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kari74 [83]
3 years ago
8

An 75-kg hiker climbs to the summit of Mount Mitchell in western North Carolina. During one 2.0-h period, the climber's vertical

elevation increases 540 m. Determine the change in gravitational potential energy of the climber-Earth system.
Physics
1 answer:
Ostrovityanka [42]3 years ago
3 0

Answer:

The change in gravitational potential energy of the climber-Earth system is  \Delta  PE  = 396900 \ J

Explanation:

From the question we are told that

    The mass of the hiker is  m = 75 \ kg

    The time  taken is  T  =  2 \ hr =  2 *  3600 =  7200 \ s

    The  vertical elevation after time  T is  H = 540 \ m

   

The  change  in gravitational potential is  mathematically represented as

         \Delta  PE  =  mgH

here g is the acceleration due to gravity with value  g =  9.8 \ m/s^2  

     substituting values  

        \Delta  PE  =  75  *  9.8  *  540

       \Delta  PE  = 396900 \ J

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A sound has a sound level of 30 dB. Its intensity is what multiple of the standard reference level for intensities?
arsen [322]
<h2>Answer:</h2>

1000th multiple of the standard reference level for intensities.

<h2>Explanation:</h2>

The sound intensity level (β), measured in decibels, of a sound with an intensity of I is defined as follows;

β = 10 log (I / I₀)       --------------------(i)

Where;

I₀ = reference intensity

Given from the question;

β = sound level = 30dB

Substitute this value into equation (i) as follows;

30 = 10 log (I / I₀)

Divide both sides by 3;

3 = log (I / I₀)

Take antilog of both sides;

10^(3) = (I / I₀)

1000 = I / I₀

Solve for I;

I = 1000I₀

Therefore the intensity of the sound is 1000 times the standard reference level for intensities (I₀)

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5 0
3 years ago
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After being struck by a bowling ball, a 1.8 kg bowling pin sliding to the right at 5.0 m/s collides head-on with another 1.8 kg
kaheart [24]

Answer:

a) v₂ = 4.2 m/s

b) v₂ = 5 m/s

Explanation:

a)

We will use the law of conservation of momentum here:

m_1u_1+m_2u_2=m_1v_1+m_2v_2

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0.8 m/s

v₂ = speed of second after before collsion = ?

Therefore,

(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0.8\ m/s)+(1.8\ kg)(v_2)\\v_2 = 5\ m/s - 0.8\ m/s

<u>v₂ = 4.2 m/s</u>

<u></u>

b)

We will use the law of conservation of momentum here:

m_1u_1+m_2u_2=m_1v_1+m_2v_2

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0 m/s

v₂ = speed of second after before collsion = ?

Therefore,

(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0\ m/s)+(1.8\ kg)(v_2)

<u>v₂ = 5 m/s</u>

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