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lbvjy [14]
4 years ago
8

A person takes a trip, driving with a constant speed of 89.5 km/h, except for a 22.0-min rest stop. If the person’s average spee

d is 77.8 km/h, (a) how much time is spent on the trip and (b) how far does the person travel?
Physics
1 answer:
DanielleElmas [232]4 years ago
7 0

Answer:

(a): It spends on the trip 2.75 hr (2 hours 45 minutes)

(b):  The person travels 213.9 km.

Explanation:

Vp= 77.8 km/h

V1= 89.5 km/h

V2= 0 km/h

t2= 0.36hr

t1= ?

Vp= (V1*t1 + V2*t2) / (t1 + t2)

clearing t1:

t1= 2.39 hr

Trip time= t1+t2

Trip time= 2.75 hr (a)

Distance traveled= V1*t1

Distace traveled= 213.9 km (b)

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Answer:

W_{earth} = 35.74 * 10^6 N : Rocket weight on earth

W_{moon} = 5.91 * 10^6 N : Rocket weight on moon

Explanation:

Conceptual analysis

Weight is the force with which a body is attracted due to the action of gravity and is calculated using the following formula:

W = m × g Formula (1)

W: weight

m: mass

g: acceleration due to gravity

The mass of a body on the moon is equal to the mass of a body on the earth

The acceleration due to gravity on a body is different on the moon and on the earth

Equivalences

1 slug = 14.59 kg

Known data

m_{earth} = m_{moon} = 250 * 10^3 slug = 250* 10^3slug * \frac{14.59kg}{1slug} = 3647.5* 10^3 kg

g_{moon}= 1.62 \frac{m}{s^2}

g_{earth}= 9.8 \frac{m}{s^2}

Problem development

To calculate the weight of the rocket on the moon and on earth we replace the data in formula (1):

W_{earth} = 3647.5* 10^3 kg * 9.8 \frac{m}{s^2} = 35.74 * 10^6 N : Rocket weight on earth

W_{moon} = 3647.5* 10^3 kg * 1.62 \frac{m}{s^2} = 5.91 * 10^6 N : Rocket weight on moon

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4 years ago
Consider a steel guitar string of initial length L=1.00L=1.00 meter and cross-sectional area A=0.500A=0.500 square millimeters.
erma4kov [3.2K]

Answer:

The extent to which it would stretch  is \Delta L = 0.015 \ m

Explanation:

From the question we are told that

    The initial length is  L = 1.00m

     The area is  A = 0.500 mm^2 = \frac{0.500}{1 *10^6} = 0.500*10^6 \ m^2

     The Young modulus of the steel is  Y = 2.0*10^{11} Pa

     The tension   is  T =1500 N

The Young modulus is mathematically represented as

       Y = \frac{\sigma}{e}

Where \sigma is the stress which is mathematically represented as

           \sigma = \frac{F}{A}  

Substituting values

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           \sigma = 3.0*10^9 N/m^2  

And  e is the strain which is mathematically represented as

            e = \frac{\Delta L}{L }

Where \Delta L The extension of the steel string

Substituting these into the equation above

             Y = \frac{3.0*10^9}{\frac{\Delta L}{L} }

Substituting values  

           2.0 *10^{11} = \frac{3.0*10^9}{\frac{\Delta L}{L} }

          \Delta L = \frac{3.0*10^9  * 1}{2.0 *10^{11}}

         \Delta L = 0.015 \ m

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3 years ago
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