Answer:
If the system consists of the block only, the work done by the gravity is negative.
If the system consists of the block and the earth the work done by the gravity is zero.
Explanation:
If the system consists of the block only, then the system experiences two external forces: one exerted by the hand that lifts the block vertically upward and other exerted by the earth (gravity), which is opposed to the movement of the system, so the work done by gravity is negative.
On the other hand, if the system consists of the block and the earth, then only exists a external force which is the exerted by the hand. So, the force exerted by gravity is zero.
Answer:
Check explanation
Explanation:
Gold - Au (Aurum)
Mercury - Hg (Hydrargyrum)
Copper - Cu (Cuprum)
Iron - Fe (Ferrum)
Lead - Pb (Plumbum)
These elements in the periodic table are some of the elements represented by letters not in line with their names.
This is because, these elements were known in ancient times and therefore, they are represented by letters from their ancient names.
(a) The equation for the work done in stretching the spring from x1 to x2 is ¹/₂K₂Δx².
(b) The work done, in stretching the spring from x1 to x2 is 11.25 J.
(c) The work, necessary to stretch the spring from x = 0 to x3 is 64.28 J.
<h3>
Work done in the spring</h3>
The work done in stretching the spring is calculated as follows;
W = ¹/₂kx²
W(1 to 2) = ¹/₂K₂Δx²
W(1 to 2) = ¹/₂(250)(0.65 - 0.35)²
W(1 to 2) = 11.25 J
W(0 to 3) = ¹/₂k₁x₁² + ¹/₂k₂x₂² + ¹/₂F₃x₃
W(0 to 3) = ¹/₂(660)(0.35)² + ¹/₂(250)(0.65 - 0.35)² + ¹/₂(105)(0.89 - 0.65)
W(0 to 3) = 64.28 J
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Responder:
20πrads ^ -1; 24πrads ^ -1; 0,1 seg; 10 Hz
Explicación:
Dado lo siguiente:
Radio (r) del círculo = 120 cm
600 revoluciones por minuto en radianes por segundo
(600 / min) * (2π rad / 1 rev) * (1min / 60seg)
(1200πrad / 60sec) = 20π rad ^ -1
Velocidad angular (w) = 20πrads ^ -1
Velocidad lineal = radio (r) * velocidad angular (w)
Velocidad lineal = (120/100) * 20πrad
Velocidad lineal = 1.2 * 20πrads ^ -1 = 24πrads ^ -1
C.) Período (T):
T = 2π / w = 2π / 20π = 0.1 seg
D.) Frecuencia (f):
f = 1 / T = 1 / 0.1
1 / 0,1 = 10 Hz
Answer:
refractive index of the unknown material is 1.33.
Explanation:
μ₁ = 1.21
incidence angle (i) = 41.9°
refraction angle (r) = 37.3°
Let us assume μ be the refractive index of the unknown material
according to snell's law of refraction.
μ₁ sin i = μ₂ sin r
1.21 × sin 41.9° = μ × sin 37.3°
μ = 1.33
hence the refractive index of the unknown material comes out top be 1.33