Answer:
The puck moves a vertical height of 2.6 cm before stopping
Explanation:
As the puck is accelerated by the spring, the kinetic energy of the puck equals the elastic potential energy of the spring.
So, 1/2mv² = 1/2kx² where m = mass of puck = 39.2 g = 0.0392 g, v = velocity of puck, k = spring constant = 59 N/m and x = compression of spring = 1.3 cm = 0.013 cm.
Now, since the puck has an initial velocity, v before it slides up the inclined surface, its loss in kinetic energy equals its gain in potential energy before it stops. So
1/2mv² = mgh where h = vertical height puck moves and g = acceleration due to gravity = 9.8 m/s².
Substituting the kinetic energy of the puck for the potential energy of the spring, we have
1/2kx² = mgh
h = kx²/2mg
= 59 N/m × (0.013 m)²/(0.0392 kg × 9.8 m/s²)
= 0.009971 Nm/0.38416 N
= 0.0259 m
= 2.59 cm
≅ 2.6 cm
So the puck moves a vertical height of 2.6 cm before stopping
Answer:
Night vision is when you can see in the dark or at night like owls
Explanation:
Answer:
a) -5.40 rad/s
b) -2.842 rad/s²
Explanation:
The direction is important in dealing with such questions. Clockwise is considered negative and counterclockwise is considered positive
a) Δω = final angular velocity - initial angular velocity
= -2.70 rad/s - 2.70 rad/s
= -5.40 rad/s
b) ∝ = Δω/Δt = (-5.40 rad/s)/1.90s = -2.842 rad/s²
Answer:
South = 1.5m
West =4.2m
Explanation:
Kindly see attached a rough draft of the situation
Step one
Given data
From the sketch the direction of the player is along the resultant of the triangle, corresponding to the Hypotenuse
Step two:
Hence for an opponent to tackle him towards the south, he must be at
sin θ= opp/hyp
sin 20=x/4.5
x=sin 20*4.5
x=0.342*4.5
x= 1.5m
Also, for an opponent to tackle him towards the south, he must be at
cos θ= adj/hyp
cos 20=y/4.5
y=cos 20*4.5
y=0.93*4.5
y= 4.2m
