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Marta_Voda [28]
3 years ago
11

On a topographic map, contour lines create a group of concentric, closed loops. Which of the following features could this indic

ate?
a. a stream channelb. hilltopc. depressiond. a cliff
Physics
1 answer:
posledela3 years ago
3 0

Answer:

b. Hill top

Explanation:

On a topographic map, the closed circles are meant to represent a hill. So if the contour lines are creating a group of concentric closed loops then it must be an indication of a hill.

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Which best describes why Keplers observation of planetary motion is a law instead of a theory
svet-max [94.6K]

Kepler's first law - sometimes referred to as the law of ellipses - explains that planets are orbiting the sun in a path described as an ellipse. An ellipse can easily be constructed using a pencil, two tacks, a string, a sheet of paper and a piece of cardboard. Tack the sheet of paper to the cardboard using the two tacks. Then tie the string into a loop and wrap the loop around the two tacks. Take your pencil and pull the string until the pencil and two tacks make a triangle (see diagram at the right). Then begin to trace out a path with the pencil, keeping the string wrapped tightly around the tacks. The resulting shape will be an ellipse. An ellipse is a special curve in which the sum of the distances from every point on the curve to two other points is a constant. The two other points (represented here by the tack locations) are known as the foci of the ellipse. The closer together that these points are, the more closely that the ellipse resembles the shape of a circle. In fact, a circle is the special case of an ellipse in which the two foci are at the same location. Kepler's first law is rather simple - all planets orbit the sun in a path that resembles an ellipse, with the sun being located at one of the foci of that ellipse.


5 0
3 years ago
Read 2 more answers
If an object 18 millimeters high is placed 12 millimeters from a diverging lens and the image is formed 4 millimeters in front o
tangare [24]
The correct answer is letter A. 6 millimeters. <span>If an object 18 millimeters high is placed 12 millimeters from a diverging lens and the image is formed 4 millimeters in front of the lens, the height of the image is 6 millimeters.
</span>
Solution:
18 / x = 12 / 4 
12x = 72
x = 6mm
7 0
3 years ago
If a car accelerates uniformly from rest to 15 meters
Talja [164]

Answer:

1.125m/s^2

Explanation:

Since acceleration is defined as the rate of change in velocity with respect to time. Mathematically

v^2= u^2+2as

Where a,v,u and s are the acceleration, final velocity, initial velocity and distance respectively.

a = ?

u = 0m/s

v = 15m/s

s = 100m

Substituting the values into the formula above

v^2= u^2+2as

15^2=0^2+2×a×100

225= 0+200a

225= 200a

Divide both sides by 200

225/200 = 200a/200

a= 1.125m/s^2

Hence the acceleration of the car is 1.125m/s^2.

Note that the car accelerated uniformly from rest, that was why the initial velocity was 0m/s

8 0
3 years ago
Read 2 more answers
A jet airliner moving initially at 406 mph (with respect to the ground) to the east moves into a region where the wind is blowin
astraxan [27]

Answer:

966 mph

Explanation:

Using as convention:

- East --> positive x-direction

- North --> Positive y-direction

The x- and y- components of the initial velocity of the jet can be written as

v_{1x} = 406 mph\\v_{1y} = 0

While the components of the velocity of the wind are

v_{2x} = (568)(cos 15^{\circ})=548.6 mph\\v_{2y} = (568)(sin 15^{\circ})=147.0 mph

So the components of the resultant velocity of the jet are

v_x = v_{1x}+v_{2x}=406+548.6=954.6 mph\\v_y = v_{1y}+v_{2y}=0+147.0=147.0 mph

And the new speed is the magnitude of the resultant velocity:

v=\sqrt{v_x^2+v_y^2}=\sqrt{(954.6)^2+(147.0)^2}=965.8 mph \sim 966 mph

6 0
3 years ago
A 1750-kilogram cars travels at a constant speed of 15.0 meters per second around a horizontal, circular track with a radius of
bulgar [2K]

m = mass of the car moving in horizontal circle = 1750 kg

v = Constant speed of the car moving in the horizontal circle = 15 m/s

r = radius of the horizontal circular track traced by the car = 45.0 m

F = magnitude of the centripetal force acting on the car

To move in a circle . centripetal force is required which is given as

F = m v²/r

inserting the above values in the formula

F = (1750) (15)²/(45)

F = (1750) (225)/(45)

F = 1750 x 5

F = 8750 N

6 0
3 years ago
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