The answer would be physical and chemical
Ans: a = 2.50 m/s^2
Explanation:
First convert the mass in its standard unit i.e. kilogram(kg):
2250 lbs = 1020.583kg
Next use Newton's Second law:
F = ma
Where F = 2552N
m = 1020.583kg
=> a = (2552/1020.583)
a = 2.50 m/s^2
Answer:
Explanation:
When the apple is held submerged in water , it experiences a buoyant force due to which it floats in water . One has to apply downward force to keep it submerged. The lower the buoyant force , lower the force needed to submerge it in water.
When apple is held at much deeper point , it experience greater pressure due to column of water around it . So its size or its volume decreases . But its weight remains the same . Due to less volume , buoyant force also decreases ( buoyant force is equal to weight of displaced volume of water. )
Due to buoyant force becoming less , force needed on apple in downward direction will also be less.
Answer:
20 N/m
Explanation:
From the question,
The ball-point pen obays hook's law.
From hook's law,
F = ke............................ Equation 1
Where F = Force, k = spring constant, e = compression.
Make k the subject of the equation
k = F/e........................ Equation 2
Given: F = 0.1 N, e = 0.005 m.
Substitute these values into equation 2
k = 0.1/0.005
k = 20 N/m.
Hence the spring constant of the tiny spring is 20 N/m
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
