Answer:
Explanation:
I'm not sure you can do this without just a bit more information. I can tell you what the mass of the water is when the rocks are removed. When we know that, we know the volume of the water that was displaced. whether or not this is enough information to determine the volume of the box, I'm not sure.
400 grams raises the box 1 cm.
The density of water = 1 gm / cm^3
400 grams of water = 400 mL or 400 cm^3
The volume of the displaced water = 400 cm^3
The volume a slice from the square prism is B*h
B = 400 cm^2
h = 1 cm
If the base is 400 cm^2 then each side is
s^2 = 400
sqrt(s^2)= sqrt(400)
s = 20
The volume of the box is 20^3 = 8000 cm^3
Answer:
a. A = 0.735 m
b. T = 0.73 s
c. ΔE = 120 J decrease
d. The missing energy has turned into interned energy in the completely inelastic collision
Explanation:
a.
4 kg * 10 m /s + 6 kg * 0 m/s = 10 kg* vmax
vmax = 4.0 m/s
¹/₂ * m * v²max = ¹/₂ * k * A²
m * v² = k * A² ⇒ 10 kg * 4 m/s = 100 N/m * A²
A = √1.6 m ² = 1.26 m
At = 2.0 m - 1.26 m = 0.735 m
b.
T = 2π * √m / k ⇒ T = 2π * √4.0 kg / 100 N/m = 1.26 s
T = 2π *√ 10 / 100 *s² = 1.99 s
T = 1.99 s -1.26 s = 0.73 s
c.
E = ¹/₂ * m * v²max =
E₁ = ¹/₂ * 4.0 kg * 10² m/s = 200 J
E₂ = ¹/₂ * 10 * 4² = 80 J
200 J - 80 J = 120 J decrease
d.
The missing energy has turned into interned energy in the completely inelastic collision
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Answer and Explanation:
(a) The attached image below shows the diagram of a CRT in the helmholtz coils and well labelled with details
(b). The electron will follow a circular path which travels along under constant magnetic field

where m = mass of electron, V = velocity of electron, q = charge of the electron and B = magnetic field strength
The work done on the puck is 96 J
Explanation:
According to the work-energy theorem, the work done on the hockey puck is equal to the change in kinetic energy of the puck.
Mathematically:
where
is the final kinetic energy of the puck, with
m = 2 kg being the mass of the puck
v = 10 m/s is the final speed
is the initial kinetic energy of the puck, with
u = 2 m/s being the initial speed of the puck
Substituting numbers into the equation, we find the work done by the player on the puck:
Learn more about work and kinetic energy:
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