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kramer
3 years ago
10

Two​ vehicles, a car and a​ truck, leave an intersection at the same time. The car heads east at an average speed of 20 miles pe

r​ hour, while the truck heads south at an average speed of 50 miles per hour. Find an expression for their distance apart d​ (in miles) at the end of t hours.

Physics
1 answer:
leva [86]3 years ago
7 0

Answer:

d = 10\times t\sqrt{29}miles

Explanation:

Given:

't' hour be the time taken for travel by  both the vehicles and 'd'  be the distance between then

then

Distance traveled by the car = 20 × t miles

and

Distance traveled by the truck  = 50 × t miles

now, using the Pythagoras theorem

d = \sqrt{(20t)^2+(50t)^2}

or

d = \sqrt{400t^2+2500t^2}

or

d = \sqrt{2900t^2}

or

d = 10\times t\sqrt{29}

thus, the equation relating the distance 'd' with the time 't' comes as

d = 10\times t\sqrt{29}miles

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(a) Calculate the acceleration due to gravity on the surface of the Sun.
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<h2>a)Acceleration due to gravity on the surface of the Sun is 274.21 m/s²</h2><h2>b) Factor of increase in weight is 27.95</h2>

Explanation:

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Substituting,

                g=\frac{6.67259\times 10^{-11}\times 1.989\times 10^{30}}{(6.957\times 10^8)^2}\\\\g=274.21m/s^2

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8 0
3 years ago
Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore f
tatiyna

Answer:

a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

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final point. Lowest point

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           w = v / r

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            Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

             Em₀ = Em_{f}

             mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)

             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

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             I_{cm} = ⅔ m r²

we substitute

            h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

            h = ½ v_{cm }^{2}/g    5/3

             h = 5/6 v_{cm }^{2} / g

           

let's calculate

           h = 5/6 6.1 2 / 9.8

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          v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

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we substitute

             v_{cm } = √ [2gh / (1 + ½)]

             v_{cm } = √(4/3 gh)

let's calculate

             v_{cm } = √ (4/3 9.8 3.16)

             v_{cm }^ = 6.43 m / s

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