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stira [4]
3 years ago
9

How long does it take a plane that is traveling at 350 km/h to travel 1750 km?

Physics
1 answer:
olga2289 [7]3 years ago
3 0
S=d/t
T=d/s
= 1750/350
= 5 hours
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The objective lens of a microscope has a focal length of 5.5mm. Part A What eyepiece focal length will give the microscope an ov
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Complete Question

The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 mm .

What eyepiece focal length will give the microscope an overall angular magnification of 300?

Answer:

The  eyepiece focal length is  f_e  = 0.027 \ m

Explanation:

From the question we are told that

    The focal length is  f_o =  5.5 \ mm =  -0.0055 \ m

This negative sign shows the the microscope is diverging light

     The  angular magnification is m = 300

     The  distance between the objective and the eyepieces lenses is  Z =  19 \ cm  = 0.19 \ m

Generally the magnification is mathematically represented as

        m  =  [\frac{Z - f_e }{f_e}] [\frac{0.25}{f_0} ]

Where f_e is the eyepiece focal length of the microscope

  Now  making f_e the subject  of the formula

         f_e  = \frac{Z}{1 - [\frac{M  *  f_o }{0.25}] }

substituting values

        f_e  = \frac{ 0.19 }{1 - [\frac{300  *  -0.0055 }{0.25}] }

         f_e  = 0.027 \ m

     

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3 years ago
Which energy changes take place when a pedalling cyclist uses a generator (dynamo) to light his bicycle lamp?
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Answer: chemical → kinetic → electrical → light

Explanation: At the same time when the generator lights up the bicycle lamp the lamp lights up using electrical energy so mechanical energy is also transformed into electrical energy.

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Is pulling a dog a contact force or a non contact force
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On average, both arms and hands together account for 13% of a person's mass, while the head is 7.0% and the trunk and legs accou
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Answer:

Angular velocity, N_f = 242.36 rpm

Explanation:

The mass of the skater, M = 74.0 kg

Mass of each arm, m_{a} = 0.13 * \frac{M}{2} ( since it is 13% of the whole body and each arm is considered)

m_{a} = 0.13 * 37\\m_a = 4.81 kg

Mass of the trunk, m_{t} = M - 2m_{a}

m_t = 74 - 2(4.81)\\m_{t} = 64.38 kg

Total moment of Inertia = (Moment of inertia of the arms) + (Moment of inertia of the trunks)

(I_{T} )_i = 2(\frac{m_{a}L^2 }{12} + m_a(0.5L + R)^2) + 0.5 m_t R^2

(I_{T} )_i = 2(\frac{4.81 * 0.7^2 }{12} + 4.81(0.5*0.7 + 0.175)^2) + 0.5 *64.38* 0.175^2\\(I_{T} )_i = 3.052 + 0.986\\(I_{T} )_i = 4.038 kgm^2

The final moment of inertia of the person:

(I_{T} )_f = \frac{1}{2} MR^{2} \\(I_{T} )_f = \frac{1}{2} * 74*0.175^{2}\\(I_{T} )_f = 1.133 kg.m^2

According to the principle of conservation of angular momentum:

(I_{T} )_i w_{i} = (I_{T} )_f w_{f}\\w_{i} = 68 rpm = (2\pi * 68)/60 = 7.12 rad/s\\4.038 * 7.12 =1.133* w_{f}\\w_{f} = 25.38 rad/s\\w_{f} = \frac{2\pi N_f}{60} \\25.38 = \frac{2\pi N_f}{60}\\N_f = (25.38 * 60)/2\pi \\N_f = 242.36 rpm

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