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-BARSIC- [3]
2 years ago
6

Latitude and longitude are the two coordinates that determine a specific point on the Earth’s surface. How does knowing the loca

tion of a point help us make maps?

Physics
1 answer:
DedPeter [7]2 years ago
6 0

Answer:

hey sandra! i hope this will help you!

Explanation:

The geographic coordinate system is a reference system that uses the two angular latitude (north or south) and longitude (east or west) coordinates to determine the positions of the land surface points. These two angular coordinates measured from the center of the Earth are from a spherical coordinate system that is aligned with its axis of rotation. These coordinates are usually expressed in sexagesimal degrees:

Latitude measures the angle between any point and the equator. Latitude lines are called parallel and are circles parallel to the equator on the Earth's surface. Latitude is the distance between any point and Ecuador, measured on the meridian that passes through that point.

• All points located on the same parallel have the same latitude.

• Those who are in the north of Ecuador receive the denomination North (N).

• Those who are in the south of Ecuador receive the denomination South (S).

• It is measured from 0º to 90º.

• The latitude of 0º corresponds to Ecuador.

• The North and South poles have latitude 90º N and 90º S respectively.

The length measures the angle along the equator from any point on Earth. It is accepted that Greenwich in London is length 0 in most modern societies. The lines of length are maximum circles that pass through the poles and are called meridians.

• All points located on the same meridian have the same length.

• Those who are east of Meridian Zero receive the designation East (E).

• Those who are west of Meridian Zero receive the denomination West (O).

• It is measured from 0º to 180º.

• The Greenwich meridian corresponds to the length 0º.

By combining these two angles, the position of any point on the Earth's surface can be expressed.

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3 A rocket of mass 1200 kg is travelling at 2000 m/s. It fires its engine for 1 minute. The forward thrust provided by the rocke
natita [175]

The impulse shared by the object equals the difference in momentum of the object. In equation form,

F • t = m • Δ v. In a collision, objects experience an impulse; the impulse causes and is equal to the difference in momentum.

<h3>How to calculate  thrust provided by the rocket engines is 10 kN (10 000 N).?</h3>

a)There is this impulse-momentum change equation.

where m$ is the mass of a body, $F$ is a force acting to the body, $t$ is time and $D E L A T A N\}=V_{2}-V_{1}$ is the change of velocity.We consider everything is happen along a straight line, and gravitation does not participate.So, the increase of momentum is $\mathrm{F}^{*} \mathrm{t}=10000 \mathrm{~N} * 60$ seconds $=600000 \mathrm{~N}^{*} \mathrm{~s}=600000\left(\mathrm{~kg}^{*} \mathrm{~m}\right)^{*} \mathrm{~s} / \mathrm{s}^{\wedge} 2=600000 \mathrm{~kg}{ }^{*} \mathrm{~m} / \mathrm{s}$.

We consider everything exits happen along a straight line, and gravitation does not participate.

So, the increase of momentum is F×t = 10000 N × 60 seconds = 600000 N*s = 600000 (kg*m)*s/s^2 = 600000 kg*m/s.

$$\Delta(\mathrm{V})=\frac{\mathrm{F.t}}{\mathrm{m}}=\frac{600000}{1200}=500 \mathrm{~m} / \mathrm{s} .$$

New velocity after  engine was firing during 60 seconds is 2000 + 500 = 2500 m/s.

To learn more about Impulse-momentum, refer

brainly.com/question/20586658

#SPJ9

5 0
1 year ago
A book 15.6 cm tall is 28.7 cm to the left of a plane mirror. What is the height of the image formed?
Ira Lisetskai [31]

Answer:

15.6cm

................

3 0
3 years ago
The tape in a videotape cassette has a total
xxMikexx [17]

Answer:

w=19.76 \ rad/s

Explanation:

<u>Circular Motion </u>

Suppose there is an object describing a circle of radius r around a fixed point. If the relation between the angle of rotation by the time taken is constant, then the angular speed is also constant. If that relation increases or decreases at a constant rate, the angular speed is given by:

w=w_o+\alpha \ t

Where \alpha is the angular acceleration and t is the time. If the object was instantly released from the circular path, it would have a tangent speed of:

\displaystyle v_t=w.r

We have two reels: one loaded with the tape to play and the other one empty and starting to fill with tape. They both rotate at different angular speeds, one is increasing and the other is decreasing as the tape goes from one to the other. We'll assume the tangent speed is constant for both (so the tape can play correctly). Let's call w_1 the angular speed of the loaded reel and w_2 that from the empty reel. We have

w_1=w_{o1}+\alpha_1 \ t

w_2=w_{o2}+\alpha_2 \ t

If r_f=35\ mm=0.035\ m is the radius of the reel when it's full of tape, the angular speed for the loaded reel is computed by

\displaystyle w_{01}=\frac{v_t}{r_f}

The tangent speed is computed by knowing the length of the tape and the time needed to fully play it.

t=1.8\ h=1.8*3600=6480\ sec

\displaystyle v_t=\frac{x}{t}=\frac{249\ m}{6480\ sec}=0.0384 \ m/s

\displaystyle w_{01}=\frac{0.0384}{0.035}=1,098 \ rad/s

If r_e=10\ mm=0.001\ m is the radius of the reel when it's empty, the angular speed for the empty reel is computed by

\displaystyle w_{02}=\frac{0.0384}{0.001}=38.426 \ rad/s

The full reel goes from w_{01} to w_{02} in 6480 seconds, so we can compute the angular acceleration:

\displaystyle \alpha_1=\frac{w_{02}-w_{01}}{6480}

\displaystyle \alpha_1=\frac{38.426-1.098}{6480}=0.00576 \ rad/sec^2

The empty reel goes from w_{02} to w_{01} in 6480 seconds, so we can compute the angular acceleration:

\displaystyle \alpha_2=\frac{1.098-38.426}{6480}=-0.00576 \ rad/sec^2

So the equations for both reels are

w_1=1.098+0.00576 \ t

w_2=38.426-0.00576 \ t

They will be the same when

1.098+0.00576 \ t=38.426-0.00576 \ t

Solving for t

\displaystyle t=\frac{38.426-1.098}{0.0115}

t=3240 \ sec

The common angular speed is

w_1=1.098+0.00576 \ 3240=19.76 \ rad/s

w_2=38.426-0.00576 \ 3240=19.76 \ rad/s

They both result in the same, as expected

\boxed{w=19.76 \ rad/s}

6 0
3 years ago
The density of a certain type of plastic is 0.75 g/cm3 . If a sheet of this plastic is 10.0 m long, 1.0 m wide, and 1 cm thick,
Oksanka [162]

Answer:

7.5 g

Explanation:

8 0
2 years ago
Draw a stationary wave and show the position of node and antinode​
Aneli [31]

I hope that the attachment helps you..

3 0
2 years ago
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