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julsineya [31]
3 years ago
8

If a topographic map included a 6,000 ft. mountain next to an area of low hills, which would best describe the contour lines on

the map?
a.
The contour lines would be dark blue.
b.
The contour lines around the mountain would be very close together.
c.
The contour lines would cross near the top of the mountain.
d.
The contour lines around the rolling hills would be very close together.
Physics
1 answer:
Stella [2.4K]3 years ago
4 0

Answer:

ITS B

Explanation:B IS THE ANSWER

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I think that ball  a hit the ground because it says that it went straight down.
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Hummingbirds may seem fragile, but their wings are capable of sustaining very large forces and accelerations. (Figure 1) shows d
gulaghasi [49]

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<h3>What is acceleration?</h3>

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The question is incomplete so we can not arrive at the final answer. However, acceleration is mathematically defined as; ΔV/t.

Learn more about acceleration: brainly.com/question/2437624

5 0
1 year ago
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6 0
3 years ago
You are traveling on an interstate highway at the posted speed limit of 70 mph when you see that the traffic in front of you has
vovikov84 [41]

Answer:8.75 s,

136.89 m

Explanation:

Given

Initial velocity=70 mph\approx 31.29 m/s

velocity after 5 s is 30 mph\approx 13.41 m/s

Therefore acceleration during these 5 s

a=\frac{v-u}{t}

a=\frac{13.41-31.29}{5}=-3.576 m/s^2

therefore time required to stop

v=u+at

here v=final velocity =0 m/s

initial velocity =31.29 m/s

0=31.29-3.576\times t

t=\frac{31.29}{3.576}=8.75 s

(b)total distance traveled before stoppage

v^2-u^2=2as

0^2-31.29^2=2\times (-3.576)\cdot s

s=136.89 m

3 0
2 years ago
The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.32 with the floor. If t
coldgirl [10]

Answer:

The shortest braking distance is 35.8 m

Explanation:

To solve this problem we must use Newton's second law applied to the boxes, on the vertical axis we have the norm up and the weight vertically down

On the horizontal axis we fear the force of friction (fr) that opposes the movement and acceleration of the train, write the equation for each axis

    Y axis

     N- W = 0

     N = W = mg

  X axis

     -Fr = m a

     -μ N = m a

     -μ mg = ma

     a = μ g

     a  = - 0.32 9.8

     a =  - 3.14 m/s²

We calculate the distance using the kinematics equations

    Vf² = Vo² + 2 a x

     x = (Vf² - Vo²) / 2 a

When the train stops the speed is zero (Vf = 0)

 Vo = 54 km/h (1000m/1km) (1 h/3600s)= 15 m/s

     x = ( 0 - 15²) / 2 (-3.14)

     x=  35.8 m

The shortest braking distance is  35.8 m

7 0
2 years ago
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