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3 years ago

If a scale of the solar system was built where 1 mm equaled 1 mile, could the model be practically built in a city?

Physics

2 answers:

Y_Kistochka [10]3 years ago

7 0

No, it couldn't be.
On that scale, Neptune would be almost 1,740 MILES from the sun.
ON THAT SCALE !

Irina18 [472]3 years ago

3 0

Answer:

Explanation:

Given the relative scale (1mm = 1mile) lets calculate the size of the Sun and distance of planets from the Sun in mm.

Diameter of Sun = 864938 miles = 864938 mm = 864.938 meters

Distance of Earth from the Sun = 92955810 miles = 92955810 mm = 92.955 km = 57.75 miles

Distance of Jupiter from Sun = 480000000 miles = 480000000 mm = 480 km = 298 miles

As can be seen, the magnitude of scaled down distances is too huge to fit inside a city.

It will not be practically possible to make such a model.

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Tom kicks a soccer ball on a flat, level field giving it an initial speed of 20 m/s at an angle of 35 degrees above the horizont

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Answer:

(a) 2.34 s

(b) 6.71 m

(d) 20 m/s

Explanation:

u = 20 m/s, theta = 35 degree

(a) The formula for the time of flight is given by

$T = \frac{2 u Sin\theta }{g}$

$T = \frac{2 \times 20 \times Sin35 }{9.8}$

T = 2.34 second

(b) The formula for the maximum height is given by

$H = \frac{u^{2} \times Sin^{2}\theta }{2g}$

$H = \frac{20^{2} \times Sin^{2}35 }{2 \times 9.8}$

H = 6.71 m

$R = \frac{u^{2} \times Sin 2\theta }{g}$

$R = \frac{20^{2} \times Sin 2 \times 35}{9.8}$

R = 38.35 m

(d) It hits with the same speed at the initial speed.

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3 years ago

. Reem took a wire of length 10 cm. Her friend Nain took a wire of 5 cm of the same material and thickness both of them connecte

Umnica [9.8K]

Given :

Reem took a wire of length 10 cm. Her friend Nain took a wire of 5 cm of the same material and thickness both of them connected with wires as shown in the circuit given in figure. The current flowing in both the circuits is the same.

To Find :

Will the heat produced in both the cases be equal.

Solution :

Heat released is given by :

H = i²Rt

Here, R is resistance and is given by :

$R = \dfrac{\rho L}{A}$

So,

$H = i^2\times \dfrac{\rho L}{A} t\\\\H = \dfrac{i^2\rho Lt}{A}$

Now, in the question every thing is constant except for the length of the wire and from above equation heat is directly proportional to the length of the wire.

So, heat produced by Reem's wire is more than Nain one.

Hence, this is the required solution.

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2 years ago

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I= $\frac{P_{avg}}{A}$