The object will continue moving in a straight line at constant speed.
Let the angle be Θ (theta)
Let the mass of the crate be m.
a) When the crate just begins to slip. At that moment the net force will be equal to zero and the static friction will be at the maximum vale.
Normal force (N) = mg CosΘ
μ (coefficient of static friction) = 0.29
Static friction = μN = μmg CosΘ
Now, along the ramp, the equation of net force will be:
mg SinΘ - μmg CosΘ = 0
mg SinΘ = μmg CosΘ
tan Θ = μ
tan Θ = 0.29
Θ = 16.17°
b) Let the acceleration be a.
Coefficient of kinetic friction = μ = 0.26
Now, the equation of net force will be:
mg sinΘ - μ mg CosΘ = ma
a = g SinΘ - μg CosΘ
Plugging the values
a = 9.8 × 0.278 - 0.26 × 9.8 × 0.96
a = 2.7244 - 2.44608
a = 0.278 m/s^2
Hence, the acceleration is 0.278 m/s^2
The appropriate answer is 4. The formation of Mid-Ocean ridges. There are two types of divergent margins: continental separation and oceanic floor spreading. Where the sea floor separates magma upwells and new oceanic crust is formed. Mid ocean ridges are home to hydrothermal vents. These support a unique ecosystem of marine organisms. The Mid- Atlantic Ridge where the African and Eurasian plates separate from the North and South American plates.
P = m*v
7.5 = m*15
m = 7.5/15 = 0.5 kg
We have to calculate the impulse of a hockey puck.
Imp = m * ( v 1 - v 2 ) = m * Δ v
v 1 = - 10 i m/s,
v 2 = ( 20 * cos 40° ) i + ( 20 * sin 40° ) j =
= ( 20 * 0.766 ) i + ( 20 * 0.64278 ) j = ( 15.32 i + 12.855 j ) m/s
Δ v = ( 15.32 i + 12.855 j ) - ( - 10 i ) =
= 15.32 i + 12.855 j + 10 i = 25.32 i + 12.855 j
| Δv | = √ ( 25.32² + 12.855²) = √806.35 = 28.4 m/s
Imp = 0.2 kg * 28.4 m/s = 5.68 N-s
Answer: D ) 5.68 N-s.