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The answer is C) the density of the rock
Density of rock is the dependent variable, because it depends on the temperature. The temperature can’t be the dependent variable because ,the density of a rock does not have magical powers that can change temperature of a room. However changing the temperature of the room ,will change the density of the rock. Hope this helps !
Answer:
W₃ = 3310.49 J
, W3 = 3310.49 J
Explanation:
We can solve this exercise in parts, the first with acceleration, the second with constant speed and the third with deceleration. Therefore it is work we calculate it in these three sections
We start with the part with acceleration, the distance traveled is y = 5.90 m and the final speed is v = 2.30 m / s. Let's calculate the acceleration with kinematics
v2 = v₀² + 2 a₁ y
as they rest part of the rest the ricial speed is zero
v² = 2 a₁ y
a₁ = v² / 2y
a₁ = 2.3² / (2 5.90)
a₁ = 0.448 m / s²
with this acceleration we can calculate the applied force, using Newton's second law
F -W = m a₁
F = m a₁ + m g
F = m (a₁ + g)
F = 69 (0.448 + 9.8)
F = 707.1 N
Work is defined by
W₁ = F.y = F and cos tea
As the force lifts the man, this and the displacement are parallel, therefore the angle is zero
W₁ = 707.1 5.9
W₁ = 4171.89 J W3 = 3310.49 J
Let's calculate for the second part
the speed is constant, therefore they relate it to zero
F - W = 0
F = W
F = m g
F = 60 9.8
F = 588 A
the job is
W² = 588 5.9
W2 = 3469.2 J
finally the third part
in this case the initial speed is 2.3 m / s and the final speed is zero
v² = v₀² + 2 a₂ y
0 = vo2₀² + 2 a₂ y
a₂ = -v₀² / 2 y
a₂ = - 2.3²/2 5.9
a2 = - 0.448 m / s²
we calculate the force
F - W = m a₂
F = m (g + a₂)
F = 60 (9.8 - 0.448)
F = 561.1 N
we calculate the work
W3 = F and
W3 = 561.1 5.9
W3 = 3310.49 J
total work
W_total = W1 + W2 + W3
W_total = 4171.89 +3469.2 + 3310.49
w_total = 10951.58 J
Answer:
Explanation:
It transmits energy into the medium through its vibration. More energetic vibration corresponds to larger amplitude. The molecules move back and forth more vigorously. The loudness of a sound is also determined by the sensitivity of the ear.
