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Alenkinab [10]
3 years ago
6

Question 3 You see the coordinates 5 E longitude, 10 N latitude. You do not need to look at a map in order to deduce that this l

ocation is a. near both the equator and the prime meridian Selected: b. near the International Date Line as well as the North PoleThis answer is incorrect. c. near the International Date Line and the prime meridian d. near the equator but quite far from the prime meridian e. near both the equator and the International Date Line
Physics
2 answers:
olchik [2.2K]3 years ago
8 0

Answer:

a. near both the equator and the prime meridian

Explanation:

Lets first understand what is equator and prime meridian.

We have drawn an imaginary grid on the Earth for navigational and other purposes. This grid has two coordinates : Latitude and Longitude. Equator is the largest latitude circle that divides the Earth in two equal hemispheres. This is taken as 0° for latitude. Latitudes increase as we move towards North/South pole which lies at 90°N and 90° S respectively. Longitudes are the lines running from north pole to south pole. The longitude that crosses from Greenwich is taken as 0°. This longitude is also known as prime meridian. The value of longitudes vary from 0° to 180° East and 180° West.

Now, the given coordinates are: 5° E and 10° N. It is quite evident that this location will be close to equator and prime meridian.

san4es73 [151]3 years ago
5 0

Answer:

a. Near both the equator and the prime meridian.

Explanation:

The  equator is at 0 degrees latitude and the prime meridian is 0 degrees longitude.

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Bill is farsighted and has a near point located 119 cm from his eyes. Anne is also farsighted, but her near point is 70.0 cm fro
Gennadij [26K]

Answer:

Explanation:

For Bill's glasses)

d_o=25 cm-2 cm=23.2 cm

d_i=129 cm-2 cm=127 cm

We know that

1 / f = 1 / d_o + 1 / d_i


f=(1 / d_o + 1 / d_i)^-1


so plug the numbers in

f = (1 / (23) + 1 / (127))^-1


= 28.08 cm

For Anna's glasses)

d_o=25 cm-2 cm=23 cm

d_i=70.0 cm-2 cm=68 cm

Same thing

(1 / f = 1 / d_o + 1 / d_i
)

f=(1 / do + 1 / di)^-1


so plug the numbers in

f = (1 / (23) + 1 / (68))^-1


= 17.21 cm

5 0
3 years ago
A solid metal sphere with radius 0.430 m carries a net charge of 0.270 nC . Part A Find the magnitude of the electric field at a
rodikova [14]

Answer:

8.46 N/C

Explanation:

Using Gauss law

E=\frac {kQ}{r^{2}}

Gauss's Law states that the electric flux through a surface is proportional to the net charge in the surface, and that the electric field E of a point charge Q at a distance r from the charge

Here, K is Coulomb's constant whose value is 9\times 10^{9} Nm^{2}/C^{2}

r = 0.43 + 0.106 = 0.536 m

E=\frac {9\times 10^{9}\times 0.270\times 10^{-9}}{0.536^{2}}=8.4581755402094007\approx 8.46 N/C

8 0
4 years ago
A massless string connects a 1.00 kg mass to a 3.00 kg cart which is resting on a frictionless horizontal surface. The mass hang
Romashka-Z-Leto [24]

Both masses will have the same acceleration. The cart accelerates to the right with a magnitude of 4.9 m/s^{2}. The correct answer is 4.90 m/s^{2}

Given that a massless string connects a 1.00 kg mass to a 3.00 kg cart which is resting on a frictionless horizontal surface.

Let M = 1kg and m = 3 kg

Since the horizontal surface is frictionless, the tension in the string will be the same. when the mass is hanged over a frictionless pulley, the tension will also be the same.

When the mass is released, the cart accelerates to the right can be calculated  from Newton' second law of motion. That is,

M( g + a) = m(g - a)

1(9.8 + a) = 3( 9.8 - a)

9.8 +a = 29.4 - 3a

collect the like terms

4a = 19.6

a = 19.6/4

a = 4.9 m/s^{2}

Therefore, the cart accelerates to the right with a magnitude of 4.9 m/s^{2}. The correct answer is 4.90 m/s^{2}

Learn more about dynamics here: brainly.com/question/24994188

5 0
3 years ago
Light reflects off many objects. Which model of light behavior best helps explain this effect? A :Particle model B: Wave model C
Varvara68 [4.7K]

Answer: B: Wave model

Explanation:

7 0
3 years ago
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How long would it take the Earth to complete 4 orbits around the Sun?
yuradex [85]

Each orbit is approximately 365.24 days ... the period we call a "year".

Four of those = 1,461 days  =  4 years .

6 0
3 years ago
Read 2 more answers
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