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Naily [24]
3 years ago
7

West of a city, a certain eastbound route is straight and makes a steep descent toward the city. The highway has a 14% grade, wh

ich means that its slope is − 14 100 . Driving on this road, you notice from elevation signs that you have descended a distance of 1000 ft. What is the change in your horizontal distance in miles
Physics
1 answer:
vova2212 [387]3 years ago
6 0

Answer:

1.35 miles

Explanation:

Since the slope is 14% grade = -14/100. This is a negative slope.

The distance of 1000 ft descended is a vertical descent, so assuming our initial position is 0, and Δx is the change in horizontal distance.

tanθ = -14/100 = (0 - 1000)/Δx = - 1000)/Δx

-14/100 = - 1000)/Δx

Δx = -1000 × -100/14= 7142.86 ft

5280 ft = 1 mile

7142.86 ft = 7142.86/5280 miles = 1.353 miles ≅ 1.35 miles

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A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 35 m/s2 for 35 s , then ru
r-ruslan [8.4K]

Answer:

T = 295.57 s

Explanation:

given,

mass of the rocket = 200 Kg

mass of the fuel = 100 Kg

acceleration = 35 m/s²

time, t = 35 s

time taken by the rocket to hit the ground, = ?

Final speed of the rocket when fuel is empty

using equation of motion

v = u + a t

v = 0 + 35 x 35

v = 1225 m/s

height of the rocket where fuel is empty

v² = u² + 2 a s

1225² = 0 + 2 x 35 x h₁

h₁ = 21437.5 m

After 35 s the rocket will be moving upward till the final velocity becomes zero.

Now, using equation of motion to find the height after 35 s

v² = u² + 2 g h₂

0² = 1225² + 2 x (-9.8) h₂

h₂ = 76562.5 m

total height = h₁ + h₂

          = 76562.5 m + 21437.5 m = 98000 m

now, time taken by before the rocket hit the ground

using equation of motion

s = u t +\dfrac{1}{2}at^2

-13500 = 1225 t -\dfrac{1}{2}\times 9.8 \times t^2

negative sign is used because the distance travel by the rocket is downward.

4.9 t² - 1225 t - 13500 = 0

t = \dfrac{-(-1225)\pm \sqrt{1225^2 - 4\times 4.9 \times (-13500)}}{2\times 4.9}

t = 260.57 s

neglecting the negative sign

total time the rocket was in air

T = t₁ + t₂

T = 35 + 260.57

T = 295.57 s

Time for which rocket was in air is equal to 295.57 s.

6 0
3 years ago
Any help would be great! Thank you x<br><br><br> Giving brainliest answer xoxo
garri49 [273]

Answer:

A vacuum would have been created. I hope this helps have a great day

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The position of a certain airplane during takeoff is given by x=1/2 *bt2, where b = 2.0 m/s2 and t = 0 corresponds to the instan
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So the total mass combination of the plane and the people inside it is

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