ω = 2πf, where f is frequency of revolution
1 rps = 6.283 rad/s, π = 3.142
ω = 2 * 3.14 * 0.25 * 6.28
ω = 9.86 rad/s
b) centripetal acceleration, a = rω²
where r is radius in meters; r = 200 cm or 2 m
a = 2 * 9.86²
a = 194. 4 m/s²
c) µs = frictional force/ normal force
frictional force = centripetal force = ma; where a is centripetal acceleration
normal force = mg; where g = 9.8 m/s²
µs = ma/mg = a/g
µs = 194.4 ms⁻²/9.8 ms⁻²
c) minimum coefficient of static friction, µs = 19.8
Answer:
speed of puck acc. to the radar gun = 138 km/h
speed of player = 15 km/h
since the player is in motion when he shoots, the speed of the puck will be the sum of the speed of the player and the speed at which he shot. so,
speed of puck = speed of player + speed of puck acc. to player
138 = 15 + speed of puck acc. to player
speed of puck acc. to player = 138 -15
speed of puck acc. to player = 123 km/h
Brainly this answer if you think it deserves it
A
method of procedure that has characterized natural science since the
17th century, consisting in systematic observation, measurement, and
experiment, and the formulation, testing, and modification of
hypotheses
V = 340 m/s
f = 256 Hz
lambda (wavelength)
v = f*lambda
340 = 256 * lambda
340/256 = lambda
lambda = 1.328 m
The y-component of the stone's velocity when it is 8 m below the hand is 14.86 m / s
v² = u² + 2 a s
s = Displacement
u = Initial velocity
a = Acceleration
u = 8 m / s
s = 8 m
v² = 8² + 2 * 9.8 * 8
v² = 64 + 156.8
v = √ 220.8
v = 14.86 m / s
The equation used to solve the problem is an equation of motion. These equations are designed to locate an object in motion using components such as velocity, displacement, acceleration and time.
Therefore, the y-component of the stone's velocity is 14.86 m / s
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