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Kipish
1 month ago
6

A physics student stands on the rim of the canyon and drops a rock. The student measures the time for it to reach the bottom to

be 3.2 seconds. How deep is the canyon?
Physics
1 answer:
inna [3.1K]1 month ago
5 0

Answer:

The canyon measures 50.176 meters deep.

Explanation:

The student drops a rock from the rim of the canyon, requiring us to ascertain the depth of the canyon—name how far the ground is below the cliff.

The data we have:

Time = t = 3.2 s

Initial velocity = v_{i} = 0 m/s

Gravitational acceleration = g = 9.8 m/s²

Height = h =?

According to the second equation of motion

h = v_{i}t + \frac{1}{2}gt^{2}

Given the initial velocity is zero, the right-hand side of the equation simplifies to zero

h = \frac{1}{2}gt^{2}

h = (0.5)(9.8)(3.2)²

h = 50.176 m

This calculation indicates that the rock dropped a distance of 50.176 meters to reach the canyon's base. Thus, the canyon depth is 50.176 meters.
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A bathroom scale operates under gravitational influence. Typically, a reading is captured when your body applies force onto the scale. Yet in this scenario, as both you and the scale move downwards, your body ceases to press against the scale. Consequently, the result is:

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2 months ago
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A power cycle operates between a lake’s surface water at a temperature of 300 K and water at a depth whose temperature is 285 K.
inna [3103]

Response: a) 0.04 kW = 40 W

b) 0.05

Explanation:

A)

The thermal efficiency of the power cycle is calculated as Input / Output

Input = 10 kW + 14,400 kJ/min which translates to 10 kW + 14,400 kJ/(60s) = 10 kW + 14,400/60 kW.

Output equals 10 kW

Thus, Thermal Efficiency = Output / Input = 10 kW / 250 kW = 0.04 kW = 40 W

B)

Maximum Thermal Efficiency of the power cycle is defined as 1 - T1/T2

where T1 = 285 Kelvin

and T2 = 300 Kelvin

Thus, Maximum Thermal Efficiency = 1 - T1/T2 = 1 - 285/300 = 0.05

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1 month ago
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5. The speed of a transverse wave on a string is 170 m/s when the string tension is 120 ????. To what value must the tension be
Sav [3153]

Answer:

The tension in the string when the speed increased is 134.53 N

Explanation:

Given;

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initial speed of the transverse wave, v₁ = 170 m/s

final speed of the transverse wave, v₂ = 180 m/s

The wave speed is expressed as;

v = \sqrt{\frac{T}{\mu} }

where;

μ represents mass per unit length

v^2 = \frac{T}{\mu} \\\\\mu = \frac{T}{v^2} \\\\\frac{T_1}{v_1^2} = \frac{T_2}{v_2^2}

The new tension T₂ will be computed as;

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Consequently, the tension in the string when the speed was increased is 134.53 N

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1 month ago
Newton’s law of cooling states that dx dt = −k(x−A) where x is the temperature,t is time, A is the ambient temperature, and k &g
Maru [3345]

Answer:

a) X= kA_o\dfrac{1}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+Ce^{-kt}

b) D does not influence the long-term results.

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\dfrac{dx}{dt}=-k(x-A)

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I=e^{kt}Now using the characteristics of linear equations

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1 month ago
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