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Schach [20]
4 years ago
9

Which of the following is the BEST example of increasing the intensity of a workou

Physics
1 answer:
Paladinen [302]4 years ago
8 0

<u>Question: </u>

Which of the following is the BEST example of increasing the intensity of a workout?

a. running one mile further than normal

b. running one mile faster than normal

c. running one mile more every other workout

d. jogging for three miles

<u>Answer:</u>

Running one mile faster than normal is the best example of increasing the intensity of a workout.

<u>Explanation: </u>

In terms of workout, the intensity is measured on the basis of rate of heart beat. The intensity of workout is determined to be directly proportional to the rate of heart beat. This means the increase in pace of heartbeat or the faster the heartbeat rate, the stronger will be the intensity of the workout.

So in order to increase the rate of heartbeat, one can run one mile faster than the normal speed to increase the workout intensity. As running with faster speed can lead to increase in the rate of heartbeat which will lead to the increase in intensity of heartbeat. Thus, running one mile faster will lead to increase in the intensity of workout.

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A 2.7-cm-tall object is 20 cm to the left of a lens with a focal length of 10 cm . A second lens with a focal length of 48 cm is
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Answer

given,

height of object = 2.7 cm

distance left of lens (u₁)= 20 cm

focal length of lens(f₁)= 10 cm  

the distance of image

\dfrac{1}{f_1}=\dfrac{1}{u_1}+\dfrac{1}{v_1}

\dfrac{1}{v_1}=\dfrac{1}{f_1}-\dfrac{1}{u_1}

\dfrac{1}{v_1}=\dfrac{1}{10}-\dfrac{1}{20}

   v₁ = 20 cm

magnification of first lens

m_1= -\dfrac{v}{u}

m_1=-\dfrac{20}{20}

    m₁ = -1

distance of object from the second lens

   u₂ = 52-20 = 32 cm

   f₂ = 48 cm

now,

\dfrac{1}{f_2}=\dfrac{1}{u_2}+\dfrac{1}{v_2}

\dfrac{1}{v_2}=\dfrac{1}{f_2}-\dfrac{1}{u_2}

\dfrac{1}{v_2}=\dfrac{1}{48}-\dfrac{1}{52}

   v₁ = 624 cm

magnification of first lens

m_1= -\dfrac{v}{u}

m_1=-\dfrac{624}{52}

    m₁ = -12

total magnification

m = m₁ m₂

m = (-1)(-12)

m = 12

height of image

m =-\dfrac{h'}{h}

12=-\dfrac{h'}{2.7}

h' = -32.4 cm

a) distance between image and second lens is equal to 624 cm

b) height of image is equal to 32.4 cm

       

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