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3 years ago

What are the contour lines on a contour map?

2 answers:

7 0

In cartography, a contour line (often just called a "contour") joins points of equal elevation (height) above a given level, such as mean sea level. A contour map is a map illustrated with contour lines, for example a topographic map, which thus shows valleys and hills, and the steepness of slopes.-hope that helped:)

dalvyx [7]3 years ago

5 0

Contour lines are lines that signifies the elevation on a mountain or hill

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Some chemical reactions release energy. Others store energy. What important chemical reaction stores energy?

Thepotemich [5.8K]

Chemical energy energy stored in the bonds of chemical compounds. Chemical energy maybe realized during a chemical reaction of in the form of heat

8 0

3 years ago

A light spring obeys Hooke's law. The spring's unstretched length is 33.5 cm. One end of the spring is attached to the top of a

ZanzabumX [31]

Answer:

807.88N/m

Explanation:

The question has some missing details in it, nevertheless, based on the given data we want to find the spring constant K

Step one

given data

Unstretched length = 33.5 cm

Final length of the spring = 42.0 cm

Δx= 42-33.5

Δx=8.5cm to m= 0.085m

mass m= 7kg

The force on the spring

F=mg

F= 7*9.81

F=68.67N

Step two:

From Hooke's law, we can make k subject of formula and find the spring constant k, we have

F=kΔx---------1

make k subject of the formula

k=F/Δx

k= 68.67/ 0.085

k=807.88N/m

4 0

3 years ago

True or False: A 24 year old motorcycle driver would not be required to wear protective eyewear when riding in Florida.

n200080 [17]

False people should always wear gear

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3 years ago

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zaharov [31]

Z is present outside the solar system.

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4 years ago

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A 14.3-g bullet is fired into a 5.21 kg block of wood. The block is attached to a spring that has a spring force constant of 450

Natasha2012 [34]

Answer:

The initial speed of the bullet is $v_{o} = 889.199\,\frac{m}{s}$ .

Explanation:

The collision between bullet and block is inelastic and let suppose that motion occurs on a horizontal surface, so that changes in gravitational potential energy can be neglected. Initially, the intial speed of the bullet-block system can be determined with the help of the Work-Energy Theorem and the Principle of Energy Conservation:

$K = U_{k} + W_{loss}$

$\frac{1}{2}\cdot (5.224\,kg)\cdot v^{2} = \frac{1}{2}\cdot \left(450\,\frac{N}{m}\right)\cdot (0.22\,m)^{2} + (0.35)\cdot (5.224\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right) \cdot (0.22\,m)$

The initial speed of the bullet-block system is:

$v \approx 2.383\,\frac{m}{s}$

Now, the initial speed of the bullet is determined by applying the Principle of Momentum Conservation:

$(0.014\,kg)\cdot v_{o} = (5.224\,kg)\cdot \left(2.383\,\frac{m}{s} \right)$

$v_{o} = 889.199\,\frac{m}{s}$

6 0

3 years ago