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denis23 [38]
3 years ago
8

what are the variables in an experiment that tests the distance honey flows at different temperatures? identify the independent

variable (the one we can control) and the dependent variable (the one we cannot control).
Physics
2 answers:
Dmitriy789 [7]3 years ago
8 0

Answer:

The distance honey flows is the dependent variable and temperature is the independent variable.

Explanation:

soldier1979 [14.2K]3 years ago
6 0

Answer:

The distance that the honey flowed would be the dependent or outcome variable and the temperature of the honey would be the independent variable.

The dependent variable is what is being measured in an experiment. You can remember it by thinking “it depends on what you’re changing.”

The independent variable in an experiment is what is being changed. You can remember this by thinking “the Independent variable is what I as the scientist change.”

Explanation:

mark me brainliest plz

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A small fish is dropped by a pelican that is rising steadily at 0.50 m/s.
natima [27]

Answer:

(a)  24.025 m/s. downward.

(b)  31 m

Explanation:

From Newton's equation of motion,

(a)

v = u + gt ................... Equation 1

Where v = final velocity, u = initial velocity, g = acceleration due to gravity, t = time.

Note: Let upward velocity be negative and downward be positive

Given: u = -0.5 m/s (upward), t = 2.5 s

Constant : g = 9.81 m/s²

Substitute into equation 1

v = 0.5+9.81(2.5)

v = -0.5+24.525

v = 24.025 m/s. downward.

(b) using

s₁ = ut + 1/2gt²......................... Equation 2

Where s₁ = distance at which the fish fall after being dropped by the pelican

Given: u = - 0.5 m/s, t = 2.5 s, g = 9.81 m/s²

Substitute into equation 2

s₁ = -0.5(2.5) + 1/2(9.81)(2.5)²

s₁ = -1.25+30.656

s₁ = 29.41 m

also,

s₂ = vt ................ Equation 3

Where s₂ = the distance by which the pelican rise during this time.

Given: v = 0.5 m/s, t= 2.5 s

s₂ = 0.5(2.5)

s₂ = 1.25 m.

Note: Distance between the pelican and fish = s₁ + s₂

Distance between the pelican and fish  = 29.41+1.25

Distance between the pelican and fish  = 30.66

Distance between the pelican and fish ≈ 31 m

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Answer:

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